explicit physics solutions part i: mechanics
TRANSCRIPT
Explicit Physics Solutions Part I: Mechanics
Gregory G. WoodCalifornia State University Channel Islands
September 7, 2005
Abstract
The need for a large selection of explicit solutions to physics prob-lems is clear to any who teach the subject. Text books give few exampleproblems. Often as many as half are trivial. Class time is limited andstudents frequently ask for more examples. Existing solution manualsfor text books are terse in the extreme. Hand written solutions are wellreceived by students, but do not reproduce well. Further, solutions areonly made available after homework is turned in. Thus this work wasbegun.
For best results, try each problem before reading the solution.
Contents
1 Constant acceleration in one dimension 2
2 Non-standard Problems 7
3 Vectors 9
4 Projectile Motion 10
5 Force and Newton’s Laws 12
6 Work and Energy 18
7 Conservation of Energy 20
8 Momentum and Collisions 23
9 Angular Motion 27
1
10 Torque 29
11 Angular Momentum and Angular Kinetic Energy 31
12 Static Equilibrium 32
13 Fluids 34
14 Oscillations 36
1 Constant acceleration in one dimension
Problems
1. A greyhound accelerates from rest to 22 m/s in 2.0 s and then continuesat this constant speed for 10.0 s more. How far does the greyhound travelin total?
2. Huck Fin tosses a water balloon straight up at 4.5 m/s from a treelimb 2.3 m off the ground. With what velocity does the balloon strike theground? How long is it in flight?
3. Robin Hood seeks to knock unconscious one of King John’s men-at-arms. Robin knows to yield a man unconscious, a rock must strike the headat about 28 m/s.1 How high up must Robin be in order so that if he dropsa rock from rest, it will strike the soldier unconscious? Assume a 1.5 m tallman-at-arms.
4. Robin Hood is stuck in a tree only 8.5 m high. The same 1.5 m tallman-at-arms from problem three walks by. How can Robin strike the manunconscious?
5. A physics student atop the library, 45.5 m above ground, observes abaseball traveling straight upward at 2.3 m/s. Assuming the baseball waslaunched from the ground, what initial velocity did it have at launch?
1Do not try at home.
2
6. A truck passes by a dog at rest at a speed of 15.0 m/s. One secondlater, the dog begins to accelerate at 8.0m/s2 until he reaches his maximumspeed of 22.0 m/s. How far does the dog travel before catching the truck?Assume the truck does not accelerate.
7. A coin is dropped into a well 55.0 m deep. If sound travels at 343 m/s,how long before the splash is heard?
Solutions
1. A greyhound accelerates from rest to 22 m/s in 2.0 s and then continuesat this constant speed for 10.0 s more. How far does the greyhound travelin total?
This problem has two parts, each with a different constant acceleration.The first has a rate of acceleration which can be found using
vf = vo + at (1)
(the second has an acceleration of zero). Plugging in numbers, and solvingfor a, we find vf/t = a and thus a = 22m/s/2.0s = 11m/s2. Find thedistance traveled during each part and sum them. The first distance, d1,can be found using:
d1 = vot +12at2. (2)
Plugging in we find: d1 = 0+1/211m/s2(2s)2 and solving we find d1 = 22m.
Now we must find the second distance, d2, and sum them. For part two,the initial velocity is 22 m/s and so is the final: the dog is not accelerating.Thus the distance is easily found using:
d = vot + 1/2at2 (3)
and setting a = 0, we find: d2 = 22m/s10.0s = 220m.2
Summing we find the total distance is dtotal = d1 + d2 = 22m + 220m =242m but we should round off to two significant figures so dtotal = 240m.
2Note we only have two significant figures
3
2. Huck Fin tosses a water balloon straight up at 4.5 m/s from a treelimb 2.3 m off the ground. With what velocity does the balloon strike theground? How long is it in flight?
For this problem, we assume Huck is on earth, where the acceleration dueto gravity is 9.8m/s2. We choose positive to mean up, so the accelerationis negative, and the origin of coordinates to be on the ground, so the initialposition of the balloon is positive 2.3 m.3 Now we can solve for final velocityusing:
v2f = v2
o + 2ad,where both a and d are negative. Plugging in we find:v2f = (4.5m/s)2 +2(−9.8m/s2)(−2.3m) = (20.25m2/s2)+45.08m2/s2 =
65.32m2/s2.Thus we find vf = ±8.1m/s.4 Now we must choose the correct sign
by logic. Since the balloon must be traveling down to strike the earth, wechoose the minus sign.
Next, we must find the time of flight. As we have already found the finalvelocity, this is easily achieved via
vf = vo + at.Using the appropriate signs, we find: −8.1m/s = 4.5m/s − 9.8m/s2t.
Solving for t, we find:t = (−8.1m/s−4.5m/s)/(−9.8m/s2). Careful cancellation of units leads
to: t = 0.46s.
3. Robin Hood seeks to knock unconscious one of King John’s men-at-arms. Robin knows to yield a man unconscious, a rock must strike the headat about 28 m/s. How high up must Robin be in order so that if he dropsa rock from rest, it will strike the soldier unconscious? Assume a 1.5 m tallman-at-arms.
We are given the initial velocity (zero), the final velocity (28 m/s down-ward), the acceleration and the final position (1.5 m above ground). We areasked to find the initial position. Let us choose the ground to be the originand upward to be the positive y-direction. Now we can use:
3Also, the initial velocity is positive as the balloon is thrown upward.4Again, we must round as we only have two significant figures.
4
v2f = v2
o + 2a(y − yo) to find the initial y-position. Our acceleration isnegative (downward), our final y-position is positive (above ground) andour initial velocity is neither (zero). Plugging in we find:
(−28m/s)2 = 0 + 2(−9.8m/s2)(1.5m− yo).
Solving for yo we find: −40.m = 1.5m− yo. Thus yo = 41.5m. Rounding,we revise our answer to yo = 42.m5
4. Robin Hood is stuck in a tree only 8.5 m high. The same 1.5 m tallman-at-arms from problem three walks by. How can Robin strike the manunconscious?
One answer is that Robin can jump to another tree which is at least 42.m high unnoticed. However, he might be noticed. Far safer to stay wherehe is and use physics. In problem three, we were specifically told that Robin”dropped” the rock. In this problem, he could ”toss” it - i.e. impart anygiven initial velocity to it. Let us solve for the initial velocity, vo. Let ususe the same co-ordinates at in problem three. The origin is on the ground,and upward is positive. Using:
v2f = v2
o + 2a(yf − yo), we find:(−28m/s)2 = (vo)2 + 2(−9.8m/s2)(1.5m− 8.5m).6
Solving for vo, we find: (vo)2 = 5.71m2/s2. Taking the square root, wefind vo = ±2.4m/s. Either velocity will work. In either case, the man-at-arms will be struck with the rock descending at 28 m/s.
5. A physics student atop the library, 45.5 m above ground, observes abaseball traveling straight upward at 2.3 m/s. Assuming the baseball waslaunched from the ground, what initial velocity did it have at launch?
Using v2f = v2
o+2a(yf−yo), we find: (2.3m/s)2 = (vo)2+2(−9.8m/s2)(45.5m−0m). Solving, we find v2
o = 887m2/s2 and thus vo = ±30m/s. As the base-ball must have been launched up, we choose the plus sign.
6. A truck passes by a dog at rest at a speed of 15.0 m/s. One secondlater, the dog begins to accelerate at 8.0m/s2 until he reaches his maximumspeed of 22.0 m/s. How far does the dog travel before catching the truck?Assume the truck does not accelerate.
5If you solved for only 40. m then you neglected to add in the height of the man-at-arms.
6here we have assumed Robin is atop the tree. Other positions within the tree will leadto other velocities.
5
Both truck and dog will travel equal distances, in equal times in order tobe in the same place at the same time. Writing the truck’s position versustime is simple:
dtruck = votruckt.The dog is more complicated. The dog can either catch the truck before
reaching maximum speed, or after having already reached maximum speed.Let us discern between the two possibilities. It will take the dog:
t = v/a = (22.0m/s)/(8.0m/s2) = 2.75sto achieve maximum speed. Find where both dog and truck are at this
time. If the dog has passed the truck, the dog is accelerating until he reachesthe truck. Otherwise we must employ other expressions for the dog. At thistime, the truck, with it’s one second head start, has traveled:
d = vt = 15m/s ∗ 3.75s = 56.25m,while the dog has traveled:d = 1/2 ∗ at2 = 1/2 ∗ 8.0m/s2(2.75s)2 = 30.25m.As the dog has not yet passed the truck at this time, but is 26 m behind.
To find when the dog catches the truck, we can employ the concept ofrelative velocity. (We can only do this because both dog and truck are notaccelerating.) The dog travels at 22 m/s and the truck at 15 m/s thus thedog is gaining on the truck at:
22m/s− 15m/s = 7m/s.Thus to find the time we divide the distance to be made up, 26m, by the
rate at which the dog catches up, 7 m/s to get:t2 = 26m/(7m/s) = 3.71s.The total time since the truck passed the dog is:3.71s + 3.75s = 7.46s.We are asked how far the dog has traveled. The easy way to find this is
to note that the dog and truck meet at this point, thus we can simply usethe distance the truck has traveled, which is easy to compute:
dtruck = vtruck ∗ ttotal = 15m/s ∗ 7.46s = 111.9m.We can also compute the distance the dog has traveled:ddog = 1/2at2 + vdog ∗ t2 = 1/2 ∗ a ∗ (2.75s)2 + 22m/s ∗ 3.71 = 111.9m.Note there are three times involved here.7. A coin is dropped into a well 55.0 m deep. If sound travels at 343
m/s, how long before the splash is heard?First we find t1, the time for the coin to strike water, next we find t2, the
time for sound to return. Summing these times gives the total time beforesound is heard. First, the coin is dropped, thus it’s initial velocity is zero.We use:
d = 1/2 ∗ at2 to find t1.
6
Plugging in we find:55.0m = 1/2 ∗ 9.80m/s2 ∗ t21.Solving for t1, we find: t1 = 3.35s. Next we find t2. The sound travels
at constant speed, thus:d = vsoundt2 = 55.0m = 343m/st2.Solving for t2, we find: t2 = 0.160s. Summing we find the total time is
t = 3.35s + 0.160s = 3.51s. (Rounding to three significant figures).
2 Non-standard Problems
1. Ant one crawls clockwise from the one o’clock hand of a giant clock(radius 1.00 meter) at 0.25 m/s. At what speed does ant two have to crawlto meet ant one at the 6 o’clock hand if ant two starts at the same timefrom the 9 o’clock hand and crawls counter clock wise?
2. A Caltrans truck is dropping orange cones onto the road at 5.0 meterintervals.
(a) If the truck is driving at 22 m/s, how long is the time interval betweendrops?
(b) If the minimum time interval between drops is 0.75 seconds, what isthe fastest speed the truck can drive and still maintain the 5.0 meter intervalbetween cones?
(c) The truck begins to accelerate at 1.0m/s2 from 3.0 m/s up to 15.0m/s. Write down an equation for the drop interval as a function of time.
3. The quad is a 100.0 m by 50.0 m rectangular patch of grass. If Padmebegins to walk North (along the long East edge of the quad) at 0.70 m/s,at what speed does Anakin have to walk to meet her if he starts at theNorth-West corner (and traverses the short North edge).
4. Giancoli’s book has dimensions of 30.0 cm high, 12.0 cm wide and4.0 cm deep. If the book stands upright, and an ant is dropped off the top(30.0 cm off the ground) of the book, at what speed does a ladybug haveto travel to catch the falling ant if the ladybug starts off at each of the fourcorners of the book which rest on the ground? Assume the ladybug travelsat constant speed, but can change direction instantaneously.
5. A boat is traveling at 4.0 m/s down river relative to the river. Theriver is moving at 1.5 m/s relative to the shore. A boy on the boat hurls arock off the back side of the boat at 15.0 m/s upriver. What is the relativevelocity of the rock with respect to the shore? What would change if theboy were running toward the back of the boat at 6.6 m/s?
7
Non-standard Problem Solutions
1. Ant one crawls clockwise from the one o’clock hand of a giant clock(radius 1.00 meter) at 0.25 m/s. At what speed does ant two have to crawlto meet ant one at the 6 o’clock hand if ant two starts at the same timefrom the 9 o’clock hand and crawls counter clock wise?
First, find the time needed for ant one to arrive at the 6 o’clock hand.Next find the distance ant two must travel to arrive there at the same time.Third, divide distance by time to get speed.
First, ant one travels from the one o’clock hand to the six o’clock hand.This is five hours (of a total of twelve) and thus is 5/12 of the entire cir-cumference, which is given by C = 2πr. Thus the distance ant one travelsis 5/12 ∗ 2 ∗ π ∗ 1.0meter = 2.62m. Traveling at 0.25 m/s, ant one takest = d/v = 2.62m/0.25m/s = 10.5s to arrive.
Ant two travels only three numbers of the clock face, thus 3/12 of theentire surface, which is 1.57 m. It must do so in the same time, 10.5s andthus must travel at 1.57m/10.5s = 0.150m/s.
2. A Caltrans truck is dropping orange cones onto the road at 5.0 meterintervals.
(a) If the truck is driving at 22 m/s, how long is the time interval betweendrops?
The truck travels 22 m/s, which means to travel 5.0 m, it takes 5.0 m /22 m/s = 0.23 s.
(b) If the minimum time interval between drops is 0.75 seconds, what isthe fastest speed the truck can drive and still maintain the 5.0 meter intervalbetween cones?
The truck can travel 5.0 m in 0.75s which is 5.0 m / 0.75s = 6.7 m/s.(c) The truck begins to accelerate at 1.0m/s2 from 3.0 m/s up to 15.0
m/s. Write down an equation for the drop interval as a function of time.First, let us write down v(t). Starting at t=0, v(t) = 3.0m/s+1.0m/s2t.
To find the drop interval, T, we divide the 5.0 m distance by the velocity atany time, t: T = 5.0m/v(t) = 5.0m
3.0m/s+1.0m/s2 .3. The quad is a 100.0 m by 50.0 m rectangular patch of grass. If Padme
begins to walk North (along the long East edge of the quad) at 0.70 m/s,at what speed does Anakin have to walk to meet her if he starts at theNorth-West corner (and traverses the short North edge).
Padme covers 100.0 m at a speed of 0.70 m/s, thus it takes her t =100.0m/0.70m/s = 140s to do so. Anakin has to cover 50.0 m in an equaltime to meet her, thus he should walk at 50.0m/140s = 0.36m/s.
4. Giancoli’s book has dimensions of 30.0 cm high, 12.0 cm wide and
8
4.0 cm deep. If the book stands upright, and an ant is dropped off the top(30.0 cm off the ground) of the book, at what speed does a ladybug haveto travel to catch the falling ant if the ladybug starts off at each of the fourcorners of the book which rest on the ground? Assume the ladybug travelsat constant speed, but can change direction instantaneously.
First we find the time for the ant to fall. Next we consider each of thefour corners of the book which rest on the ground. Dividing the distance bythe time we solve for each velocity.
The time for the ant to fall can be found using: d = vo ∗ t + 1/2 ∗ a ∗ t2.Since the ant is dropped, vo = 0. We call this time t=0, and we can easilysolve for t. Substituting we find: 30.0cm = 1/2 ∗ 980cm/s2 ∗ t2 and solvingwe find: t2 = 60cm/980cm/s2 = 0.0612s2 and thus t = ±0.247s. We choosethe plus sign.
Consider the ladybug at each of the four corners of the book which reston the ground. At the corner directly under the ant, the ladybug does nothave to move, so v1 = 0. Next consider the width of the book, 12.0 cm. Ifthe ladybug started there she would have to travel at v2 = 48.6cm/s. Nextif the ladybug starts at the depth of the book, 4.0cm, she has to travel onlyv3 = 16.2cm/s. Last, if the ladybug is at the ”far” corner of the book, shemust travel a total of 12.0 cm + 4.0 cm to catch the ant. Thus her speedmust be v4 = 16.0cm/0.247s = 64.8cm/s.
5. A boat is traveling at 4.0 m/s down river relative to the river. Theriver is moving at 1.5 m/s relative to the shore. A boy on the boat hurls arock off the back side of the boat at 15.0 m/s upriver. What is the relativevelocity of the rock with respect to the shore? What would change if theboy were running toward the back of the boat at 6.6 m/s?
Since the boat is traveling down river, along with the current, it is movingat 4.0 m/s + 1.5 m/s = 5.5 m/s with respect to the shore. A rock hurledoff the back of the boat (upriver) will seem slower as viewed from the shore.Thus we subtract: vrock = 5.5m/s − 15.0m/s = −9.5m/s where the minussign indicates a direction opposite to the boat.
3 Vectors
1. Buddy, the golden retriever, runs 105.0 m North, then 55.0 meters south.The total trip takes 33.0 seconds. Find each of the following quantities: (a)distance traveled (b) displacement (c) average speed (d) average velocity
2. Buddy runs 90.0 m North, then 44.0 m due West. Find Buddy’s finaldisplacement in rectangular and polar co-ordinates.
9
3. Buddy runs 38.0 m at 22.0◦ North of West. Then he runs 83.0 m at44.0◦ East of North. Find Buddy’s final displacement.
4 Projectile Motion
1. A cannon is fired imparting an initial speed of 430 m/s to a shell. Thecannon is aimed at 33◦ above the horizon on flat ground. Find: (a) range(b) time of flight (c) maximum height above flat ground reached (d) time toreach maximum height (e) final velocity vector just before the shell strikesthe ground
2. A shell is fired from a cannon on a hill that is 15.0 m above flatground. The cannon is fixed to fire at an angle of 45◦ . What initial speedmust be imparted to the shell for it to travel 200.0 meters?
3. A naughty child hurls a rock at a suspended power line strung 6.0 mabove ground. The rock is released 1.0 m above ground at 60◦ above thehorizontal and the child is 4.8 m (horizontally) away from the cable. Thechild imparts a speed of 8.5 m/s to the rock. Does the rock go over or underthe power line? By how much?
Projectile Motion Solutions
1. A cannon is fired imparting an initial speed of 430 m/s to a shell. Thecannon is aimed at 33◦ above the horizon on flat ground. Find: (a) range(b) time of flight (c) maximum height above flat ground reached (d) time toreach maximum height (e) final velocity vector just before the shell strikesthe ground
There are two problems here. The first deals with the entire trajectoryof the shell. The other deals with only half (going up). Since the projectileis fired on flat ground, the time of flight for going up equals the time forcoming down.7 We will prove this.
Part (a): to find the range, we first need the time of flight. The flightis y-limited, meaning the flight ends when a condition along the y-axis (upand down) is satisfied.8 We choose the initial position of the shell to be theorigin, so x0 and y0 are zero. Write down a y-equation: y = y0+voy−1/2gt2.Plugging in we find: 0 = 0+430m/s∗sin(33)t−1/2∗9.8m/s2∗t2. One powerof time cancels and simplifying: 0 = 430m/s ∗ sin(33)− 1/2 ∗ 9.8m/s2 ∗ t2.Now we can solve for time and we find t = 47.9s. This answers part (b).
7Strictly speaking, we must neglect friction for this to be true.8Compare with problem 3, which is x-limited.
10
To solve part (a), we plug into an x-equation: x = x0 + voxt + ax ∗ t2, butthere is no acceleration in the x-direction, leaving: x = x0 + voxt. Pluggingin we find: x = 0 + 430m/s ∗ cos(33) ∗ 47.8s = 17200m and rounding we getx = 17000m = 17km.
Part (c): This begins the second half of the problem, which asks onlyabout the first half of the flight. Although we can do part (c) first, let us firstfind the time and see that it equals exactly half of the full. The projectilereaches maximum height when the y-velocity is zero.9 Using this fact, weapply: vyf = vyo − gt. Plugging in we find: 0 = 430m/ssin(33)− 9.8m/s2t,solving we find t = 23.9s, which is half of the previous time. Now we caneasily find the maximum height via: y = y0 + voy − 1/2 ∗ g ∗ t2. Pluggingin we find: y = 0 + 430m/s ∗ sin(33) ∗ 23.9s − 1/2 ∗ 9.8m/s2 ∗ (23.9s)2 =5597m− 2800m = 2798m = 2.8km.
Part (e): This asks about the full flight. We know the time alreadyhaving solved part (b). Thus we plug in the time to the simplest equationsto find the final velocity components: vxf = vxo+ax∗t and vyf = vyo+ay ∗t.Since ax = 0 and ay is known, this is simple. Solving we find: vxf = 430m/s∗cos(33) = 360m/s and vyf = 430m/s∗sin(33)−9.8m/s2∗47.9s = −235m/s.This final y-velocity is negative indicating the shell is heading downward,and it is the negative of the initial velocity.10
2. A shell is fired from a cannon on a hill that is 15.0 m above flatground. The cannon is fixed to fire at an angle of 45◦ . What initial speedmust be imparted to the shell for it to land 200.0 meters, horizontally, fromlaunch?
Calling flat ground y = 0, then y0 = +15m and yf = 0m. Callingthe starting point x = 0, we know x0 = 0 and xf = 200.0m. Knowingthe position information, we apply the equation with maximum positioninformation in each direction:
x = x0 + v0cos(45)t + 1/2axt2
andy = y0 + v0sin(45) + 1/2ay ∗ t2.Each equation has two unknowns: v0 and time. Since the x equation is
9If the y-velocity is positive, then the shell is rising and a larger time will yield a largerheight. Conversely, if the y-velocity were negative, the shell is falling and an earlier timewill yield a larger height. Only if the y-velocity is exactly zero is the shell motionless,neither rising nor falling, at the maximum height.
10The reason for this will be explained later by conservation of energy. Consider: ne-glecting friction, objects come down at exactly equal and opposite to the speed the roseup, since during the flight no energy was added to or taken from the object. If the objectcame down at a different speed (faster or slower) there must have been a transfer of energy.It would not have undergone constant acceleration.
11
simpler, we solve it for time and plug into the y equation, thus eliminationtime altogether:
x = v0cos(45)tsolving for t:t = x
v0∗cos(45) .Now plug into the second equation:0 = y0 + v0 ∗ sin(45) ∗ x
v0∗cos(45) + 1/2 ∗ ay ∗ ( xv0∗cos(45))
2.Simplifying, we find:−y0 = x− 1/2 ∗ g ∗ v2
0 ∗ cos2(45).11
Plugging in and solving for v0 we find: −215m = −4.9m/s2∗400m2/(v20∗
1/2) and solving: v0 = 42.7m/s.3. A naughty child hurls a rock at a suspended power line strung 6.0 m
above ground. The rock is released 1.0 m above ground at 60◦ above thehorizontal and the child is 4.8 m (horizontally) away from the cable. Thechild imparts a speed of 8.5 m/s to the rock. Does the rock go over or underthe power line? By how much?
Find the vertical position of the rock when it has traveled 4.8 m in the x-direction. If it is above 6.0 m, the rock crosses over the power line. To begin,write an x-equation since we evaluate the flight after a certain x-conditionis met. First:
x = x0 + voxt + 0.We know all but time so solving for time:t = x−x0
vox= 4.8m/(8.5m/s ∗ cos(60)) = 1.13s.
Now we evaluate the y-position at this time:y = y0 + voyt− 1/2 ∗ g ∗ t2.Plugging in:y = 1.0m + 8.5m/s ∗ sin(60)− 4.9m/s2 ∗ (1.13s)2 = 1.53m.12
Thus the rock travels under the power cable, and misses by 6.0m−1.5m =3.5m, thus the child must throw faster or stand closer or try a different angleto succeed.
5 Force and Newton’s Laws
1. A 10.0 kg box is slid across flat ground by a 5.0 N force acting at 30◦
above the horizontal. Find all forces acting on the box and it’s acceleration.11This simplification notes that cos(45)=sin(45) to cancel them. If the angle were not
45◦ , the expression would be more complex.12Note the initial position is 1.0m since the rock is hurled from 1.0 m above ground,
assumedly due to the size of the child.
12
(Neglect friction).2. A 150.0 g statue of Isaac Newton is launched straight upward via
a small rocket which imparts a force such that the rocket has an initialacceleration of 2.2m/s2. Find the force exerted by the rocket.
3. A mother pulls a child in a little red wagon such that it acceleratesat 0.375m/s2. A second child clings to the back of the red wagon and glidesalong on roller skates. Given: the mass of the wagon/child system is 30.0 kg,the mass of the roller skating freeloader is 35.0 kg (older child), and Momis pulling at an angle of 60.0◦ above the horizontal, how much force is Momexerting? With how much force does the older child have to cling onto thewagon?
Solutions
1. A 10.0 kg box is slid across flat ground by a 5.0 N force acting at 30◦
above the horizontal. Find all forces acting on the box and it’s acceleration.(Neglect friction).
To solve, we sum the force vectors in each direction, x and y, individually.First in the x-direction, only a component of the pulling force is applied,Fpcos(30). Summing this one force we find:Fpcos(30) = max. Thus we cansolve for acceleration:
5.0N∗cos(30)10.0kg = ax = 0.43m/s2.
Secondly, in the y-direction three forces act: pulling, gravity and thenormal force. These sum to zero thus we write: Fpsin(30) + FN − Fg =may = 0. The only unknown force is the normal force. Solving for it wefind:
FN = Fg − Fpsin(30) = 10.0kg ∗ 9.8m/s2 − 5.0N ∗ sin(30) = 980N .Note that the y-component of the pulling force is so small that, when
rounded properly, the normal force equals gravity.2. A 150.0 g statue of Isaac Newton is launched straight upward via
a small rocket which imparts a force such that the rocket has an initialacceleration of 2.2m/s2. Find the force exerted by the rocket.
Summing the forces in the y-direction we find: Fr − Fg = may. Solvingfor Fr, we find:
Fr = may + Fg = m(ay + ag) = 0.150kg ∗ (2.2m/s2 + 9.8m/s2) = 1.8N .3. A mother pulls a child in a little red wagon such that it accelerates
at 0.375m/s2. A second child clings to the back of the red wagon and glidesalong on roller skates. Given: the mass of the wagon/child system is 30.0 kg,the mass of the roller skating freeloader is 35.0 kg (older child), and Momis pulling at an angle of 60.0◦ above the horizontal, how much force is Mom
13
exerting? With how much force does the older child have to cling onto thewagon?
The total mass Mom is pulling is 65.0kg. Thus in the x-direction:Fpcos(60.0) = (m1 + m2)a. From this we can find Fp = 48.8N . The olderchild has to pull herself forward only: F = ma = 35.0kg ∗ 0.375m/s2 =13.1N .
Friction Problems
1. A mass m slides down a plane inclined at angle θ above the horizontal.Find the minimum coefficient of friction such that the mass will not slowdown or stop.
2. A mass M is pulled horizontally across flat, frictionless surface. Atopmass M is a second mass m and a coefficient of friction µ exists between thetwo masses. Find the maximum force with which the masses can be pulledand remain together.
3. A 0.522 kg mass package rests on the roof of a house. The staticcoefficient of friction is 0.40 and the kinetic is 0.20. The roof is pitched at 25◦
above the horizontal. Does the package move? If so, find it’s acceleration.
Solutions to Friction Problems
1. A mass m slides down a plane inclined at angle θ above the horizontal.Find the minimum coefficient of friction, µ, such that the mass will not slowdown or stop.
Drawing a picture of the incline, choose coordinates down the plane (x)and normal to the surface (y). Drawing all the forces acting. Three forcesact: the normal force in the +y direction, friction in the -x direction, andgravity, which acts down. Breaking gravity into its components, we find:mgsinθ in the +x direction and mgcosθ in the -y direction. Sum forcesin each direction. In the y-direction the forces must sum to zero, thusFN −mgcosθ = 0. In the x-direction, the forces must also sum to zero: themass is not accelerating. Summing in the x-direction we find:
mgsinθ − Ff = 0.Knowing friction is µ time the normal force, expanding this expression:mgsinθ − µmgcosθ = 0Solving for µ:µ = mgsinθ
mgcosθ = sinθ/cosθ = tanθ.This is a kinetic coefficient of friction because the block is sliding. It
only depends on the angle of incline and is independent of gravity, mass,
14
and speed. Taking the limiting cases: if θ = 0, then any friction would slowthe block, thus µ = 0. If θ = 90o, the plane is vertical, thus an infiniteamount of friction would be needed to hold the block on. Thus µ =∞.
2. A mass M is pulled horizontally across flat, frictionless surface. Atopmass M is a second mass m and a coefficient of friction µ exists between thetwo masses. Find the maximum force with which the masses can be pulledand remain together.
First, draw a force diagram for each block. In the x-direction: for M,force Fp pulls it forward. For mass m, friction pulls it forward. However,since friction acts of m, there is an equal and opposite force which actson mass M.13 No friction slows block M. Setting friction on mass m to amaximum, we find:∑
Fx = max = Ff = µmg.This µis the static coefficient of friction because mass m is not slipping
(or sliding) across mass M. They are moving together.14
Now we sum forces on mass M (in the x-direction):∑Fx = Max = Fp − Ff = Fp − µmg.
Note we have the same acceleration for each block, ax, because they aremoving together. Now we have two equation with two unknowns: ax andFp are unknown. We solve for ax in the m equation and plug into the Mequation.
ax = µg from the m equation.Plugging into the M equation:Mµg = Fp − µmg.Solving for Fp:Fp = Mµg + mµg = (M + m)µg.As an exercise, you can re-do this problem with a second coefficient of
kinetic friction acting on mass M. The resulting expression for Fp will bemore complex.
3. A 0.522 kg mass package rests on the roof of a house. The staticcoefficient of friction is 0.40 and the kinetic is 0.20. The roof is pitched at25 degrees above the horizontal. Does the package move? If so, find it’sacceleration.
First we must find the component of gravity pulling the package downthe roof and compare it with the static friction. If gravity is greater, the
13Remember, for every force there is an equal and opposite force but they act on differentobjects.
14Thus friction does not always oppose motion. It opposes slipping. In this example,friction is causing motion.
15
package slips. If it is indeed slipping, we use the kinetic coefficient of frictionand sum forces to find the resulting acceleration.
Again, the component of friction down the plane is mgsinθ. The staticforce of friction is equal to µsFN . The normal force is equal and oppositeto the vertical component of gravity, which is mgcosθ. Thus we compare:mgsinθ to µsmgcosθ. Plugging in we find gravity is greater: 2.17N > 1.85N .Thus the package is sliding.
To find it’s acceleration, we sum the forces in the x-direction:∑Fx = max = mgsinθ − Ff .
This friction is the kinetic friction, which uses the smaller, kinetic coef-ficient. Solving for ax:
ax = gsinθ − µsgcosθ = g(sinθ − µscosθ).Which is independent of mass. Plugging in values we find: ax = 2.4m/s2.
Circular Motion Problems
1. A bucket, mass M, is swung in a vertical circle in uniform circular motionat a speed v. Find the tension force in the rope of length l holding thebucket at three points:
(a) when the bucket is directly below the pivot (bottom of its motion)(b) when the bucket is directly above the pivot(c) when the bucket is level with the pivot (the rope is completely hori-
zontal).2. A car, mass M, is traveling over a locally circular hill of radius R.
Find the maximum speed the car can travel over the hill without leavingthe ground.
3. We are all traveling in uniform circular motion about the center of theearth. The earth has a radius of 6400 km and revolves on its axis once perday. Consider the earth a perfect sphere (which is not true, the earth bulgesat the center) and find the difference in apparent weight of two identicalpeople: one at the north pole and one at the equator. Their masses are 70.0kg. Take the acceleration due to gravity to be 9.80m/s2.
Circular Motion Solutions
1. A bucket, mass M, is swung in a vertical circle in uniform circular motionat a speed v. Find the tension force in the rope of length l holding thebucket at three points:
(a) when the bucket is directly below the pivot (bottom of its motion)(b) when the bucket is directly above the pivot
16
Beware of the signs in this problem. (a) Tension acts upward, gravitydownward, and the result (the centripetal acceleration) is upward thus theequation is:∑
F = ma = mv2/r = Ft − Fg
Solving for Ft:Ft = mv2/l + mg.(b) In this part, both gravity and tension act downward and the result
is downward (negative):∑F = ma = −mv2/l = −Ft − Fg
Ft = mv2/l −mg.2. A car, mass M, is traveling over a locally circular hill of radius R.
Find the maximum speed the car can travel over the hill without leavingthe ground.
Drawing all forces acting on the car, we have the normal force actingupward, gravity downward, and the resulting acceleration is downward thus:∑
F = Ma = −Mv2/R = FN − Fg.To find the maximum safe speed, we set the normal force to zero. If the
car were to travel any faster, it would depart from the road:−Mv2/R = 0− Fg = −Mg.Canceling the M’s and solving for v, we find:v =
√Rg.
3. We are all traveling in uniform circular motion about the center of theearth. The earth has a radius of 6400 km and revolves on its axis once perday. Consider the earth a perfect sphere (which is not true, the earth bulgesat the center) and find the difference in apparent weight of two identicalpeople: one at the north pole and one at the equator. Their masses are 70.0kg. Take the acceleration due to gravity to be 9.80m/s2.
The forces acting on each person are the normal force and gravity. Atthe north pole, they are equal and opposite. At the equator, they are not.They sum to the centripetal force, which acts inward. Thus the normal forceis less than gravity and the person appears lighter. The difference in theweights is the centripetal force: mv2/r. To find the speed v, we divide thedistance traveled by the person at the equator: 2πr, where r is the radiusof the earth, by the period of revolution (T), one day:
v = 2πrT .
Substituting, we find the force is: 4π2rm/T 2. Converting the period toseconds and substituting we find:
4 ∗ π2 ∗ 6.4x106m ∗ 70.0kg/(8.64x104s)2 = 2.37N .Since the actual weight of each person is: mg = 70.0kg ∗ 9.80m/s2 =
686N , this is a small difference, about 0.3%.
17
6 Work and Energy
1. A sled of mass M is dragged across the quad at constant speed by astudent pulling at an angle θ (above the horizontal ground). There is acoefficient of friction, µ, between the sled and ground. Find the work doneby each and every force acting on the sled if the sled is pulled through adistance d in terms of the given variables and g, the acceleration of gravity.
2. Estimate the work needed to fill a bookcase with 1.20 kg physicsbooks. The case has 3 shelves which are located 10.0, 40.0 and 70.0 cm offthe ground. The books are 20.0 cm tall and ten books fit on each shelf.Initially the books are standing upright on the ground.
3. A bus accelerates from rest to 22.2 m/s in 9.4 seconds. Given a busof 6650 kg mass and a coefficient of friction of 0.25 find the work done bythe bus’s tires on the ground. Next, assuming the engine is 19.0% efficient,estimate the work done by the engine.
Work and Energy Solutions
1. A sled of mass M is dragged across the quad at constant speed by astudent pulling at an angle θ (above the horizontal ground). There is acoefficient of friction, µ, between the sled and ground. Find the work doneby each and every force acting on the sled if the sled is pulled through adistance d in terms of the given variables and g, the acceleration of gravity.
Four forces act on the sled: friction, normal, gravity and pulling force.The normal force and gravity do no work as they act perpendicular to thesurface.15 Friction and the pulling force must do equal but opposite amountsof work as the sled is moving at constant speed (not accelerating). All weneed to do is find either the frictional force or the pulling force in terms ofthe given variables and the problem is solved. To do so, we sum forces inthe x and y directions:∑
Fx = max = 0 = Fpcosθ − Ff = Fpcosθ − µFN
and:∑Fy = may = 0 = Fpsinθ + FN −Mg.
We have introduced a third unknown, FN , and removed friction. Nowwe have two equations and two unknowns. Since we want to solve for Fp,we will first solve for FN and substitute into the other equation (eliminating
15Many physics students believe the normal force can never do work. In fact, it can.For example, consider a rubber ball dropped vertically to the ground. The normal forcecauses the ball to rebound.
18
FN ). Looking at the equations, the y-equation appears easier to solve forFN :
FN = Mg − Fpsinθ.Substituting into the x-equation we find:0 = Fpcosθ − µ(Mg − Fpsinθ).Solving for Fp we collect terms together:Fpcosθ + µFP sinθ = µMg.Factoring out Fp we find:Fp(cosθ + µsinθ) = µMg.Dividing we find:Fp = µMg
cosθ+µsinθ .Thus the work done by the pulling force is:WF p = Fpdcosθ = dcosθµMg
cosθ+µsinθ .The work done by friction is just equal and opposite to this.2. Estimate the work needed to fill a bookcase with 1.20 kg physics
books. The case has 3 shelves which are located 10.0, 40.0 and 70.0 cm offthe ground. The books are 20.0 cm tall and ten books fit on each shelf.Initially the books are standing upright on the ground.
The work done is just overcoming gravity mgδh, but the thirty books tobe lifted are lifted to various heights. The height through which a book islifted is simply the height of each shelf.16 Thus there are three terms:
W = 10 ∗ 1.20kg ∗ 9.8m/s2 ∗ 0.100m + 10 ∗ 1.20kg ∗ 9.8m/s2 ∗ 0.400m +10 ∗ 1.20kg ∗ 9.8m/s2 ∗ 0.700m
Factoring out common terms we find:W = 10 ∗ 1.20kg ∗ 9.8m/s2 ∗ (0.100m + 0.400m + 0.700m) = 141J .This is only an estimate because of the extra force needed to lift the
books (exceeding gravity) and the loss of energy when the book strikes thebookshelf at non-zero speed.
3. A bus accelerates from rest to 22.2 m/s in 9.4 seconds. Given a busof 6650 kg mass and a coefficient of friction of 0.25 find the work done bythe bus’s tires on the ground. Next, assuming the engine is 19.0% efficient,estimate the work done by the engine.
To find the work done by the tires, we compute: W = Fdcosθ. As-suming the tires are pushing directly backward, θ = 0 and thus cosθ = 1.First we will find the distance, then the force. Finding the distance is a
16Do not be tempted to include half the height of each book into the change in height.Each book is initially sitting upright on the ground, thus the book’s center of mass ismoved only the height of the shelf.
19
constant acceleration problem. Given the speeds and time, we first find theacceleration via:
vf = vo + at.Which leads to: a = vf/t. Now we can find the distance traveled via:v2f = v2
o + 2a(x− xo).Solving we find:d = x− xo = v2
f/2a = vf t/2.17
Now we can find the force via summing forces. Already knowing theacceleration, and knowing the normal force is equal to gravity, we find:∑
F = ma = Ftires − Ff = Ftires − µmg.Solving for Ftires, we find:Ftires = µmg + ma = m(µg + a).Substituting for the known acceleration we find:Ftires = m(µg + vf/t).Substituting both force and distance into the work equation we find:Wtires = Ftires ∗ d = vfm(µg + vf/t)/2.Now substituting number we find:Wtires = 22.2m/s∗6650kg(0.25∗9.8m/s2+22.2m/s/9.4s)/2 = 3.55x105J .The second part of the question asks how much work the engine did if
it is 19.0% efficient. This means only 19.0% of the work done by the enginegoes to the tires (the rest to waste heat and friction in various bearings).Thus:
0.19 ∗Wengine = Wtires.Solving for work done by the engine:Wengine = Wtires/0.19 = 1.87x104J .
7 Conservation of Energy
1. Alice, entering wonderland, leaps down the hole after the white rabbit.However, the white rabbit has placed a giant spring at the bottom of thehole for Alice to land on. If Alice’s downward speed is vo at the top of thehole, how far does she rise up above the top of the hole after reboundingfrom the spring?
2. Joe Montana throws a football at a 22◦ angle above the horizontal at14 m/s and Jerry Rice catches the ball just millimeters off the ground. Find
17One way to think of this result is via average velocity. Since the bus accelerated fromrest to vf , it’s average speed was vf/2, thus the distance traveled is simply the averagespeed times time.
20
the angle at which the football makes with the ground when it is caught.Assume the ball is thrown from 1.8m off the ground.
3. Happy Fun Land, an amusement park, has a new ”attraction” inwhich people on ice skates are shot from a giant, horizontal, spring acrossa frozen lake. Each person is shot at 60 miles per hour (28 m/s), and thegiant spring has a spring constant of 55000 N/m. The coefficient of frictionbetween the ice and the skates is 0.10 (rather rough ice). The spring hasto be compressed a different amount for each visitor. For a 60.0 kg visitor,find: (a) the amount the spring should be compressed and (b) the distancethe person travels over ice before coming to rest.
4. A spring with constant k and length l is placed on the ground, orientedvertically. On the ceiling above it is an identical spring. The room, floor toceiling, is height h. A ball is placed on spring one and compressed a certaindistance, x, such that when it rises up toward the ceiling and encounters thesecond spring, the spring compresses exactly half x. Find an expression forx.
Conservation of Energy Solutions
1. Alice, entering wonderland, leaps down the hole after the white rabbit.However, the white rabbit has placed a giant spring at the bottom of thehole for Alice to land on. If Alice’s downward speed is vo at the top of thehole, how far does she rise up above the top of the hole after reboundingfrom the spring?
Irregardless of how deep the hole is, or what the spring constant is,Alice’s energy is conserved. At the top of the hole Alice has gravitationalpotential energy (let’s call the bottom of the hole y=0) and kinetic energy.In the final situation, she has only gravitational potential energy:
mgho + 1/2mv2o = mghf .
Solving for hf − h− o, which is the distance Alice rose up above the topof the hole we find:
hf − ho = v2o
2g .2. Joe Montana throws a football at a 22◦ angle above the horizontal at
14 m/s and Jerry Rice catches the ball just millimeters off the ground. Findthe angle at which the football makes with the ground when it is caught.Assume the ball is thrown from 1.8m off the ground.
There is no acceleration in the x-direction so the final x velocity is equalto the initial. Further, we can find the final speed via conservation of energy:
1/2mv2o + mgh = 1/2mv2
f .
21
Plugging in and canceling the masses, we find: vf = ±15.2m/s. Theplus or minus will not matter because we will only use v2
f (see below). Next,we need to find the y component of final velocity using the Pythagoreantheorem:
v2f = v2
x + v2y .
Where vx and vy refer to the final x and y velocities. We know the xvelocity is given by vocosθ = 14m/scos(22 ◦ ) = 13.0m/s. Solving for theunknown, vy, we find:
(±15.2m/s)2 = (13.0m/s)2 + v2y
Which leads to:vy = ±7.9m/s.Since we know the ball must be traveling downward to strike the ground,
we choose the minus sign.18 Having both, we can use the tangent functionto find the final angle via:
tanθ = vy/vx = −7.9m/s/13.0m/s = −0.608.Thus taking the arctangent19 we find: θ = −31.3◦ . The minus sign
means the ball is heading down, 31.3◦ below the horizontal.3. Happy Fun Land, an amusement park, has a new ”attraction” in
which people on ice skates are shot from a giant, horizontal, spring acrossa frozen lake. Each person is shot at 60 miles per hour (28 m/s), and thegiant spring has a spring constant of 55000 N/m. The coefficient of frictionbetween the ice and the skates is 0.10 (rather rough ice). The spring hasto be compressed a different amount for each visitor. For a 60.0 kg visitor,find: (a) the amount the spring should be compressed and (b) the distancethe person travels over ice before coming to rest.
First, we can find the amount of compression via conservation of energy.The initial form of energy is elastic potential the final is kinetic (very littleis lost over this short distance). We write:
Eo = 1/2kx2 = Ef = 1/2mv2.Solving for x, we find:x2 = mv2/k.Plugging in our values, we find: x = 0.855m or 85.5 cm.Next we find how far the person travels over the frozen lake. We could
sum forces and find an acceleration, or we could use the work-energy theo-rem. The only non-conservative force acting on the person is friction, thus
18The other solution has a physical meaning, too. It corresponds to the speed withwhich the ball would have had to leave the ground (somewhere behind Joe Montana) toarrive in Joe’s hand with that exact speed and angle.
19Beware: calculators in degree mode will return an answer in degrees. Calculators inradians mode will return answers in radians.
22
the initial kinetic energy is all dissipated into frictional heating (melting) ofthe ice. We know: Ff = µFN and the normal force is equal and oppositeto gravity (in this case): FN = mg. The work done by friction is given by:Wf = Ffd. The work-energy theorem yields:
Wf = µmgd = EK = 1/2mv2o .
Canceling the m’s and solving for d, we find:d = v2
o2µg = 400.0m.
Note this result is independent of mass thus all riders will arrive at thesame spot on the ice.
4. A spring with constant k and length l is placed on the ground, orientedvertically. On the ceiling above it is an identical spring. The room, floor toceiling, is height h. A ball is placed on spring one and compressed a certaindistance, x, such that when it rises up toward the ceiling and encounters thesecond spring, the spring compresses exactly half x. Find an expression forx.
The initial energy is gravitational potential plus elastic potential. Thefinal is gravitational potential plus elastic potential. Writing an expressionwe find:
mg(l − x) + 1/2kx2 = mg(h− l + 1/2x) + 1/2k(x/2)2.Grouping terms we find:3/8kx2 = mg(h− 2l) + 3/2mgx.Applying the quadratic equation we find:
x = 3/2mg±√
9/4(mg)2+3/2kmg(h−2l)
3/4k .This is not pretty, but it is an expression for the amount of compression
needed. If, when plugging in numbers, a complex or imaginary answer isthe result (the square root of a negative number is taken), then no physicalvalue of x is attainable for those numbers. Not all physics problems havenice, neat answers.
8 Momentum and Collisions
1. Ice skaters of mass m are shot from a giant spring (see above) at speed vo
over ice which has a coefficient of friction of µ. However, to allow them tocome to rest naturally takes too long and too much ice (hundreds of metersof ice). Thus, the management of Happy Fun Land has decided to have theskaters travel only a distance d over the ice before colliding with a giant -but soft - sphere of mass M (initially at rest). After this (inelastic) collision,the skaters are supposed to cling to the sphere. Find: (a) the speed at whichthe skaters strike the sphere, (b) velocity of the skater-sphere system just
23
after the collision, and (c) the distance they travel together before comingto rest.
2. Two bumper cars have an elastic collision. The first is initially trav-eling at speed v1 west and has mass m1. The second is initially travelingeast at speed v2 and has mass m2. Find the final velocity of each car.
3. A giant beer advertisement is floating, serenely, across a frozen, fric-tionless, lake when it is struck with a giant paint pellet traveling at rightangles to the advertisement. (Thus if the ad was moving North, the pelletwas traveling eeither East or West). The mass of the giant pellet is 50.0g. The velocity of the pellet is 400.0 m/s and the velocity of the ad was aglacial 2.0 m/s. After the collision, they are found to be moving off at anangle of 10◦ from the original direction of motion of the ad. Find: the massof the ad and the final speed of the re-painted ad.
Momentum Solutions
1. Ice skaters of mass m are shot from a giant spring (see above) at speed vo
over ice which has a coefficient of friction of µ. However, to allow them tocome to rest naturally takes too long and too much ice (hundreds of metersof ice). Thus, the management of Happy Fun Land has decided to have theskaters travel only a distance d over the ice before colliding with a giant -but soft - sphere of mass M (initially at rest). After this (inelastic) collision,the skaters are supposed to cling to the sphere. Find: (a) the speed at whichthe skaters strike the sphere, (b) velocity of the skater-sphere system justafter the collision, and (c) the distance they travel together before comingto rest.
First, to find the speed at which skaters strike the sphere, we can use ei-ther acceleration of energy. Let us choose acceleration. Summing horizontalforces, we find only friction acting thus:∑
Fx = max = Ff = µFN .To find the normal force, we sum forces in the y direction and find the
normal force equals gravity: FN = Fg = mg. Solving for ax, we find:ax = µFN/m = µmg/m = µg.Next, we apply an equation of constant acceleration over a distance d to
find the final velocity using:v2f = v2
o + 2a(x− xo).We are not going to call this final velocity final because there are later
changes in velocity. This is only an intermediate velocity so we shall call itv1. Thus: vf = v1. Plugging in our acceleration and distance we find:
v21 = v2
o + 2µgd.
24
Taking the square root we find:v1 = ±
√v2o + 2µgd.
Next, we find the velocity of the skater-sphere system just after thecollision by using conservation of momentum. It is an inelastic collision,thus we cannot easily apply conservation of energy.20
pi = mv1 = pf = (m + M)v2.Solving for v2, we find:v2 = m
M+mv1 = mM+m
√v2o + 2µgd.
Lastly, we are asked to find the distance the skater clinging to the spheretravels before coming to rest. Let us call this distance x, since d is alreadyused. Using the work-energy theorem, we find:
Wf = Ffd = µ(M + m)gx = ∆E = 1/2(m + M)v22.
Solving for the distance, x, we find:x = v2
22µg .
Substituting our expression for v2 from above we find:x = m2(v2
o+2µgd)2µg(M+m) .
Notice that the square root and the square canceled out, thus the plusor minus did not come into play.
2. Two bumper cars have an elastic collision. The first is initially trav-eling at speed v1 west and has mass m1. The second is initially travelingeast at speed v2 and has mass m2. Find the final velocity of each car.
We have two equations: conservation of energy and momentum, andwe need to solve for two unknowns, v′1 and v′2. Writing conservation ofmomentum, and choosing west to be the positive direction:
v1m1 − v2m2 = v′1m1 + v′2m2.Note we choose the plus sign in the final momentum because we are al-
lowing v′1 and v′2 to be either positive or negative. If positive, they indicatewestward travel, if negative then imply eastward travel. Writing conserva-tion of energy:
1/2m1v21 + 1/2m2v
22 = 1/2m1v
′21 + 1/2m2v
′22 .
Note all the factors of one half cancel from this equation. To solve firstfor v′1, we will eliminate v′2 from our equations. We must solve one of theequation for v′2 and substitute into the second. The momentum equationlooks easier, so let us begin there:
v1m1 − v′1m1 − v2m2 = v′2m2.Solving for v2, we find:v′2 = m1
m2(v1 − v′1)− v2.
20Energy is always conserved, however some of it goes into forms which are hard tomeasure, like heat, light and sound.
25
Plugging into the energy expression we find:m1v
21 + m2v
22 = m1v
′21 + m2(m1
m2(v1 − v′1)− v2)2.
Dividing by m1 and expanding the square we find:v21 + (m2/m1)v2
2 = v′21 + (m1/m2)(v1− v′1)2 + (m2/m1)v2
2 − 2v2(v1− v′1).The v2
2 terms cancel and, expanding further we are left with:v21 = v′21 − 2v2v1 + 2v2v
′1 + (m1/m2)v2
1 − 2(m1/m2)v1v′1 + (m1/m2)v′21 .
We will need to apply the quadratic equation. To do so, we collect termsof like powers of v′1:
0 = −v21 − 2v2v1 + v′1(2v2 − 2(m1/m2)v1) + v′21 (1 + m1/m2).
Plugging in to the quadratic formula we find:
v′1 = 2(m1/m2)v1−2v2±√
(2v2−2(m1/m2)v1)2−4(−v21−2v2v1)(1+(m1/m2))
−2v21−4v2v1
.Extensive simplification of this expression is possible. One possible result
of such simplification is:
v′1 = m2v2−m1v1±√
v22m2
2+v21(m1+m2)2+v1v2(2m2
2−m1m2)
m2(v21−v2v1)
.This is not the best expression, only one of many. To solve for v′2, we
now apply one of our previous expressions:v′2 = m1
m2(v1 − v′1)− v2.
Since we now know v′1, we also know v′2. The problem is solved, despitethe algebraic complexity of the remaining expressions.
3. A giant beer advertisement is floating, serenely, across a frozen, fric-tionless, lake when it is struck with a giant paint pellet traveling at rightangles to the advertisement. (Thus if the ad was moving North, the pelletwas traveling eeither East or West). The mass of the giant pellet is 50.0g. The velocity of the pellet is 400.0 m/s and the velocity of the ad was aglacial 2.0 m/s. After the collision, they are found to be moving off at anangle of 10◦ from the original direction of motion of the ad. Find: the massof the ad and the final speed of the re-painted ad.
First, we determine this is an inelastic collision. Thus we have conser-vation of momentum in the x and y directions independently. Second, thepaint sticks to the ad, thus the two move off together. Thus we have twoequations with two unknowns. Writing conservation of momentum:
px = MVo = (M + m)vfcosθ,py = mvo = (M + m)vfsinθ.Where M is the mass of the ad and m is the mass of the pellet, Vo is the
initial speed of the ad and vo is the initial speed of the pellet. Solving thesecond equation for the final speed:
vf = mvo(M+m)sinθ .
Plugging back into the first equation we find:
26
MVo = (M + m) mvo(M+m)sinθ cosθ.
Simplifying we find:MVo = mvo
tanθ .Solving for M, we find:M = mvo
Votanθ .Plugging in numbers we find:M = 56.7kg.Be sure to convert the grams to kilograms. Next we plug this mass back
into an equation above to find the final velocity:vf = 2.03m/s.Which is almost exactly the initial velocity.
9 Angular Motion
1. A car rounds a locally circular turn at 22.2 m/s. From the center of theturn to the center of the car is 15.0 m. The driver is 40 cm further out andthe passenger is 40 cm closer in to the center of the turn. Find the followingquantities: the angular speed, angular acceleration, and linear accelerationof both the driver and the passenger.
2. At Happy Fun Land, visitors can step onto a giant revolving diskof radius R and experience a force of twice gravity. The edge of the disk(where visitors stand) is elevated at an angle θ above the horizontal. Findthe speed necessary such that the visitors will stand comfortably upright onthe disk, at a right angle to the disk.
3. A small steel ball of mass m revolves about a roulette wheel (notmoving) of radius r. The ball completes N revolutions every second. Findthe linear and angular speed of the ball, the linear acceleration of the ball,and how far the ball travels in 1.00 seconds, and how large an angle the balltravels in that time interval.
Angular Motion Solutions
1. A car rounds a locally circular turn at 22.2 m/s. From the center of theturn to the center of the car is 15.0 m. The driver is 40 cm further out andthe passenger is 40 cm closer in to the center of the turn. Find the followingquantities: the angular speed, angular acceleration, and linear accelerationof both the driver and the passenger.
The angular quantities are the same for the driver and passenger. Theangular acceleration is zero in both cases. The angular velocity is given byv/r = 22.2m/s/15.0m = 1.48 radians per second. The linear acceleration of
27
each is given by v2/r - where r is the distance from the center of the turnto each person thus we find:
ad = (22.2m/s)2/15.4m = 32m/s2
ap = (22.2m/s)2/14.6m = 34m/s2.2. At Happy Fun Land, visitors can step onto a giant revolving disk
of radius R and experience a force of twice gravity. The edge of the disk(where visitors stand) is elevated at an angle θ above the horizontal. Findthe speed necessary such that the visitors will stand comfortably upright onthe disk, at a right angle to the disk.
There are just two forces acting on the visitor, gravity and the normalforce. Drawing a picture, we sum forces in each direction:∑
Fx = max = mv2/r = FNcosθand:∑
Fy = 0 = FNsinθ −mg.The y forces must sum to zero since the visitor is not rising off the disk.
The x-forces must exactly equal the centripetal force so that the visitor canstand straight upward without having to lean one way or the other. Solvingthe y equation for the normal force we find:
FN = mgsinθ .
Plugging in to the x equation we find:mv2/r = mg
sinθ cosθ.Canceling the masses and making a simple trig substitution:v2 = rg/tanθ.Thus the speed is given by:v = ±
√rg
tanθ .3. A small steel ball of mass m revolves about a roulette wheel (not
moving) of radius r. The ball completes N revolutions every second. Findthe linear and angular speed of the ball, the linear acceleration of the ball,and how far the ball travels in 1.00 seconds, and how large an angle the balltravels in that time interval.
The speed of the ball is given by distance over time. The total distancetraveled per second is: 2πr ∗N . During which time it travels an angle givenby: θ = d/r in radians, thus θ = 2πrN/r = 2πN radians. The angularspeed is θ/t = 2πN radians per second and the linear speed is given by:2πrN meters per second. The linear acceleration of the ball is given by:v2/r = (2πrN)2/r = (2πN)2r.
28
10 Torque
1. Two masses, m1 and m2, hang vertically from a pulley of radius r andmoment of inertia I by a massless cord. Find the angular acceleration of thesystem. (You may assume m2 > m1 if it helps).
2. A group of 86 gram mice decide to lift Giancoli’s 2.2 kg book byplacing a long, thin lever arm underneath it. They use a small rock as apivot. The distance from the pivot to the book is 9.5 cm and from the pivotto the end of the rod is 24.4 cm. How many mice must cling to the end ofthe rod to lift the book?
3. A set of five nuts of radius r hold a tire on a car. Each needs a torqueτabsurd to release them. To apply this torque, the car manufacturer providesa tire iron with a lever arm of radius R. Find a formula for the force neededto be applied to the end of the tire iron to loosed each lug nut. Next, if a75 kg man standing on the 35 cm radius lever arm cannot loosed the 2.0 cmradius lug nuts, what does this say about τabsurd? Finally, what does it sayabout the force acting on each of the six points of the lug nuts?
Torque Solutions
1. Two masses, m1 and m2, hang vertically from a pulley of radius r andmoment of inertia I by a massless cord. Find the angular acceleration of thesystem. (You may assume m2 > m1 if it helps).
Two different tensions act in the two different cords. Let T1 be the ten-sion in the cord to mass one and T2 be the tension to mass two. Define up aspositive for each mass, thus they will have equal, but opposite, accelerations.Summing forces on mass one:
m1a1 = T1 −m1g.Summing forces on mass two:m2a2 = T2 −m2g.Yet we know the two accelerations are linked - if one mass moves up
five centimeters, mass two must move down five centimeters. Just as thedisplacements are equal but opposite, so are the velocities and accelerations.Assuming m2 is lighter, it will rise and thus we choose a2 as positive:
a2 = −a1 = a.Now there is just one acceleration in the problem. Next we write the
sum of torques on the pulley:∑τ = Iα = T1r − T2r.
The angular acceleration, α, is also related to the one acceleration21 via21This assumes the rope does not slip over the pulley.
29
α = a/r. Now we solve equation one for T1:T1 = m1(g − a),and equation two for T2:T2 = m2(a + g),and substituting into the torque equation we find:Ia/r = m1(g − a)r −m2(a + g)r.The last unknown is acceleration, a. Grouping terms we find:a(I/r + m1r + m2r) = m1gr −m2gr = (m1 −m2)gr.Solving for a, we find:a = (m1−m2)r2
I+m1r2+m2r2 .Thus we see the acceleration depends on the difference in masses. If they
are equal, no acceleration will occur - the system can still move at constantspeed. Further, the denominator is effectively a sum of moments of inertia.Since the masses hang from the radius r, it is as if they are point massesstuck to the edge of the disk.
2. A group of 86 gram mice decide to lift Giancoli’s 2.2 kg book byplacing a long, thin lever arm underneath it. They use a small rock as apivot. The distance from the pivot to the book is 9.5 cm and from the pivotto the end of the rod is 24.4 cm. How many mice must cling to the end ofthe rod to lift the book?
Balancing the torques about the pivot, we have a number, n, mice onone side and Giancoli’s book on the other:∑
τ = 0 = 2.2kg9.8m/s20.095m = n0.086kg9.8m/s20.244m.Solving for n we find:n = 9.96 mice.Thus when the tenth mouse leaps onto the end of the rod, they will
overcome the weight of the book and lift it.3. A set of five nuts of radius r hold a tire on a car. Each needs a torque
τabsurd to release them. To apply this torque, the car manufacturer providesa tire iron with a lever arm of radius R. Find a formula for the force neededto be applied to the end of the tire iron to loosed each lug nut. Next, if a75 kg man standing on the 35 cm radius lever arm cannot loosed the 2.0 cmradius lug nuts, what does this say about τabsurd? Finally, what does it sayabout the force acting on each of the six points of the lug nuts?
To find the torque, we use τ = RFsinθ. Assuming the force is appliedat the optimal angle we find:
τabsurd/R = F .Next, we set F to the weight of the man and plug in numbers to set a
limit on τ :
30
τabsurd > 0.35m75kg9.8m/s2 = 260mN .Next, to compute the force on each individual point of the lug nut, we set
the torque due to these six forces equal to the torque from the man standingon the tire iron:
6rFnut = 260mN .Solving for Fnut = 2200N . This is not a limit, it is the force being
applied currently. More force will need to be applied before the nut beginto rotate, if the metal does not deform first (possibly stripping the nut).
11 Angular Momentum and Angular Kinetic En-ergy
1. A bullet, mass m is shot horizontally at the center of a vertically hangingrod (free to pivot about the top end) at a speed vo. It exists from the otherside of the rod (which has mass M and length L) at half it’s initial speed.Find the angular speed of the rod just after the bullet passes through it.
2. A ball of radius R is rolling without slipping at linear speed v. Findthe radio of translational to rotational kinetic energy.
3. A disk of mass m and radius r is rotating with period T on a friction-less surface. An identical disk is dropped on top of the first one and theyeventually rotate together. Find the final angular speed of the two disks.
Angular Momentum and Angular Kinetic Energy Solutions
1. A bullet, mass m is shot horizontally at the center of a vertically hangingrod (free to pivot about the top end) at a speed vo. It exists from the otherside of the rod (which has mass M and length L) at half it’s initial speed.Find the angular speed of the rod just after the bullet passes through it.
This is a conservation of angular momentum problem. Using the pivotatop the hanging rod as the origin, we compute the initial and final angularmomentums as:
Li = mvoL/2 = Lf = mv0/2L/2 + Iω.Solving for ω, we find:ω = mvoL
4I .2. A ball of radius R is rolling without slipping at linear speed v. Find
the radio of translational to rotational kinetic energy.The translational energy is given by: 1/2mv2 and the rotational by:
1/2Iω2. We need to find both I and ω. First, we look up the moment of
31
inertia of a sphere and find: I = 2/5mR2. Next, we relate the angular tolinear speed: ω = v/R. Substituting we find:
Erot = 1/2 ∗ 2/5 ∗mR2 ∗ (v/R)2.Forming the ratio of the two, we find many terms cancel leaving:Et/Er = 5/2.3. A disk of mass m and radius r is rotating with period T on a friction-
less surface. An identical disk is dropped on top of the first one and theyeventually rotate together. Find the final angular speed of the two disks.
This is an angular collision problem where angular momentum (only) isconserved. To find the initial angular momentum, we must find the angularspeed, ω. Since the period is given, we can find the linear velocity via:v = 2πr/T and the angular velocity via: ω = v/r = 2π/T .
Next we find the moment of inertia of a disk is 1/2mr2 and now we areready to apply conservation of angular momentum:
Lo = Iωo = Lf = 2Iωf
From this we note the final angular speed must be half the initial thus:ωf = 1/22π/T = π/T .
12 Static Equilibrium
1. A lamp, mass m, burns at the end of a horizontal rod, mass M length L,connected by a pin connection to a (vertical) wall. Assisting in the supportof the rod/lamp system is a thin (massless) cord from the wall to the endof the rod. This cord makes a 30◦ angle to the rod (thus a 60◦ angle to thewall to form a 30/60/90 triangle). Find the tension in the cord and the fullforce vector applied by the pin.
Static Equilibrium Solutions
1. A lamp, mass m, burns at the end of a horizontal rod, mass M length L,connected by a pin connection to a (vertical) wall. Assisting in the supportof the rod/lamp system is a thin (massless) cord from the wall to the endof the rod. This cord makes a 30◦ angle to the rod (thus a 60◦ angle to thewall to form a 30/60/90 triangle). Find the tension in the cord and the fullforce vector applied by the pin.
Drawing a force diagram of the rod we find four forces acting: gravityon the rod, gravity on the lamp, tension and the pin force. Since the rod isconnected via a pin, the pin connection can apply this force at an arbitraryangle. We call this angle θ above the horizontal, and the pin force Fp.
32
The tension will act directly along the cable and we call it’s magnitude T .Summing force in each direction we find:∑
Fy = may = 0 = Tsin30 + Fpsinθ −Mg −mg,and∑
Fx = max = 0 = Fpcosθ − Tcos30.Yet we have three unknowns, so we need a third equation. This is the
torque equation. We can sum the torques about any axis we wish. Letus choose summing about the pin connection, which will eliminate two un-knowns. Summing here we find:∑
τ = Iα = 0 = L/2Mg + Lmg − LTsin30.Here we have chosen a particular direction for the torques to be positive.
Our choice is that torques causing the rod to move down are positive andtorques countering this motion are negative.22 Solving for the one unknownwe find:
T = g(M/2+m)sin30 .
Knowing sin30=1/2, and substituting this result into our x and y equa-tions above to reveal:
g(M/2 + m) + Fpsinθ = g(M + m)and:Fpcosθ = g(M/2+m)
tan30 .Simplifying the first equation we find:Fpsinθ = gM/2.To solve this system of equations, we can divide one equation by the
other to find:tanθ = gM/2tan30
g(M/2+m) ,which simplifies to:tanθ = tan30
1+2m/M .To solve for Fp, we can square both above equations and sum them,
eliminating the sin and cos terms via:sin2θ + cos2θ = 1.We find:F 2
p sin2θ + F 2p cos2θ = (gM/2)2 ∗ (1+2m/M)2
tan230,
simplifying we find:F 2
p = (gM/2)2 ∗ 3(1 + 2m/M)2
and taking the square root we find:Fp = ±
√3g(M/2 + m).
The plus or minus indicates the angle could be off by 180◦ .22This is a completely arbitrary choice as is easily seen since the equation can be mul-
tiplied by minus one without changing the solutions.
33
13 Fluids
1. A simple hollow toy ball (think ping pong ball) will implode when thetotal pressure reaches 2.2 atmospheres on it’s surface. Such a ball is placedin a vat of unknown fluid and submerged until it implodes, at a depth of1.82 meters. Find the density of the fluid and it’s specific gravity.
2. A spherical bead is placed in the center of a glass of pure water andreleased. The bead has a radius of 1.0 cm and a mass of 7.6 grams. Willthe bead move up or down? What is it’s (initial) acceleration? Fixing theradius, what would the mass of the bead have to be for it not to move upor down but just float freely?
3. A barrel is full of water. It is an upright cylinder with a diameter of30 cm and a height of 42 cm. At a point 7 cm from the bottom, a smallhole is drilled, horizontally, into the barrel and the water comes pouringout. Find the initial (horizontal) velocity of the water and where the streamwill land (how far from the base of the barrel). Next, assume the hole is1 mm in diameter and find the rate at which the water level in the barrelis decreasing (at t=0, right as the hole is drilled - we can find the rate ingeneral later).
Fluids Solutions
1. A simple hollow toy ball (think ping pong ball) will implode when thetotal pressure reaches 2.2 atmospheres on it’s surface. Such a ball is placedin a vat of unknown fluid and submerged until it implodes, at a depth of1.82 meters. Find the density of the fluid and it’s specific gravity.
Pressure as a function of depth is given by: P (d) = ρgd, where ρ isthe fluid’s density, g is the acceleration due to gravity, and d is the depth.However, the ball is under not only the fluid, but the atmosphere (pressureone atmosphere) - thus the total pressure on the ball is:
1 atm + ρgdwhich we set equal to 2.2 atm. Thus:2.2 atm - 1 atm = ρgdThus:ρ = 1.2atm
gd .Converting atmospheres to Pascals, we find:ρ = 1.22×105
9.8×1.82 = 6815 kg/m3.The specific gravity is the density divided by that of pure water (1000
kg/m3) so the specific gravity of this unknown fluid is 6.815. (Pure number,no units).
34
2. A spherical bead is placed in the center of a glass of pure water andreleased. The bead has a radius of 1.0 cm and a mass of 7.6 grams. Willthe bead move up or down? What is it’s (initial) acceleration? Fixing theradius, what would the mass of the bead have to be for it not to move upor down but just float freely?
There are two forces acting on the bead: gravity and the buoyant force.Buoyancy is equal to the weight of the displaced fluid, in this case waterand given by:
FB = ρgVwhere ρ is the density of the fluid, g is the acceleration due to gravity
and V is the submerged volume, in this case the entire volume of the bead.The volume of a sphere is given by: V = 4
3πr3 which, in the case of ourbead is 4.19x10−6m3 (about four millionths of a cubic meter).
Summing forces we find:ΣF = maFB −mg = maa = FB−mg
m = ρgV−mgm = (0.041− 0.0745)/0.0076 = −4.4m/s2.
The minus sign means it accelerates downward: gravity exceeds thebuoyant force.
3. A barrel is full of water. It is an upright cylinder with a diameter of30 cm and a height of 42 cm. At a point 7 cm from the bottom, a smallhole is drilled, horizontally, into the barrel and the water comes pouringout. Find the initial (horizontal) velocity of the water and where the streamwill land (how far from the base of the barrel). Next, assume the hole is1 mm in diameter and find the rate at which the water level in the barrelis decreasing (at t=0, right as the hole is drilled - we can find the rate ingeneral later).
We apply Bernoulli’s equation: P + ρgh + 1/2ρv2 = C where C is someconstant. Consider two points: one atop the barrel and the other just outsidethe barrel where the water is pouring out:
P1 + ρgh1 + 1/2ρv21 = P2 + ρgh2 + 1/2ρv2
2
At both points 1 and 2, the pressure is merely atmospheric. Let us setthe height equal to zero on the ground, thus h1 = 42 cm, the top of thebarrel, and h2 = 7 cm (up from the ground). The velocity of the water atopthe barrel is very near zero - since the hole is very small very little water iscoming out: v1 = 0 and it is v2 that we wish to find thus:
0 + ρgh1 + 0− 0− ρgh2 = 1/2ρv22.
The ρ cancels out in all terms and multiplying by 2 we find:2g(h1 − h2) = v2
2
35
which is: 6.68m2/s2. Solving for v2 we get 2.62 m/s. The plus or minussign we naturally get from the speed squared is a clue that this calculationis not giving us the direction, only the speed of the stream.
Assuming the hole is 1mm in diameter, we can use continuity to find thespeed at which the top level of the barrel is descending: A1v1 = A2v2 withv2 given above.
14 Oscillations
1. Given the equation y(t) = 4 sin(3t + 1) where y is in meters and t is inseconds, find the amplitude, period, frequency and a time at which y is amax, a min and zero.
2. A 2.2 kg mass oscillates horizontally on a frictionless surface whileconnected to a 292 N/m spring constant spring. It is initially displaced by5.0 cm from it’s rest position and released from rest. Find:
a. Amplitudeb. Periodc. Write down the equation x(t) for the mass.3. A 235 gram mass oscillates horizontally attached to a spring, with
constant k = 310 N/m. The system has 19 J of energy, and at t=0, allthe energy is kin kinetic, and the mass moves in the positive x-direction.(Neglect friction).
(a) Find the amplitude of oscillation(b) Find the frequency of oscillation(c) Write down an equation for x(t), set x=0 at t=0.(d) Find a time at which x is a maximum.(e) Find the equation for velocity of time. Find a time at which v=0.4. Tom Sawyer runs horizontally, grabs a hanging vine from a tree, and
swings back and forth across the ground. Initially, the vine is 3.5 m longand hangs vertically and Tom is running at 4.2 m/s and has a mass of 72kg. Assume Tom’s center of mass is even with the end of the vine (e.g. 3.5meters below the branch). Find the period and amplitude of Tom swingingand find his equation of motion and maximum speed and acceleration.
5. A 2.2 kg mass is dropped from 0.57 m onto a vertical spring with anunstretched length of 0.13 m and a spring constant of 321 N/m. The masshooks onto the spring and oscillates after striking. Find the maximum andminimum positions of the oscillating mass-spring system, find the frequency.
36
Oscillation Solutions
1. Given the equation y(t) = 4 sin(3t + 1) where y is in meters and t is inseconds, find the amplitude, period, frequency and a time at which y is amax, a min and zero.
We can read off the amplitude as 4 meters and angular frequency ω as3 radians/s. Since ω = 2πf , the frequency, f=0.478 Hz.
To find a time at which y is a max, set sin(3t + 1) = 1, since sine variesfrom -1 to 0 to +1. Taking the arcsine of each side, with calculator in radianmode23 revels: 3t+1 = π/2 and thus t = 0.19s. Note that this is the earliestpositive time at which y is a maximum. Any number of periods can be addedto this to give another time at which y is a maximum (you can verify thiswith a calculator) thus there are an infinite number of such answers all canbe written as: t = 0.19 + nT where n is any integer and T is the period,T = 1/f = 2.09s. This is because of the nature of the sine function - thereare many times which will satisfy sin(3t + 1) = 1 and there must be forin simple harmonic motion the mass reaches the maximum position many(infinite) times. You calculator will give one value - in fact it doesn’t haveto give the same value as listed above - but it should give a correct valuemeaning that when you plug the time back in sin(3t + 1) really does equalone.
Similar procedure is followed to find a time at which y is zero or aminimum. For the minimum, we set sin(3t + 1) = −1 and find t = −0.86s.All such times are: t = −0.86 + nT .
To find a time when y is zero, we set sin(3t+1) = 0 and find t = −1/3s.More solutions are given by t = −1/3 + nT however these are not all thepossible solutions, in fact they are only half. Plotting either sine or cosineover one full cycle (from zero to two pi, let’s say) will yield two crossingsof the origin. For sine these occur at zero and pi24, for cosine at π/2 and3π/2. The zeros are half a period apart (π radians). Thus in this case, allsuch times are given by t = −1/3 + nT/2 where n is any integer.
2. A 2.2 kg mass oscillates horizontally on a frictionless surface whileconnected to a 292 N/m spring constant spring. It is initially displaced by5.0 cm from it’s rest position and released from rest. Find:
a. Amplitude23By convention, wave and oscillation equations assume radians. This is equivalent to
saying that the 3 before the t has units of radians per second and thus is an angularfrequency, an ω. Were we to use degrees instead, many other equations would be alteredand various factors of 360 and 360/2π would arise. Best to avoid.
24And two pi, of course, but physically this is the same point as zero in the same waythat if you turn in one full circle you are facing the same direction
37
b. Periodc. Write down the equation x(t) for the mass.3. A 235 gram mass oscillates horizontally attached to a spring, with
constant k = 310 N/m. The system has 19 J of energy, all of which at t=0,is kinetic; the mass begins to move in the positive x-direction. (Neglectfriction).
(a) Find the amplitude of oscillationAt an extrema, all energy is elastic potential Ue = 1/2kx2, thus we set
19 J = 1/2kx2 and solve for x; this will be the amplitude. Solving we find:A = .35m = 35cm.
(b) Find the frequency of oscillationFor a mass on a spring, this will be f = 1
2π
√k/m =5.78 Hz.
(c) Write down an equation for x(t), set x=0 at t=0.x(t) = A sin(ωt) where A = amplitude = .35m and ω = sqrtk/m =36.3
rad/s.(d) Find a time at which x is a maximum.At one quarter of a full period it will be a maximum. T = 1/f = .173s,
thus the time t = 0.0432s will yield a maximum in x. This should satisfysin(ωt) = 1.
(e) Find the equation for velocity of time. Find a time at which v=0.v(t) = dx(t)/dt = ωA cos(ωt). At t=0 this is a maximum (cosine is a
maximum there). The leading ω is from the chain rule.4. Tom Sawyer runs horizontally, grabs a hanging vine from a tree, and
swings back and forth across the ground. Initially, the vine is 3.5 m longand hangs vertically and Tom is running at 4.2 m/s and has a mass of 72kg. Assume Tom’s center of mass is even with the end of the vine (e.g. 3.5meters below the branch). Find the period and amplitude of Tom swingingand find his equation of motion and maximum speed and acceleration.
For small amplidutes, the period, T = 2π√
l/g = 2π√
3.5m/9.8m/s2 =3.75s.
The amplitude can be found via conservation of energy. Tom’s initialkinetic energy converts to gravitational potential. This will yield a heightabove his starting height. However, it is more common to speak in termsof the angle the vine makes from the vertical. This we can find from theheight via geometry.
Conserving energy:Ek = 1/2mv2
o = mgh yields:h = 0.90m. Huck’s center of gravity rises 0.9 m off the ground, at most.Draw the vine hanging vertically, and draw Huck on the vine at max-
imum height. This is a triangle. It is an isocolies triangle (two sides are
38
the vine, thus have equal length) but we can construct a right triangle bypassing a line parallel to the ground. This will intersect the vertical hangingvine at a right angle. Now we know the hypotenuse is L, and the adjacentside lenght is L-h, we can find θ, the angle the vine makes, at most, fromthe vertical.
cos(θ) = L−hL = 0.733rad.
This is not a really small angle (compared to one radian), thus the periodwe found above is suspect. Including the next term in the expansion, wecan find the period should be:
T = 2π√
L/g(1 + 14 sin2( θmax/2
) ) = 3.87s.It is common to not see this more exact form for the period of a pendu-
lum. However, you should be aware the simple form used above (to give the3.75s number) is an approximation, good only when the maximum angle,in radians, is much smaller then one. In this case, it is pretty close to one(0.733) but it is only off by about 5%.
Using our more exact period, we find the equation of motion is given by:θ(t) = 0.733 sin(2πt/3.87).
Huck’s speed is given by v(t) = Ldθ/dt, since dθ/dt is the angular speedandspeed is given by25: v = rθ. Thus v(t) = 4.2 cos(1.62t).
Similarly: a(t) = dv(t)/dt = −6.8 sin(1.62t).The maximum speed is 4.2 m/s, and the maximum acceleration is 6.8
m/s2.5. A 2.2 kg mass is dropped from 0.57 m onto a vertical spring with an
unstretched length of 0.13 m and a spring constant of 3210 N/m. The masshooks onto the spring and oscillates after striking. Find the maximum andminimum positions of the oscillating mass-spring system, find the frequency.
Using conservation of energy, and setting our zero of gravitational po-tential to be on the ground, we find: Ug = Ugf + Ue. Gravitational goes toelastic, plus a bit of gravitational. Symbolically, let H be the initial height,L be the lenght of the spring and x be the compression:
mgH = mg(L− x) + 1/2kx2.This is a quadratic equation in x. Plugging in numbers we find:12.29 = 2.80− 21.56x + 1605x2
x = 0.084m = 8.4 cm.Thus the minimum position is L-x = 4.6 cm above ground.The maximum position is not L+x. To find it, we again need to conserve
energy. The equation is identical, but the minus sign before the 21.56 turnsto a plus:
25This is one of the many equations easier in radians
39
mgH = mg(L + x) + 1/2kx2
Leading to x = 0.0705m = 7.05cm thus the maximum position aboveground is L+x = 20.05cm.
The frequency is still 12π
√K/m = 6.08 Hz.
40