ch05 circular motion 5 02 2015(2)

Chapter 5

Dynamics of Uniform Circular Motion

Chapter 5

Dynamics of Uniform Circular Motion

Tutorial Problems Chapter 51, 4, 15, 18, 21, 23, 24, 25, 27, 28, 30

Topic for Discussion: Speed Limit around a curve

1. What is the meaning of the sign indicated?2. How did they come to say the speed indicated should be 100 km/h?.3. Do you think the small car and the big truck or bus should obey the

same speed limit when passing through the indicated sign of 100 km/h4. What do you think can happen if one travels more or less than the

indicated speed limit?

A ball is thrown vertically upwards from the surface of the earth. Consider the following quantities based on the motion of the ball.1: Speed; 2: velocity; 3: acceleration

Speed, velocity and acceleration activities 5 02 2015

1.1 Which of these is (are) zero when the ball has reached the maximum height at position B? Select the correct one

A: 1 and 2 only, B: 1 and 3 only, C: 1 only, D: 2 only, E: 1, 2 and 3

1.2 What do you think will be the magnitude and direction of acceleration of an object at the following points? Give reasons for your answer.

1.2.1 Point A moving up magnitude_____direction_______(UP or DOWN)

1.2.2 Point B at maximum height: magnitude___direction__(UP or DOWN)

1.2.3 Point C moving down: magnitude___direction_____ (UP or DOWN)

A ball is thrown vertically upwards from the surface of the earth. Consider the following quantities based on the motion of the ball.1: Speed; 2: velocity; 3: acceleration

Speed, velocity and acceleration activities 9 02 2015

1.3 What do you think will happen to the velocities of an object at the following points? (Only write, increase, decrease, zero or remains the same and give reason(s))

1.3.1 At position A : increase, decrease, zero or remains the same

1.3.2 At Position B: increase, decrease, zero or remains the same1.3.3 At position C: increase, decrease, zero or remains the same

What about the accelerations at the points above?

1.2.1 At position A : increase, decrease, zero or remains the same

1.2.2 At Position B: increase, decrease, zero or remains the same1.3.3 At position C: increase, decrease, zero or remains the same

Newton’s Laws of MotionState and apply Newton’s Laws of Motion

Conditions for equilibrium: How do we know if objects are in equilibrium?

• Sum of the forces is equal to zero (along the x and y independently)• Objects move at the same velocity• The acceleration is zero

What is ACCELERATION?Do you think an object moving at constant speed around a circular track is accelerating? Claim Evidence and reasoning format of an answer.

5.1 Uniform Circular Motion

DEFINITION OF UNIFORM CIRCULAR MOTION

Uniform circular motion is the motion of an object traveling at a constant speed on a circular path.

5.1 Uniform Circular Motion

Let T be the time it takes for the object totravel once around the circle.

vr

T

2r

What is the formula to calculate the circumference of the circle?

Then the speed v around the circular path will be given by

Do example 1 as a homework,

𝐶=2𝜋 𝑟

R the radius of the circle and T the period in seconds

5.2 Centripetal Acceleration

In uniform circular motion, the speed is constant, but the direction of the velocity vector is not constant.

90

90

Similarities of Triangles (back 2000)

Two triangles are similar if:(a) The corresponding angles are equal(b) The corresponding sides are in proportionNote: -Corresponding sides are sides opposite equal angles-Corresponding angles are angles opposite sides in proportion

Your Turn

Consider Prove that the two triangles are similar

What are equal angles in the two triangles?What are the sides in proportion between the two triangles?

Lastly we derive the formula for centripetal acceleration using mathematical geometry


rtv

vv

rv

tv 2

rvac

2

Diagram a and b (sides in proportion are equal)

Centripetal acceleration


The direction of the centripetal acceleration is toward the center of the circle; in the same direction as the change in velocity.

rvac

2

Note that the velocity is constant and only time is changing

Your Turn Which Way Will the Object Go?

An object is in uniform circular motion. At point O it is released from its circular path. Does the object move along the straight path between O and A or along the circular arc between points O and P ?

Claim:

Evidence:

Reasoning:

The Impact of Radius on Centripetal Acceleration

The bobsled track contains turns with radii of 33 m and 24 m. Without doing any calculation which of the curve will have greater acceleration?Claim:Evidence:Reasoning:

Justifications using calculations

gac 6.3sm35m 33

sm34 22

gac 9.4sm48m 24sm34 2

2

5.3 Centripetal Force

aF

mm

F

a

Recall Newton’s Second Law

When a net external force acts on an objectof mass m, the acceleration that results is directly proportional to the net force and has a magnitude that is inversely proportional to the mass. The direction of the acceleration isthe same as the direction of the net force.


Thus, in uniform circular motion there must be a netforce to produce the centripetal acceleration.

The centripetal force is the name given to the net force required to keep an object moving on a circular path.

The direction of the centripetal force always points towardthe center of the circle and continually changes direction as the object moves.

rvmmaF cc

2


Example 5: The Effect of Speed on Centripetal Force

The model airplane has a mass of 0.90 kg and moves at constant speed on a circle that is parallel to the ground. The path of the airplane and the guideline lie in the samehorizontal plane because the weight of the plane is balancedby the lift generated by its wings. Find the tension in the 17 mguideline for a speed of 19 m/s.

rvmTFc

2

N 19m 17sm19kg 90.0

2

T


Check Conceptual Example 6: A Trapeze Act

In a circus, a man hangs upside down from a trapeze, legsbent over and arms downward, holding his partner. Is it harder for the man to hold his partner when the partner hangs straight down and is stationary of when the partner is swinging through the straight-down position?

Personal Activity: Due: 23 Feb 2015

To negotiate an unbanked curve at faster speed, a driver puts a couple of sand bags in his van to increase the force of friction between the tires and the road. Will the sand bags help?

Claim: (1)

Scientific Evidence: (2)

Reasoning: (2)

5.4 Banked Curves

Note:On an unbanked curve, the static frictional forceprovides the centripetal force. Centripetal force andSafe Driving Check example page 141

Highway Curves: Banked & UnbankedCase 1 - Unbanked Curve: When a car rounds a curve, there MUST be a net force toward the circle center (a Centripetal Force) of which the curve is an arc. If there weren’t such a force, the car couldn’t follow the curve, but would (by Newton’s 1st Law) go in a straight line.

On a flat road, this Centripetal Force is the static friction force.

No static friction? No Centripetal Force The Car goes straight!

=

Note:

If the friction force isn’t sufficient, the car will tend to move more nearly in a straight line (Newton’s 1st Law) as the skid marks show.

As long as the tires don’t slip, the friction is static. If the tires start to slip, the friction is kinetic, which is bad

1. The kinetic friction force is smaller than the static friction force.

2. The static friction force can point toward the circle center, but the kinetic friction force opposes the direction of motion, making it difficult to regain control of the car & continue around the curve.

5.4 Banked Curves

On a frictionless banked curve, • the centripetal force is the horizontal component of

the normal force. • The vertical component of the normal force balances the

car’s weight.

5.4 Banked Curves

rvmFF Nc

2

sin )1(cos mgFN

)2(sin2

rvmFN

:)2()1( bydividergv2

tan Conclusion:The speed around the curveIs independent of the mass of the car or bus

5.4 Banked Curves

rvmFN

2

sin

mgFN cosrgv2

tan

5.4 Banked CurvesExample 8: The Daytona 500

The turns at the Daytona International Speedway have a maximum radius of 316 m and are steely banked at 31degrees. Suppose these turns were frictionless. At what speed would the cars have to travel around them?

rgv2

tan tanrgv

mph 96 sm4331tansm8.9m 316 2 v

Satellites

5.5 Satellites in Circular Orbits

There is only one speed that a satellite can have if the satellite is to remain in an orbit with a fixed radius. What are the names and formula of the forces acting ?


rvm

rmMGF E

c

2

2

rGMv E

Hence the speed that a satellite is independent of its mass but its radius (r) from the Earth.


Example 9: Orbital Speed of the Hubble Space Telescope

Determine the speed of the Hubble Space Telescope orbitingat a height of 598 km above the earth’s surface.

hmi16900 sm1056.7

m10598m1038.6kg1098.5kgmN1067.6

3

36

242211

v


Tr

rGMv E 2

EGMrT

232


Global Positioning System

hours 24TEGM

rT232

Section 5.2 no 1 p 1491. Two cars are traveling at the same constant speed v. Car A is moving along a straight section of the road, while B is rounding a circular turn. Which statement is true about the acceleration of the cars?(a) The acceleration of both cars is zero, since they are traveling at a constant speed.(b) Car A is accelerating, but car B is not accelerating. (c) Car A is not accelerating, but car B is accelerating. (d) Both cars are accelerating.Explanation :The velocity of car A has a constant magnitude (speed) and direction. Since its velocity is constant, car A does not have an acceleration. The velocity of car B is continually changing direction during the turn. Therefore, even though car B has a constant speed, it has an acceleration (known as a centripetal acceleration).

Section 5.4: Banked Curves 10. Two identical cars, one on the moon and one on the earth, have the same speed and are rounding banked turns that have the same radius r. There are two forces acting on each car, its weight mg and the normal force exerted by the road. Recall that the weight of an object on the moon is about one-sixth of its weight on earth. How does the centripetal force on the moon compare with that on the earth? (a) The centripetal forces are the same.(b) The centripetal force on the moon is less than that on the earth. (c) The centripetal force on the moon is greater than that on the earth.Explanation

The magnitude of the centripetal force is given by Fc = mv2/r. The two cars have the same speed v and the radius r of the turn is the same. The cars also have the same mass m, even though they have different weights due to the different accelerations due to gravity. Therefore, the centripetal accelerations are the same.

Section 5.5: Satellites in circular orbirts

11. Two identical satellites are in orbit about the earth. One orbit has a radius r and the other 2r. The centripetal force on the satellite in the larger orbit is ________ as that on the satellite in the smaller orbit.(a) the same (b) twice as great (c) four times as great (d) half as great (e) one- fourth as greatExplanation:

(e) The centripetal force acting on a satellite is provided by the gravitational force. The magnitude of the gravitational force is inversely proportional to the radius squared (1/r2), so if the radius is doubled, the gravitational force is one fourth as great; 1/22 = 1/4.

Question: Skidding on a curveA car, mass m = 1,000 kg car rounds a curve on a flat road of radius r = 50 m at a constant

speed v = 14 m/s (50 km/h). Will the car follow the curve, or will it skid? Assume a. Dry pavement with the coefficient of static friction μs = 0.6.

b. Icy pavement with μs = 0.25.

Free BodyDiagram

Approach

Step 2:Draw the FBD and identify all forces acting on the car The force of gravity downward The normal force The horizontal frictional force

Step 1: Criteria to follow the curve or skidding.Reasoning: The car will follow the curve if the maximum static frictional force is

greater than the centripetal force. That is where Noting that once the car skid, the static frictional force becomes kinetic frictional

force which is less than static frictional force

Answer: Compare the static frictional force with the centripetal forceIf hence the car will follow the curveIf hence the car will skid

a. Dry pavement with the coefficient of static friction μs = 0.6

𝐹 𝑓=𝐹𝑓 =𝜇𝑠𝐹𝑁=(0.6)(9800)=5900𝑁>𝐹𝐶𝐻𝑒𝑛𝑐𝑒 h𝑡 𝑒𝑐𝑎𝑟𝑤𝑖𝑙𝑙 𝑓𝑜𝑙𝑙𝑜𝑤 h𝑡 𝑒𝑐𝑢𝑟𝑣𝑒b. Icy pavement with μs = 0.25

𝐹 𝑓=𝐹𝑓 =𝜇𝑠𝐹𝑁=(0.25)(9800)=2500𝑁<𝐹𝐶𝐻𝑒𝑛𝑐𝑒 h𝑡 𝑒𝑐𝑎𝑟𝑤𝑖𝑙𝑙 𝑠𝑘𝑖𝑑

Vertical circular motionQuestion: Is vertical circular motion uniform? Explain

• When the speed of travel on a circular path changes from moment to moment, the motion is said to be non-uniform.

• But, we can use the concepts that apply to uniform circular motion to gain considerable insight into the motion that occurs on a vertical circle.

Considering the second diagramThere are four points on a vertical circle where the centripetal force can be identified easily, denoted by points 1 to 4• Keep in mind that the centripetal force is not a new and separate force of nature. Instead, at each point the centripetal force is

the net sum of all the force components oriented along the radial direction, and it points toward the centre of the circle. • The drawing shows only the weight of the cycle plus rider (magnitude = mg) and the normal force pushing on

the cycle (magnitude = FN). • The magnitude of the centripetal force at each of the four points is given as follows in terms of mg and FN:• As the cycle goes around, the magnitude of the normal force changes. It changes because the speed changes

and because the weight does not have the same effect at every point

Case 1: At the bottom, the normal force and the weight opposes one another, giving a resulting centripetal force of magnitude FN1 - mg and the second law becomes: Resultant F:

Case 3: On top, the normal force and the weight are facing the same direction, giving a resulting centripetal force of magnitude FN1 +mg and hence the second law becomes:

Cases 2 and 4: In both cases, the normal force and weight are perpendicular to each other,, and weight not in the same direction as the centripetal force hence the resulting centripetal force will be caused by the normal force: second law becomes: and

The case of the minimum force to keep an object at circular on top of vertical track

• Riders who perform the loop-the-loop trick must have at least a minimum speed at the top of the circle to remain on the track. This speed can be determined by considering the centripetal force at point 3.

• The speed in the equation is a minimum when is zero. Hence it will be given by • Then, the speed is given by . • At this speed, the track does not exert a normal force to keep the cycle on the circle, because the weight mg provides

all the centripetal force.

mg

FN

Free-body diagram of the motorcycle

Do no problem no 46 on your own

47. REASONING Because the crest of the hill is a circular arc, the motorcycle’s speed v is related to the centripetal force Fc acting on the motorcycle: 2

cF mv r

where m is the mass of the motorcycle and r is the radius of the circular crest. Solving Equation 5.3 for the speed, we obtain 2

cv F r m or cv F r m .

The free-body diagram shows that two vertical forces act on the motorcycle. One is the weight mg of the motorcycle, which points downward. The other is the normal force FN exerted by the road. The normal force points directly opposite the motorcycle’s weight.

mg

FN

Free-body diagram of the motorcycleNote that the motorcycle’s weight must be greater than the normal force. The reason

for this is that the centripetal force is the net force produced by mg and FN and must point toward the center of the circle, which lies below the motorcycle.

Only if the magnitude mg of the weight exceeds the magnitude FN of the normal force will the centripetal force point downward.

Therefore, we can express the magnitude of the centripetal force as Fc = mg − FN. With this identity, the relation cv F r m becomes

When the motorcycle rides over the crest sufficiently fast, it loses contact with the road. At that point, the normal force FN is zero. In that case, Equation (1) yields the motorcycle’s maximum speed:

REASONING The drawing at the right shows the two forces that act on a piece of clothing just before it loses contact with the wall of the cylinder. At that instant the centripetal force is provided by the normal force and the radial component of the weight. From the drawing, the radial component of the weight is given by

FN

mg

Clothes

mg mg mg cos = cos (90 – ) = sin

Therefore, with inward taken as the positive direction, Equation 5.3 ( Fc mv2 / r ) gives

Therefore, with inward taken as the positive direction, Equation 5.3 ( Fc mv2 / r ) gives

F mg mvr

2

N sin =

At the instant that a piece of clothing loses contact with the surface of the drum, FN 0 N, and the above expression becomes

mg mvr

2

sin =

Substituting the equation of speed, v 2r /T , we get

gr Tr

rT

2

sin = ( / )2 4 2

2

Solving for T and substituting unknowns we get

T rg

rg

4 2 1 17

2

sin sin

2 0.32 m9.80 m / s sin 70.0

s2c h .

Therefore, the number of revolutions per second that the cylinder should make is

1 1

1 17T

. s0.85 rev / s

Extra Problems

REASONINGThe speed v of a satellite in circular orbit about the earth is given by , where G is the universal gravitational constant, and is the mass of the earth, and r is the radius of the orbit. The radius is measured from the center of the earth, not the surface of the earth, to the satellite. Therefore, the radius is found by adding the height of the satellite above the surface of the earth to the radius of the earth ().

Satellite A Satellite B

𝑉=7690𝑚 /𝑠 𝑉=7500𝑚 /𝑠

Vertical Circular MotionFor vertical circular motion, the motion is not uniform, as the object increases speed on the downward swing and decreases on the way up. When analysing vertical circular motion, one finds that there are two forces acting on the object.

These two forces are T, the tension in the string, and FW, the weight of the object, as shown below.The weight, FW, can be resolved into a tangential component expressed as FWsinθ and a radial component expressed as FWcosθ

𝑎𝑇=𝐹 𝑇𝑚 =

𝐹𝑊 𝑠𝑖𝑛𝜃𝑚 =

𝑚𝑔𝑠𝑖𝑛𝜃𝑚 =𝑔𝑠𝑖𝑛𝜃

𝑎 𝑅=𝐹 𝑅𝑚 =

𝑇 −𝐹𝑊 𝑐𝑜𝑠 𝜃𝑚 =

𝑇 −𝑚𝑔𝑐𝑜𝑠 𝜃𝑚 =

𝑇 −𝑚𝑔𝑐𝑜𝑠𝜃𝑚

Applying Newton's second law to express the tangential acceleration gives:

Similarly, the radial acceleration can be expressed as:

Substituting for and solving for T yields: 𝑇=𝑚 (𝑣2

𝑟 +𝑔𝑐𝑜𝑠 𝜃)

Two interesting points to consider are the top and the bottom of the circle.

At the lowest point (Bottom) of the circular path, θ = 0° and cos 0° = 1. Substituting into the equation for T yields:

𝑇=𝑚 (𝑣2𝑟 +𝑔)

At the highest point of the circular path, θ = 180° and cos 180° = -1. Substituting into the equation for T yields:

T = m( - g)

At Gold Reef City, a child of mass m rides on a Ferris wheel as shown in the diagram above. The child moves in a vertical circle of radius 10.0 m at a constant speed of 3.00 m/s.(a)Calculate the force exerted by the seat on the child

at the bottom of the ride. Express your answer in terms of the weight of the child, mg.

(b) Calculate the force exerted by the seat on the child at the top of the ride.

Test Your Understanding

(a) At the bottom, Consider the forces acting at the bottom: Force of gravity downwards and the normal force Hence

(b) At the bottom, Consider the forces acting at the bottom: Force of gravity downwards and the normal force Because the child is always in upright position.

Test your understanding:

An object with a mass of 8.0 kg is swung in a vertical circle of radius 2.4 m with a speed of 6.0 m/s. (a) Determine the maximum and minimum tension in the string. (b) The string breaks when the tension exceeds 340 N. Determine the maximum speed of the object and where the mass will be when the string breaks. Justify your answer as to where the mass will be.

For solution check: http://mmsphyschem.com/verCMAns.htm

http://mmsphyschem.com/verCMAns.htm



1. SOLUTION Since 2c /a v r and 2 /v r T , the magnitude of the car’s

centripetal acceleration is

2

2 32 22

2c 2

24 10 m4 0.86 m/s

400 s

rv rTar r T

4 SOLUTION Using Equation 5.2 for the centripetal acceleration of each boat, we have

2 2A B

cA cBA B

and v v

a ar r

Setting the two centripetal accelerations equal gives 2 2A B

A B

v vr r

Solving for the ratio of the speeds gives A A

B B

80 m 0.58240 m

v rv r

15. REASONING At the maximum speed, the maximum centripetal force acts on the tires, and static friction supplies it. The magnitude of the maximum force of static friction is specified by Equation 4.7 as MAX

s s Nf F , where s is the coefficient of static friction and FN is the magnitude of the normal force. Our strategy, then, is to find the normal force, substitute it into the expression for the maximum frictional force, and then equate the result to the centripetal force, which is

2c /F mv r , according to Equation 5.3. This will lead us to an expression for the

maximum speed that we can apply to each car. SOLUTION Since neither car accelerates in the vertical direction, we can

conclude that the car’s weight mg is balanced by the normal force, so FN = mg. From Equations 4.7 and 5.3 it follows that

2MAX

s s N s cmvf F mg F

r

Thus, we find that 2

s s or mvmg v grr

Applying this result to car A and car B gives

A s, A B s, B and v gr v gr

Some of the Tutorial Solutions

In these two equations, the radius r does not have a subscript, since the radius is the same for either car. Dividing the two equations and noting that the terms g and r are eliminated algebraically, we see that

s, B s, B s, BBB A

A s, As, A s, A

0.85 or 25 m/s 22 m/s1.1

grvv v

v gr

18. REASONING The centripetal force Fc that keeps the car (mass = m, speed = v) on

the curve (radius = r) is 2c /F mv r (Equation 5.3). The maximum force of static

friction MAXsF provides this centripetal force. Thus, we know that MAX 2

s /F mv r

, which can be solved for the speed to show that MAXs /v rF m . We can apply

this result to both the dry road and the wet road and, in so doing, obtain the desired wet-road speed.

Applying the expression MAXs /v rF m to both road conditions gives

MAX MAX

s, dry s, wetdry wet and

rF rFv v

m m

We divide the two equations in order to eliminate the unknown mass m and unknown radius r algebraically, and we remember that

MAX MAX1s, wet s, dry3

F F :

MAX

MAX

MAXMAX

s, wet

s, wetwet

dry s, drys, dry

13

rFFv m

v FrF

m

Solving for vwet, we obtain

drywet

25 m / s 14 m / s3 3

vv

What you should do

Tutorial Problems Chapter 51, 4, 15, 18, 21, 23, 24, 25, 27, 28, 30, 35, 38, 55

ch05 circular motion 5 02 2015(2)

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