1 motion of a single particle - linear momentum, work and...

1 Motion of a single particle - Linearmomentum, work and energy principle

1.1 In-class problem A block of mass m slidesdown a frictionless incline (see Fig.). The block isreleased at height h above the bottom of the loop.

(a ) What is the force of the inclined track on theblock at the bottom (point A)?Find an expression in terms of h andR.

(b ) What is the force of the track on the block at point B?

(c ) At what speed does the block leave the track?

(d ) How far away from point A does the block land on level ground?

Solution

(a) I. Choose the reference systemConsider the reference system as shown in the figure of the assignment.

II. Draw the free-body diagramThe free-body diagram at point A:

III. Apply the appropriate principle(s)Applying linear momentum principle in the y direction:

Py =∑

Fy ⇒ mv2/R = N −mg. (1.1)

Remark: make sure you re-read example I.3 in the lecture notes.Since all forces are either potential (G = mg) or do no work (N),the system is conservative, thus, we can use conservation of energy principle.Total energy at the top of the track:

Etop = Ttop + Vtop = 0 +mgh. (1.2)

Total energy at point A:

EA = TA + VA =1

2mv2 + 0. (1.3)

1-1

1 Motion of a single particle - Linear momentum, work and energy principle 1-2

Now we can calculate v from: Etop = EA

mgh =1

2mv2 ⇒ v =

√2gh. (1.4)

Substituting (1.4) into (1.1) gives:

N = mg

(1 +

2h

R

). (1.5)

(b)I. Choose the reference systemConsider the reference system as shown in the figure of the assignment.

II. Draw the free-body diagramThe free-body diagram at point B:

III. Apply the appropriate principle(s)Applying linear momentum principle in the normal direction at point B.Pn =

∑Fn ⇒ mv2

R= N −mg cos 45◦

The force of the track exerted on the block at point B is the following:

N =mv2

R+mg√

2. (1.6)

We know that the total energy according to (1.2) is Etot = mgh.Since the system is still conservative, get v by the conservation of energy.The total energy at point B is:

EB = TB + VB =1

2mv2 +mghB, (1.7)

where hB can be written as:

hB = R

(1− 1√

2

). (1.8)


Finally the conservation equation Etot = EB reads

mgh =1

2mv2 +mgR

(1− 1√

2

). (1.9)

Solving equation (1.9) for v2 gives:

2

[gh− gR

(1− 1√

2

)]= v2, (1.10)

and substituting (1.10) into (1.6) yields

N = mg

[2h

R+

(3√2− 2

)]. (1.11)

(c) From (1.10):

v =

√2

[gh− gR

(1− 1√

2

)]. (1.12)

(d) Projectile motionI. Choose the reference systemConsidering point A as the reference system origin.

II. Draw the free-body diagramThe free-body diagram.

III. Apply the appropriate principle(s)Applying linear momentum principle in the x and y direction:

Px =∑

Fx ⇒ mx = 0 and Py =∑

Fy ⇒ my = −mg

Integrating twice with respect to time these equations and substituting the initial con-ditions, namely x(0) = x0, y(0) = y0, vx(0) = vx0 , vy(0) = vy0 , we get the equation ofmotion for the projectile.

x = x0 + vx0t , (1.13)


y = y0 + vy0t−1

2gt2 , (1.14)

where g = 9.81m/s2. Substituting the already known quantities (1.13) and (1.14) became

x =R√

2+vB√

2t , (1.15)

y = hB +vB√

2t− 1

2gt2 . (1.16)

Now solve (1.16) for t when y = 0 (the block reaches the ground)

gt2 − 2hB −√

2vBt = 0 ,

t =

√2vB ±

√2v2B + 8ghB

2g. (1.17)

We discard the negative root since it gives a negative time and substitute into (1.15):

x =R√

2+vB√

2

(√2vB +

√2v2B + 8ghB

2g

). (1.18)

Using the previous expressions for vB and hB yields

x = (√

2− 1)R + h+

√h2 − 3

2R2 +

√2R2 . (1.19)


1.2 In-class problem A bead, under the influence of gravi-ty, slides along a frictionless wire whose height is given by thefunction y(x). Assume that at position (x, y) = (0, 0), the wireis vertical and the bead passes this point with a given speed v0downward. What should the shape of the wire be (y as a functionof x) so that the vertical speed remains v0 at all times?

Solution

I. Choose the reference systemConsider the reference system as shown in the figure of the assignment.

II. Draw the free-body diagramFree-body diagram:

III. Apply the appropriate principle(s)Since all forces are either potential (G = mg) or do no work (N),the system is conservative, thus, we can use conservation of energy principle.The bead’s speed at any time is given by

1

2mv2 +mgy =

1

2mv20 ⇒ v =

√v20 − 2gy . (1.20)

(Note that y is negative here).The vertical component of the velocity is

y =dy

dt= −v sin θ , (1.21)

where the angle θ is defined as shown below at every point of the curve.


Therefore sin θ = y′/√

1 + y′2 with y′ = −dy/dx being the slope of the wire.Now the requirement is y = −v0, which is equivalent to

−v sin θ = −v0 . (1.22)

Then we get the following formula by substituting (1.20) into (1.22)√v20 − 2gy

(y′√

1 + y′2

)= v0 . (1.23)

Squaring both sides and solving for y′ = −dy/dx yields

dy

dx=−v0√−2gy

. (1.24)

Separating variables and integrating gives∫ √−2gy dy = −v0

∫dx⇒ (−2gy)3/2

3g= v0x , (1.25)

where the constant of integration has been set to zero, because (x; y) = (0; 0) is a pointon the curve. Therefore the shape of the wire is:

y = −(3gv0x)2/3

2g. (1.26)


1.3 In-class problem The ball of mass m is given a speed ofvA =

√3gr at position A. When it reaches B, the cord hits the

small peg P, after which the ball describes a smaller circular path.Determine the position x of P so that the ball will just be able toreach point C.

Solution

I. Choose the reference systemNormal-tangential (n-t) coordinates are attached to, and move with, the ball.

II. Draw the free-body diagramFree-body diagram:

III. Apply the appropriate principle(s)Applying linear momentum principle in the normal direction at point C:

Pn =∑

Fn ⇒ mv2Cr − x

= mg ⇒ v2C = g(r − x) (1.27)

(Note: when the ball is just about to complete the small circular path,

the cord becomes slack at position C, i.e., T = 0, an =v2Cr−x .)

Since all forces are either potential (mg) or do no work (T ),the system is conservative, thus, we can use conservation of energy principle.With reference given in Fig.(b), the gravitational potential energy of the ball at positionsA and C are

VA = mghA = 0, VC = mghC = mg(2r − x) . (1.28)

(Note: do not confuse v with V .)


Conservation of Energy:

TA + VA = TC + VC ⇒1

2m(3gr) =

1

2mv2C +mg(2r − x) (1.29)

v2C = g(2x− r) (1.30)

Solving (1.27) and (1.30) yields:

x = 23r and vC =

√13gr .


1.4 Homework The m = 65 kg skier startsfrom rest at A. Assuming that the friction, airresistance and the size of the skier can be neglec-ted:

(a ) Determine his speed at point B as a functi-on of hAB.

(b ) Determine the distance l and the time twhen he lands in C.

(c ) Calculate the distance l and the time t when he lands in C, if α = 30◦, hAB = 15 mand hBD = 4.5 m .

Solution


II. Draw the free-body diagramFree-body diagram at point B:

III. Apply the appropriate principle(s)(a) Since all forces are either potential (G = mg) or do no work (N),the system is conservative, thus, we can use conservation of energy principle(vA = 0 and yB = y0).

1

2mvA

2 +mg(yA − y0) =1

2mvB

2 +mg(yB − y0)⇒ vB =√

2g(yA − yB) =√

2ghAB

(1.31)

(b) Kinematics.By considering the motion in the x direction:

sx = (sB)x + (vB)xt⇒ sx =√

2ghAB cos(α)t ,

and from the geometry

∆x = lx = l cosα ,


thus,

l cosα =√

2ghABt cos(α)⇒ l =√

2ghABt . (1.32)

By considering the motion in the y direction

sy = (sB)y + (vB)yt+1

2ayt

2 =√

2ghABt sin(α) + 0.5(−g)t2 ,

and from the geometry

|∆y| = hBD + ly = hBD +√

2ghABt sin(α) .

Equating the geometrical quantity to the kinematic one (i.e. the vertical position of theskier with respect the chosen coordinate frame) we get

sy = −|∆y| ,

thus,

0 = hBD + 2√

2ghABt sin(α)− 0.5gt2 . (1.33)

Therefore the solution for t is:

t =−2 sin(α)

√2ghAB −

√(2 sin(α)

√2ghAB)2 + 2hBDg

−g. (1.34)

(c) Applying (1.34) and (1.32):t = 3.743 s and l = 64.2 m .


1.5 Homework A m = 300 kg crate is dropped vertically onto a conveyor belt that ismoving at v = 1.2 m/s. A motor maintains the belt’s constant speed. The belt initiallyslides under the crate, with a friction coefficient µ = 0.4. After a short time, the crate ismoving at the speed of the belt. During the period in which the crate is being accelerated,find the work done by the motor which drives the belt.(Note that the dynamic friction is not considered here.)

Solution


II. Draw the free-body diagramFree body diagram for the belt and the crate.

III. Apply the appropriate principle(s)Since the gravity is balanced by the normal force, the only remaining, the only forces thatremain are the two Newton-pair friction forces (Fx) - of the belt on the crate, and of thecrate on the belt:

Fx = µmg .

The key insight is to realize that only the belt matters here, because it is the belt’smotor that we’re interested in. We need to calculate the work exerted on the belt. Weknow the force, and so we need the distance the belt travels. Since we know how fast thebelt is traveling, if we find how long the block takes to move with the belt, we can findhow much it traveled.

The initial velocity of the block is v0 = 0 m/s, while the final velocity is vf = 1.2 m/s.The acceleration of the block is a = Fx/m = µg. The time to reach the final velocity is

vf = v0 + at⇒ t =vfa

=vfµg

.


Given this, the distance the belt traveled is

d = vf t =v2fµg

.

Finally the work done by the motor is

W = Fxd = µmgv2fµg

= mv2f = 432J . (1.35)


1.6 Homework A ball of mass m is attached to a string oflength L. It is being swung in a vertical circle with enough speedso that the string remains taut throughout the ball’s motion.Assume that the ball travels freely in this vertical circle withnegligible loss of total mechanical energy. At the top and bottomof the vertical circle, the magnitude of the constraint force on theball is Ntop and Nbot, respectively.Find the difference Nbot −Ntop. (Find an expression in terms of m.)

Solution

I. Choose the reference systemNormal-tangential (n-t) coordinates are attached to, and move with, the ball.

II. Draw the free-body diagramFree-body diagram at the top and the bottom:

III. Apply the appropriate principle(s)Let vtop and vbot be the velocities of the ball at the top and bottom of the circle respectively.At the top of the circle, applying the linear momentum principle in the normal direction:

Pn,top =∑

Fn,top ⇒mv2topL

= Ntop +mg . (1.36)

Using the same consideration at the bottom of the circle:

Pn,bot =∑

Fn,bot ⇒mv2botL

= Nbot −mg . (1.37)

Therefore we get

Nbot −Ntop =m

L(v2bot − v2top) + 2mg . (1.38)

Since all forces are either potential (G = mg) or do no work (Nbot, Ntop), the system isconservative, thus, we can use conservation of energy principle to relate vtop and vbot

Ttop + Vtop = Tbot ⇒1

2mv2top + 2mgL =

1

2mv2bot , (1.39)

4gL = v2bot − v2top . (1.40)

Substituting (1.40) into (1.38)

Nbot −Ntop =m

L4gL+ 2mg = 6mg . (1.41)


1.7 Homework The 2 kg collar is released from rest at A andtravels along the smooth vertical guide. Determine the speed ofthe collar when it reaches position B. Find also the normal forceexerted on the collar at this position. The spring has an unstret-ched length of 200 mm.

Solution

I. Choose the reference systemNormal-tangential (n-t) coordinates are attached to, and move with, the collar.

II. Draw the free-body diagramFree-body diagram at point A:

III. Apply the appropriate principle(s)Since all forces are either potential (Fsp, mg) or do no work (N),the system is conservative, thus, we can use conservation of energy principle.With reference to the datum set in Fig.(a),

the gravitational potential energy of the collar at positions A and B are

(Vg)A = mghA, (Vg)B = mghB . (1.42)

When the collar is at positions A and B, the spring stretches sA = lA− l0 and sB = lB− l0.Thus, the elastic potential energy of the spring when the collar is at these two positionsis

(Ve)A =1

2ks2A, (Ve)B =

1

2ks2B . (1.43)

Conservation of energy:

TA + (Vg)A + (Ve)A = TB + (Vg)B + (Ve)B ,


1

2mv2A +mghA +

1

2ks2A =

1

2mv2B +mghB +

1

2ks2B . (1.44)

Since hA = 0 m and vA = 0 m/s:

vB =

√k(s2A − s2B)

m− 2ghB . (1.45)

Substituting sA = 0.3657 m , sB = 0.08284 m and hB = 0.6 m into (1.45):vB = 5.127 m/s.

Free-body diagram at point B:

Applying linear momentum principle in the normal direction at point B.

Pn =∑

Fn ⇒ man = mg + Fsp sin(θ)−NB

NB = mg + Fsp sin(θ)−man = mg + Fsp sin(θ)−mv2Br

(1.46)

Since the Fsp = k(lB − l0) = ksB, substituting the numbers we get:

NB = −208N .

Note: the negative sign indicates that NB acts in the opposite direction to that shown inthe free-body diagram.


1.8 Homework The man on the bicy-cle attempts to coast around the ellipsoidalloop without falling off the track. Determi-ne the speed he must maintain at A justbefore entering the loop in order to per-form the stunt. The bicycle and the manhave a total mass of 85 kg and a center ofmass at G. Neglect the mass of the wheels.

Hints: If the path is defined as y = f(x), the radius of curvature at the point where

the particle is located can be obtained from r = [1+(dy/dx)2]3/2

d2y/dx2

Solution

I. Choose the reference systemNormal-tangential (n-t) coordinates are attached to, and move with, the cyclist.

II. Draw the free-body diagramNow the free-body diagram is the following:

Geometry.First calculate the radius of the curvature at point A and B.

y =q

p

√p2 − x2 (1.47)

dy

dx= −q

px(p2 − x2)−1/2 (1.48)

d2y

dx2= −q

p[x2(p2 − x2)−3/2 + (p2 − x2)−1/2] (1.49)

The slope angle at point B is given by:

tan θ =dy

dx

∣∣∣∣x=0

= 0⇒ θ = 0 , (1.50)


and the radius of curvature at point B is

r =[1 + (dy/dx)2]3/2

d2y/dx2= 2.25m. (1.51)

Since the center of mass (hG) for the cyclist is 1.2 m off the track, the radius of curvaturefor the cyclist is: rb = r − hG = 1.05m .

III. Apply the appropriate principle(s)Apply now the linear momentum principle in the normal direction at point B.In order for the cyclist to pass point B without falling off the track, it is required thatNB ≥ 0, that is

Pn =∑

Fn ⇒ man = mg +NB ⇒v2Brb

= g ⇒ v2B = grb .

(Note that NB = 0.)Setting the origin of the reference frame at the center of mass of the cyclist before heenters the track, when he is at point B, ∆h with respect to his initial position (point A)is: ∆h = yB − yA − 2hG.Since all forces are either potential (G = mg) or do no work (NB),the system is conservative, thus, we can use conservation of energy principle.

TA + VA = TB + VB ⇒1

2mv2A =

1

2mv2B +mg∆h (1.52)

vA =√v2B + 2g∆h⇒ vA =

√grb + 2g∆h = 10.96m/s (1.53)


1.9 Homework The m = 0.5 kg ball of negligible size isfired up the smooth vertical circular track using the springplunger. The plunger keeps the spring compressed at ∆s =0.08 m when s = 0 m. Determine how far it must be pulledback and released so that the ball will begin to leave thetrack when θ = 135◦.

Solution

I. Choose the reference systemNormal-tangential (n-t) coordinates are attached to,and move with, the ball.

II. Draw the free-body diagramFree-Body diagram at point B:

III. Apply the appropriate principle(s)Applying linear momentum principle in the normal direction at point B:

Pn =∑

Fn ⇒ mv2Br

= mg cos(180◦ − θ) +NB ⇒ v2B = rg cos(180◦ − θ) . (1.54)

(Note that the limiting case is when NB = 0.)Since all forces are either potential (Fsp, mg) or do no work (NB),the system is conservative, thus, we can use conservation of energy principle.Here, the ball is being displaced vertically by hB = r+ r sin(θ− 90◦) and the spring forceis given by Fsp(s) = k(s+ ∆s).Applying the conservation of energy, and considering that the elastic energy of the springis totally transfered to the ball

TA + VA = TB + VB ⇒s∫

0

Fsp ds = mghB +1

2mv2B , (1.55)

s∫0

k(s+ ∆s)ds = mghB +1

2mv2B ⇒ 0.5ks2 + (k∆s)s−mghB −

1

2mv2B = 0 . (1.56)


Substituting expressions for hB and (1.54) into (1.56) leads to

s =−k∆s+

√k2(∆s)2 + 2k[mg(r + r sin(θ − 90◦)) + (0.5mrg cos(180◦ − θ))]

k.

(1.57)

Substituting values into (1.57) gives s = 0.1789 m.

1 motion of a single particle - linear momentum, work and...

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