EM 311M - Dynamics

Exam 2 - Solutions

March 7, 2012

1. i. (10 points) (a) Derive the principle of work and energy for an object of mass m.( 5 points) (b) Explain in which case the energy is conserved and derive the statement

of conservation of energy.

ii. (10 points) In terms of polar coordinates the force exerted on an object by a nonlinearspring is

~F = −[k(r − r0) + q(r − r0)3

]er

with k and q constants, and r0 the unstretched length of the spring. Find the potentialenergy V (r) associated with this nonlinear spring force.

iii. (10 points) The collar A, with weight WA, slides on the smooth horizontal bar. The non-linear spring constants k and q are given. At the instant shown, the spring is unstretchedand the block B, with weight WB, is moving downward at a velocity v0. Use conser-vation of energy to determine the expression of the velocity v of B when it has moveddownward a distance h from its current position, as a function of the given parametersand g.

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224

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h

Solution:

i) (a) (5 points) By Newton’s 2nd law

Σ~F = m~a = md~v

dt

with Σ~F the sum of external forces and ~a the acceleration of the center of mass of the object.By taking the dot product with ~v we obtain

Σ~F · ~v = md~v

dt· ~v .

We now observe that this can be written as follows

Σ~F · d~rdt

= Σ~F · ~v = md~v

dt· ~v =

1

2md(v2)

dt.

(5 points) We first obtain

Σ~F · d~r =1

2md(v2),

and by integration ∫ ~r1

~r1

Σ~F · d~r =

∫ v22

v21

1

2md(v2) =

1

2mv2

2 −1

2mv2

1,

wherev1 = magnitude of the velocity at ~r1,v2 = magnitude of the velocity at ~r2.

We then obtain the Principle of Work and Energy:∫ ~r1

~r1

Σ~F · d~r =1

2mv2

2 −1

2mv2

1 .

(b) (3 points) Suppose there is a scalar function of position V (~r) that satisfies

dV = −Σ~F · d~r,

then the energy is conserved, as we demonstrate below.(2 points) Using this expression, the work can be computed as∫ ~r1

~r1

Σ~F · d~r =

∫ V2

V1

−dV = −(V2 − V1),

with V1 = V (~r1) and V2 = V (~r2). We then obtain the Statement of Conservation of Energy:

12mv

21 + V1 = 1

2mv22 + V2 ,

so that the sum of potential and kinetic energy is conserved.

ii) (5 points) In terms of polar coordinates, the force exerted on an object by a nonlinearspring is

~F = −[k(r − r0) + q(r − r0)3

]er

with k and q constants, and r0 the unstretched length of the spring. In order to compute thepotential energy function, we use the relation

dV = −Σ~F · d~r.

We need to find the expression for d~r. In the polar coordinate system

~r = rer ⇒ ~v =d~r

dt=dr

dter + r

dθ

dteθ.

Then, we obtaind~r = drer + rdθeθ.

(5 points) We now have

dV = −Σ~F · d~r = −(−[k(r − r0) + q(r − r0)3

]er)· (drer + rdθeθ)

=[k(r − r0) + q(r − r0)3

]dr.

Integrating, we get

V − V0 =

∫ V

V0

dV =

∫ r

r0

[k(r − r0) + q(r − r0)3

]dr =

1

2k(r − r0)2 +

1

4q(r − r0)4,

where V = V (r) and V0 = V (r0). Since work only depends on differences of potential energy,we take arbitrarily V0 = 0 and obtain

V (r) = 12k(r − r0)2 + 1

4q(r − r0)4 .

iii) (5 points) The forces acting on the system are conservative: nonlinear spring and weight,thus we can use conservation of energy to obtain

((((mAgyA +1

2mAv

20 +

1

2mBv

20 =((((mAgyA +

1

2mAv

2 +1

2mBv

2 −mBgh+1

2kh2 +

1

4qh4,

where the collar A and the block B move together, i.e., they have the same velocity at alltimes, and we notice that the nonlinear spring stretch r − r0 = h, once the block movesdownward a distance h. The potential energy of the weight of A is computed for some heightyA; however, this potential energy is unchanged during the motion because the height of Aremains constant in the process and thus this expression cancels in the conservation of energyequation.

(5 points) We now reorder the terms to obtain

1

2(mA +mB)v2

0 +mBgh−1

2kh2 − 1

4qh4 =

1

2(mA +mB)v2,

giving the result

v = −√v2

0 +2g

(WA +WB)

(WBh−

1

2kh2 − 1

4qh4

)

where we used the relation W = mg, and we take the negative root for v since the block Bmoves down-ward.

2. A block of mass mB is moving with velocity vB as it hits a sphere of mass mA , which is at restand hanging from a cord attached at O. The coefficient of kinetic friction µK between theblock and the horizontal surface, and the coefficient of restitution e between the block andthe sphere are given.

i. (15 points) Determine the expression for the maximum height h reached by the sphereas a function of the given parameters and g.

ii. (15 points) Determine the expression for the distance x traveled by the block as a functionof the given parameters and g.

iii. (3 points) Given the cord length `, determine the angular momentum vector about O ofthe sphere immediately after the impact, as a function of the given parameters.(2 points) Is the sphere’s angular momentum conserved? Why?

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�vB

6

?

h

O

BA

-� x

Solution:

i) (5 points) During the impact: by conservation of linear momentum

(∗) mA��*0vA + mBvB = mAv

′A + mBv

′B,

where vA = 0 since the sphere starts at rest. By the coefficient of restitution

(∗∗) e =v′B − v′A��*

0vA − vB.

From Equation (∗∗), we obtain

v′B = v′A − vBe .

(5 points) Substitute v′B into Equation (∗) to obtain

mBvB = mAv′A +mB(v′A − vBe).

Reordering the terms we get

v′A =mB

mA +mB(1 + e)vB .

(5 points) After the impact: by conservation of energy for the sphere A

1

2��mA(v′A)2 =��mAgh⇒ h =

1

2g(v′A)2.

Therefore,

h =1

2g

(mB

mA +mB

)2

(1 + e)2v2B .

ii) (5 points) After the impact: by the Principle of Work and Energy for the block B (takinginto account friction) ∫ −x

0µKNdx =

1

2mB(vB(−x))2 − 1

2mB(vB(0))2,

where we assume the positive x-direction is to the right and x = 0 at the impact location.Furthermore, the block travels the distance x until it stops, so vB(−x) = 0, and the velocityof the block just after the impact is vB(0) = v′B.

(5 points) To find the magnitude of the normal N , we write Newton’s 2nd law in they-direction for the block B:

ΣFy = N −mBg = mBay = 0⇒ N = mBg .

(5 points) Substituting N and vB(0) into the work-energy relation we obtain

−∫ 0

−xµKmBgdx = −µKmBgx = −1

2mB(v′B)2 ⇒ x =

1

2µKg(v′B)2 .

We now need v′B, which can be obtained, using the previous results, as follows:

v′B = v′A − vBe =mB

mA +mB(1 + e)vB − vBe =

mB +���mBe−mAe−���mBe

mA +mBvB

=mB −mAe

mA +mBvB

Substituting the expression for v′B in the result for x we obtain

x =1

2µKg

(mB −mAe

mA +mB

)2

v2B .

iii) (3 points) The angular momentum vector of the sphere A about O, immediately afterthe impact, is

~HO = ~rA ×mA~v′A.

Let’s take a look at the sphere immediately after the impact:

2. A block of mass mB is moving with velocity vB as it hits a sphere of mass mA , which is at restand hanging from a cord attached at O. The coefficient of kinetic friction µK between theblock and the horizontal surface, and the coefficient of restitution e between the block andthe sphere are given.

i. (15 points) Determine the expression for the maximum height h reached by the sphereas a function of the given parameters and g.

ii. (15 points) Determine the expression for the distance x traveled by the block as a functionof the given parameters and g.

iii. (3 points) Given the cord length �, determine the angular momentum vector about O ofthe sphere immediately after the impact, as a function of the given parameters.(2 points) Is the sphere’s angular momentum conserved? Why?

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✛vB

✻

❄

h

O

BA

✲✛ x

Solution:

i) (5 points) During the impact: by conservation of linear momentum

(∗) mA✟✟vA + mBvB = mAv�A + mBv�B,

where vA = 0 since the sphere starts at rest. By the coefficient of restitution

(∗∗) e =v�B − v�A✟✟vA − vB

.

From Equation (∗∗) we obtain

v�B = v�A − vBe .

(5 points) Substitute v�B into Equation (∗) to obtain

mBvB = mAv�A + mB(v�A − vBe).

Reordering the terms we get

v�A =mB

mA + mB(1 + e)vB .

After the impact: by conservation of energy for the sphere A

1

2✘✘mA(v�A)2 =✘✘mAgh ⇒ h =

1

2g(v�A)2.

?

O

~rA

� ~v ′AA

In Cartesian coordinates, we have

~rA = −` j,~v ′A = −v′Ai.

Then

~HO = (−` j)×mA(−v′Ai),

and since j × i = −k, we obtain ~HO = −`mAv′Ak. Substituting v′A gives the result

~HO = −` mAmB

mA +mB(1 + e)vB k .

(2 points) The angular momentum of the sphere is not conserved. This can be observedsince the velocity of the sphere when its height is h is ~0. The reason why the angularmomentum is not conserve is that the weight acts in the direction −j and thus it is notparallel to ~rA at all times.

3. The man releases a ball at a height H with no vertical velocity and a small horizontal velocityv. It hits the floor, and rebounds to a height h as shown in the figure.

i. (24 points) Determine the coefficient of restitution e of the impact of the ball with thefloor.

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h

ii. (2 points) What is the height h′ the ball will reach in the next rebound?

iii. Assume now that h is unknown:

- (2 points) If e = 0, what happens with the ball? What type of impact is that?

- (2 points) What will be the height h under a perfectly elastic impact?

Solution:

i) (8 points) The initial velocity of the ball is

~v0 = vi+ 0j.

From kinematics, we know that the velocity of the ball in the x-direction doesn’t change formthe starting point until the ball hits the floor. Since the impact occurs in the y-direction, thevelocity of the ball in the x-direction is unchanged by the impact as well.

Let~v = velocity of the ball just before the impact with the floor,~v′ = velocity of the ball just after the impact with the floor.

Then, we have(~v0)x = (~v)x = (~v′)x = v.

We now determine the velocity in the y-direction just before the impact:By conservation of energy (before the impact)

������1

2m((~v0)x)2 +mgH =

����

��1

2m((~v)x)2 +

1

2m((~v)y)

2,

where we used the fact that (~v)2 = ((~v)x)2 + ((~v)y)2. We then get

(~v)y = −√2gH ,

and we take the negative root because the velocity is down-ward.

The velocity before the impact is then

~v = vi−√

2gH j.

(8 points) To find the velocity after the impact, we use the equation for the coefficientof restitution in the y-direction:

e =(~v′floor)y − (~v′)y(~v)y − (~vfloor)y

.

The velocity of the floor before the impact (~vfloor)y = 0 and the velocity of the floor after theimpact is assumed negligible, i.e., (~v′floor)y � (~v′)y. We then obtain

(~v′)y = −(~v)y e =√

2gH e .

The velocity after the impact is then

~v′ = vi+√

2gH e j.

(8 points) We now relate the coefficient of restitution to the maximum height h afterfirst rebound:By conservation of energy (after the impact)

����

��1

2m((~v′)x)2 +

1

2m((~v′)y)2 =

����

��1

2m((~v′)x)2 +mgh⇒ ((~v′)y)2 = 2gh.

Substituting the expression for (~v′)y we obtain

2gHe2 = 2gh,

from where we conclude that the coefficient of restitution is

e =

√h

H.

ii) (2 points) We observe that given the initial height H, we get the relation h = He2. Then,the maximum height in the next rebound is

h′ = he2,

and the result is

h′ =h2

H.

iii) Assume now that h is unknown.

- (1 points) If e = 0, by the relation h = He2, we obtain h = 0 .In this case, the ball gets stuck to the floor.

(1 points) This is a perfectly plastic impact.

- (2 points) For a perfectly elastic impact e = 1, and since h = He2, we obtain h = H .