Physics-Kleppner and Kollenkow- Answers

Physics 141 Problem Set 2 Corrected Solutions

Nataliya Yufa

(2.6) Gravel mixer

We want to find ω such that at least for a moment the particles are not stuck to thecylinder. This is most likely to happen at the top, where the normal force is in the directionof gravity. For the particle to leave the cylinder, N=0. Hence the y-component of the totalforce on the particle is

Fg + N = Fg =mv2

r. (1)

Using the relation v = ωR we obtain

mg = mv2

r= mω2R, (2)

⇒ ωc =

√g

R. (3)

Thus for all values of ω less than ωc the particles will not be stuck to the walls all the time.

(2.9) Particle in a cone

The particle happens to be moving in a horizontal circle inside a cone of half-angle θ.We want to find the radius r of the circle in terms of v0, θ and g. Consider the x- andy-components of the net force acting on the mass.

Since there’s no movement in the y-direction, Fy is zero, i.e.

Fy = Ny −mg = 0 ⇒ Ny = mg. (4)

From the geometry we find that tan θ = Ny

Nx, hence

Nx =Ny

tan θ=

mg

tan θ. (5)

Circular motion in the horizontal direction implies that

Fx = Nx =mg

tan θ=

mv20

r, (6)

and solving for r we have

r =v2

o tan θ

g. (7)

1

(2.12) Pulling out the tablecloth

Initially, the glass is accelerated by the force of friction due to the tablecloth for sometime tmax. In this time it will reach velocity vo

vo =Ff tmax

m= µgtmax. (8)

On the other hand, the glass will be slowed down by the friction due to the table (which,because the coefficients of friction happen to be the same for the table and the tablecloth,has the same magnitude as the frictional force of the tablecloth). Therefore, during both theacceleration and the deceleration the glass will travel exactly half of the available distance.

We want the glass to slow down in a given distance d/2, thus

v0 =√

µgd. (9)

Combine the two expressions for v0 to get

v0 = µgtmax =√

µgd (10)

⇒ tmax =

√µgd

µg=

√d

µg=

√0.5ft

0.5 · 32ft/s2=

1

4√

2s. (11)

(2.25) Shortest possible period

Intuitively, the shortest period of rotation will occur when the two solid spheres are asclose as possible to each other. Let’s try to prove this. Because gravity is responsible for thecircular motion, we have

Gm2

4r2=

mv2

r(12)

⇒ v =

√Gm

4r. (13)

The period is given by

T =2πr

v=

4π√Gm

r3/2. (14)

Note that the smaller the radius, the shorter the period. However, we’re limited by theradius of the spheres, as they cannot come closer than their radius R. Hence the shortestpossible period would be T = 4π√

GmR3/2.

(2.28) Car going around a bend

2

Let’s choose the x- and y-axis to be horizontal and vertical respectively, since the carshould be moving in a horizontal surface and therefore there should be a non-zero force solelyin the horizontal direction. Writing the components of the total force we have

Fy = 0 = N cos θ −mg ± µN sin θ (15)

and

Fx = N sin θ ∓ µN cos θ =mv2

R. (16)

Here the different signs correspond to the maximum and minimum velocities. It follows that

N =mg

cos θ ± µ sin θ. (17)

Substituting this into the expression for Fx we get an equation for v2:

v2 =gR

cos θ ± µ sin θ(sin θ ∓ µ cos θ). (18)

Writing this in a different form

gR(sin θ − µ cos θ)

cos θ + µ sin θ< v2 <

gR(sin θ + µ cos θ)

cos θ − µ sin θ. (19)

(2.33) Particle on a rotating rod

a) The trick to this problem is to realize that there is no force in the radial direction as

the rod is frictionless. Hence the total force is given by ~F = F θ. Newton said F = ma, thus

F θ = m[(r − rθ2)r + (rθ + 2rθ)θ

], (20)

where θ equals ω. It must be that the radial component of the acceleration is zero since theradial part of the force is zero:

r − ω2r = 0. (21)

Now we need to find the solution to this differential equation.Let’s try the solution r = Ae−γt + Beγt. Then

r = γ2Ae−γt + γ2Beγt = γ2r. (22)

In order to satisfy our differential equation we need to have

r = γ2r = ω2r, (23)

and therefore γ = ω.b) By inspection, notice that Beωt will tend to positive or negative infinity, unless B is

zero. Thus if we want the radius to be always decreasing without reaching zero, B must bezero. In all other cases where the particle doesn’t reach the origin, we must have a positiveB. In that limit, we see that Aeωt will tend to zero for large t, while Beωt will go to infinity.

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7

Physics 141

Problem Set 7

Eduard Antonyan

1 (7.2) Rotating flywheel on a rotating table

W

a w0 TT L0

Figure 1: Side view.

As it can be seen from the picture the torque is out of the picture and its magnitude is:

τ = 4Tαl, (1)

where we assumed that the angle α is small. The derivative of the angular momentum d−→L0dt is also out the picture.

That, and also its magnitude follows from:d−→L0

dt=−→Ω ×

−→L0 (2)

Thus τ = dL0dt = ΩL0 = ΩI0ω0, and therefore:

α =I0ω0Ω4T l

. (3)

2 (7.5) Car on a curve

(a) Let us understand first why should the car tend to roll over. As one can see from the picture,−→N1,

−→f1 and

−→f2 create

torque (about the center of mass), that’s into the picture, and−→N2 creates torque out of the picture. If the car is stable,

then the total torque should be zero. By Newton’s second law:

f1 + f2 = Ma =Mv2

rN1 + N2 = Mg (4)

1

N2

d N1 L

a Mg f1 f2

Figure 2: View from behind. The car is turning to the left.

So the faster the car is moving the larger are f1 and f2, and thus the torque into the page. To have equilibrium, N1

will have to decrease and N2 will have increase. For high enough velocities N1 will become zero and the car will startrolling over.

Now the reason putting a spinning flywheel can help is because we can put it in such way that the torque will beused to change the angular momentum of the flywheel and not of the car!

Now let’s consider the case where the car is turning to the left with angular velocity−→Ω which will point vertically

upward. Let’s put the flywheel so that its angular momentum−→L is pointing radially outward (to the right on the

picture). To make that angular momentum rotate with the car, one needs torque equal to d−→L

dt =−→Ω ×

−→L , i.e. pointing

forward with respect to the car, which is exactly what we’ve got.Note, however, that if the car is turning to the right, we should not reverse the direction of the angular momentum,

i.e. if the car is turning to the right the angular momentum should point radially inward. This is because in that case−→Ω is pointing vertically downward, and the torque is pointing out of the page (backward w.r.t. the car). Thus theright choice for

−→L would be pointing again to the right on the picture, so that

−→Ω ×

−→L is again in the direction of the

torque. (More mathematically the reason the direction of−→L doesn’t change is because torque, angular velocity and

angular momentum are axial vectors, not real vectors).

(b) If the loading on the wheels is the same, then N1 = N2 ≡ N , thus f1 = f2 ≡ f . Since we’ve already discussed thedirections of the derivative of the angular momentum and the torque, let’s just write down the scalar version of thetorque equation:

τ =dL

dt. (5)

Nowτ = f1d sinα + N1d cos α + f2d sinα−N2dcosα = 2fd sinα = 2fL = MvΩL, (6)

where in the last equality we used Newton’s second law (4). Finally for a disk-shaped flywheel using

dL

dt= ΩL = ΩIω = Ω

mR2

2ω, (7)

we have:ω =

2MvL

mR2. (8)

3 (7.8) Deflecting hoop

(a) The torque induced by the force acting on the top of the hoop will be in the direction shown in the picture. Since−→τ = d

−→L

dt that means that the angular momentum will start rotating in the horizontal plane and if we look from thetop it will be rotating counterclockwise (−→ω pointing vertically upward). So in the gyroscope approximation the forcewill not incline the plane of the hoop, but will instead deflect the line of rolling.

2

I

Mv′ Mvf

L

t I

(a) (b)

Figure 3: (a) “Normal” view. (b) View from above.

By the conservation of momentum we have:

M−→v +−→I = M−→v ′, (9)

where −→v ′ is the velocity of the hoop after the tap. Thus the line of the rolling of the hoop will change by an angle:

φ ≈ tanφ =I

Mv. (10)

(b) The gyroscope approximation is valid if dLdt Lω, i.e. if the change in the angular momentum is small compared

to the initial angular momentum. In our case this would mean that

τmax = Fb Lω = Mb2ω2 = Mv2. (11)

Thus for the gyroscope approximation to hold, we need the peak applied force to satisfy:

F Mv2

b. (12)

4 (10.2) Oscillating spring

For the damping constant we have:

γ =ω1

Q=

2π 2 Hz60

≈ 0.21 s−1. (13)

Then from

ω0 =

√k

m, (14)

we have:

k = mω20 = m

(ω2

1 +γ2

4

)= 0.3kg

[(2π 2 Hz)2 +

(0.21 s−1)2

4

]≈ 47.377 N/kg. (15)

5 (10.6) Falling masses and critical damping

(a) At the rest position we have equilibrium of forces, thus:

Mg = kx0, (16)

where m is the mass of the platform and x0 is the distance from the initial position of the platform to its final position.Thus

k =Mg

x0= 980 N/kg. (17)

3

(b) The velocity of the mass just before it hits the platform v is determined from energy conservation:

Mgh =Mv2

2, (18)

where h is the height of the mass above the platform. Then we have an inelastic collision of the mass with the platform,so by momentum conservation:

Mv = (M + m)v′, (19)

where v′ is their velocity just after they stick together.Since we want critical damping we need γ = 2ω0 = 2

√k

M+m ≈ 18.07 s−1.Now the general equation of motion for critical damping is given by:

x(t) = Ae−γ2 t + Bte−

γ2 t. (20)

Here x(t) is the height of the platform w.r.t. the final position. Since we want x(0) = x0 and dxdt |0 = v′, we get for the

coefficients A and B:

A = x0

B = v′ +γ

2=

M

M + m

√2gh +

√g

x0

M

M + m. (21)

So finally the equation of motion is:

x(t) = x0e−

√g

x0t+

(M

M + m

√2gh +

√g

x0

M

M + m

)te−

√g

x0t. (22)

4

1

2

3

4

5

6

7

8

9

MASSACHUSETTS INSTITUTE OF TECHNOLOGY

Department of Physics

Physics 8.012 Fall Term 2005

PROBLEM SET 8

Due: Friday, November 18 at 4:00pm.

Reading: Finish Kleppner & Kolenkow, Chapter 6

0. Collaboration and discussion. Please give a brief statement at the top of your homework telling

us the names of all the students with whom you discussed the homework problems.

. Kleppner & Kolenkow, Problem 6.9









10. Kleppner & Kolenkow, Problem 6.35

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