Question 1. A

Since the motion is uniform and circular, the acceleration is centripetal and its magnitude is a=v2/R=52.4 m/s2.

Question 2. B

Question 3. E

Question 4. E

We use conservation of energy. Set the zero of potential energy at the level of the nail. Then the initial mechanical energy is totally rotational. At the final position it is gravitational potential energy.

I w2 = mgR

w2 = 2mgR/I

I is found using the parallel axis theorem I = ICM + MR2= MR2 + MR2 =3/2MR2 W=8.1 rad/s

Question 5. C

We call the lighter mass m1 and the heavier mass m2. The heavier mass is m2= 1.5m1 and m1 + m2 = 20 kg. Therefore the m1=8kg and m2=12kg.

The beam is in equilibrium so torques are zero. We choose a pivot point about the lighter end. Torque =0= 20*Xcm m2*3.0=0 20*Xcm 12*3.0=0 Xcm = 1.8 m

Question 6. D

FT = mv2/r The speed doubles, but r and m stay the same.

This means the tension increases by a factor of 4

Question 7. A

wf =wi + alpha *t wf=500*2pi/60 wi=250*2pi/60 alpha=5.23 rad/s^2 at=alpha*r at=0.3 m/s^2

Question 8. D

Since the rod is doubled the CM is now at position L (taking origin on left). Using the definition of center of mass

Xcm = [m1x1 +m2x2 +m3x3]/mtot

Xcm=L =(1 (0)+2(x) +2(2L))/4

X=L/2

Question 9. C

It is in static equilibrium, so the torques must balance

Question 10. C

The moment of inertia of the yoyo can be calculated by adding the moment of inertia of the two disks and the moment of inertia of the cylindrical shaft.

I=2*(1/2)M1R12+(1/2)M2R22=2.8*10-4 kg.m2.

Question 11. D

The only two external forces acting on the yoyo are its weight (W) and the tension in the string (T), considering the friction forces between the string and the shaft of the yoyo are negligible.

The net force is not zero, since the center of mass of the yoyo is first accelerating and then decelerating downward.

The tension applies a zero torque on the yoyo with respect to its instantaneous axis of rotation, since it is applied exactly at the point. The weight of the yoyo on the other hand applies a positive torque on the yoyo, therefore the net torque is positive.

The angular momentum of the yoyo is obviously not conserved, as the yoyo is initially at rest (zero angular momentum) and then acquires an increasing angular momentum as it goes down. This is consistent with the observation that the net torque is not zero.

Question 12. D

Of the two forces applied on the yoyo, only the gravitational force, W, is conservative. The tension, on the other hand, is not conservative. But since it is applied at the instantaneous axis of rotation of the yoyo, which is immobile, the tension does not work. Therefore the mechanical energy of the yoyo.

The question of whether friction forces between the string and the yoyo may play a role is a complicated one. By considering that the force applied by the string on the yoyo is equal to the tension in the string, we in fact assume that the string is part of the system yoyo for which the energy is conserved (and this is perfectly ok as long as the string as a negligible mass and that we dont need to consider its kinetic energy). Therefore any friction between the string and the yoyo is in fact an internal force. If we do not want to consider the string as part of our system, then we can see those friction forces applied by the spring on the yoyo as the transmission of the tension in the rope: the net friction force must be equal to the tension in the string (this can be seen by applying Newtons second law to the string, for which ma=0 since it has a negligible mass, meaning that the net force applied on the string is 0; therefore the tension in the string must be equal in magnitude and opposite in direction to the friction forces applied by the yoyo on the string; therefore, because of Newtons third law, the friction forces applied by the string on the yoyo are exactly equal to the tension in the rope).

Question 13. C

Since the mechanical energy of the yoyo, E, is conserved during its fall, we can calculate it at any point of the trajectory, in particular at z=L when the yoyo is at rest. At z=L, E=M*g*L=(2*M1+M2)*g*L=0.88 J.

Question 14. D

The instantaneous rotation axis of the yoyo (immobile axis) is found at the point where the string stops being wrapped up around the yoyo, that is at a distance R2+d of the center of mass of the yoyo.

By analogy with the case of a rotating wheel we can see that the speed of the center of mass must therefore be v=(R2+d)w.

Question 15. D

The speeds at z=L/2 and z=0 can be calculated using energy conservation.

At z=L/2: 1/2*I*w2+1/2*Mv2+M*g*L/2=E

1/2*I/(R2+d)2*v2+1/2*Mv2= E/2

1/2*[M+I/(R2+d)2]* v2=E/2

v2=E/[M+I/(R2+d)2]

Using E=0.88J (question 7), I=2.8*10-4 kg.m (question 6), d=0.03 m and M=2M1+M2=0.105 kg, one finds v=1.8 m/s.

At z=0: 1/2*I*w2+1/2*Mv2=E v2=2E/[M+I/(R2+d)2]

Using the same parameters as above except for d=0.0 m, on finds: v=0.78 m/s.

Question 16. E

The speed of the center of mass of the yoyo decreases between z=L/2 and z=0 (cf. question 11 and the introduction text of the problem), and therefore the linear kinetic energy of the yoyo, KL, decreases.

Since the yoyo goes down between z=L/2 and z=0, the gravitational potential energy of the yoyo, U, decreases as well.

Since mechanical energy is conserved for the yoyo: KL+KR+U=E is a constant, therefore if both KL and U decrease, then KR must increase.

Question 17.