1. You are trying to open a door that is stuck by pulling on the doorknob in a direction perpendicular to the door. If you instead tie a rope to the doorknob and then pull with the same force, is the torque you exert increased?
1. yes
2. no
Answer: 2. Because the force you are applying is unchanged and the perpendicular
distance between the line of action and the pivot point (the
lever arm) is likewise unchanged, the torque you apply does not increase.
(Pulling at an angle only decreases the lever arm.)

2. 4. The answer depends on the rotational inertia of the dumbbell.
A force F is applied to a dumbbell for a time interval Dt, first as in (a) and then as in (b). In which case does the dumbbell acquire the greater center-of-mass speed?
1. (a)
2. (b)
3. no difference
4. The answer depends on the rotational inertia of the dumbbell.
Answer: 3. Because the force acts for the same time interval in both cases,
the change in momentum ΔP must be the same in both cases, and thus the
center-of-mass velocity must be the same in both cases.

3. 4. The answer depends on the rotational inertia of the dumbbell.
A force F is applied to a dumbbell for a time interval Dt, first as in (a) and then as in (b). In which case does the dumbbell acquire the greater energy?
1. (a)
2. (b)
3. no difference
4. The answer depends on the rotational inertia of the dumbbell.
Answer: 2. If the center-of-mass velocities are the same, so the translational
kinetic energies must be the same. Because dumbbell (b) is also rotating,
it also has rotational kinetic energy.

4. Imagine hitting a dumbbell with an object coming in at
speed v, first at the center, then at one end. Is the center-of-mass speed of the dumbbell the same in both cases?
1. yes
2. no
It is easier to start the object rotating than to make the entire mass translate. m v V’
Answer: 2. The moving object comes in with a certain momentum. If it hits the center, as in case (a), there is no rotation, and this collision is just like a one-dimensional collision between one object of inertial mass m and another of inertial mass 2m. Because of the larger inertial mass of the dumbbell the incoming object bounces back. In (b), the dumbbell, starts rotating. The incoming object encounters less “resistance” and therefore transfers less of its momentum to the dumbbell, which will therefore have a smaller center-of-mass speed than in case (a). m 2m v v’ V’

5. stick if it is balanced by a support force at the 0.25-m mark?
A 1-kg rock is suspended by a massless string from one end of a 1-m measuring stick. What is the mass of the measuring
stick if it is balanced by a support force at the 0.25-m mark? kg kg
3. 1 kg
4. 2 kg
5. 4 kg
6. impossible to determine
Answer: 3. Because the stick is a uniform, symmetric body,we can consider
all its weight as being concentrated at the center of mass at the 0.5-m mark.
Therefore the point of support lies midway between the two masses, and
the system is balanced only if the total mass on the right is also 1 kg.

6. Two wheels with fixed hubs, each having a mass of 1 kg, start from rest, and forces are applied as shown. Assume the hubs and spokes are massless, so that the rotational inertia
is I = mR2. In order to impart identical angular accelerations, how large must F2 be? N N
3. 1 N
4. 2 N
5. 4 N
Answer: 4. For each wheel, the moment of inertia times the angular acceleration
must equal the force on the wheel rim times the distance from
the rim to the center of the wheel.