Question

a force is applied to a dumbbell for a certain period of time, first as in (a) and then as in (b).
in which case does the dumbbell acquire the greater kinetic energy?
a. case (a)
b. case (b)
c. no difference
d. it depends on the moment of inertia of the dumbbell

Explanation:

Step1: Apply impulse - momentum theorem

The impulse \(J = F\Delta t\) is equal to the change in linear momentum \(\Delta p\). Since the force \(F\) and the time interval \(\Delta t\) are the same in both cases (a) and (b), the change in linear momentum \(\Delta p\) is the same for both cases. Using \(J=\Delta p = m_{total}v_{cm,f}-m_{total}v_{cm,i}\), assuming the dumb - bell starts from rest \(v_{cm,i} = 0\), so \(p = m_{total}v_{cm}\), and \(v_{cm}=\frac{F\Delta t}{m_{total}}\) (where \(m_{total}=2m\) for both cases).

Step2: Calculate the kinetic energy

The total kinetic energy \(K\) of a rigid body is \(K = K_{trans}+K_{rot}\). The translational kinetic energy \(K_{trans}=\frac{1}{2}m_{total}v_{cm}^{2}\). Since \(m_{total}\) and \(v_{cm}\) are the same for both cases (a) and (b) (from the impulse - momentum relation), \(K_{trans}\) is the same. In case (a), there is also rotational kinetic energy \(K_{rot}=\frac{1}{2}I\omega^{2}\) (because the force is applied at a distance from the center of mass, creating a torque \(\tau= rF\) and thus an angular acceleration \(\alpha=\frac{\tau}{I}\), leading to an angular velocity \(\omega\)). In case (b), the force is applied along the line connecting the two masses (passing through the center of mass), so \(\tau = 0\), \(\alpha=0\), and \(K_{rot}=0\).

Answer:

B. Case (b)