Unit 6: Energy and Momentum of Rotating Systems
Weightage : 10-15%
The uniform, rigid rod of mass \(m\), length \(L\), and rotational inertia \(I\) shown above is pivoted at its left-hand end. The rod is released from rest from a horizontal position.
What is the linear speed of the rod’s center of mass when the rod passes through a vertical position?
Correct Answer: \( \boxed{\mathrm{D}} \)
The torque on the rod is not constant during its motion, so rotational kinematics cannot be used directly. Instead, apply conservation of mechanical energy.
As the rod falls from the horizontal to the vertical position, its center of mass drops by a distance
\( \frac{L}{2} \)
Thus, the loss in gravitational potential energy equals the gain in rotational kinetic energy:
\( mg\left(\frac{L}{2}\right)=\frac{1}{2}I\omega^2 \)
The linear speed of the center of mass is related to the angular speed by
\( v=\omega\left(\frac{L}{2}\right) \)
or
\( \omega=\frac{2v}{L} \)
Substituting into the energy equation,
\( mg\left(\frac{L}{2}\right)=\frac{1}{2}I\left(\frac{2v}{L}\right)^2=\frac{2Iv^2}{L^2} \)
Solving for \(v\),
\( v^2=\frac{mgL^3}{4I} \)
Therefore,
\( v=\sqrt{\frac{mgL^3}{4I}} \)
Therefore, the correct answer is (D).
A wheel with rotational inertia \(I\) is mounted on a fixed, frictionless axle. The angular speed \(\omega\) of the wheel is increased from zero to \(\omega_f\) in a time interval \(T\).
What is the average power input to the wheel during this time interval?
Correct Answer: \( \boxed{\mathrm{B}} \)
The average power is the total work done divided by the time interval:
\(P_{\mathrm{avg}}=\dfrac{W}{T}\)
The work done on the wheel equals the increase in its rotational kinetic energy:
\(K_{\mathrm{rot}}=\dfrac{1}{2}I\omega^2\)
Since the wheel starts from rest,
\(W=\Delta K_{\mathrm{rot}} =\dfrac{1}{2}I\omega_f^2\)
Therefore,
\(P_{\mathrm{avg}} =\dfrac{\dfrac{1}{2}I\omega_f^2}{T} =\dfrac{I\omega_f^2}{2T}\)
Thus, the average power delivered to the wheel depends on the change in rotational kinetic energy divided by the elapsed time.
Therefore, the correct answer is (B).
