Unit 5: Torque and Rotational Dynamics
Weightage : 10-15%
A cylinder rotates with constant angular acceleration about a fixed axis. The cylinder’s moment of inertia about the axis is \(4\,\mathrm{kg\cdot m^2}\). At time \(t=0\), the cylinder is at rest. At time \(t=2\,\mathrm{s}\), its angular velocity is \(1\,\mathrm{rad\,s^{-1}}\).
What is the angular acceleration of the cylinder between \(t=0\) and \(t=2\,\mathrm{s}\)?
Correct Answer: \(\boxed{\mathrm{A}}\)
Since the angular acceleration is constant, use the rotational kinematics equation
\(\alpha=\dfrac{\Delta\omega}{\Delta t}\)
The cylinder starts from rest, so
\(\omega_i=0,\qquad \omega_f=1\,\mathrm{rad\,s^{-1}},\qquad \Delta t=2\,\mathrm{s}\)
Therefore,
\(\alpha=\dfrac{1-0}{2}=0.5\,\mathrm{rad\,s^{-2}}\)
The given moment of inertia is not needed because the problem asks only for the angular acceleration, which is determined directly from the change in angular velocity over time.
Therefore, the correct answer is (A).
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: \(\boxed{\mathrm{B}}\)
Average power is the rate at which work is done:
\(P_{\mathrm{avg}}=\frac{\Delta K}{T}\)
The rotational kinetic energy of the wheel is
\(K_{\mathrm{rot}}=\frac{1}{2}I\omega^2\)
Since the wheel starts from rest,
\(\Delta K=\frac{1}{2}I\omega_f^2\)
Therefore,
\(P_{\mathrm{avg}}=\frac{\Delta K}{T}=\frac{\frac{1}{2}I\omega_f^2}{T}=\frac{I\omega_f^2}{2T}\)
Alternatively, since
\(P=\tau\omega\),
\(\tau_{\mathrm{avg}}=\frac{I\omega_f}{T}\)
and
\(\omega_{\mathrm{avg}}=\frac{0+\omega_f}{2}=\frac{\omega_f}{2}\),
giving
\(P_{\mathrm{avg}}=\tau_{\mathrm{avg}}\omega_{\mathrm{avg}}=\left(\frac{I\omega_f}{T}\right)\left(\frac{\omega_f}{2}\right)=\frac{I\omega_f^2}{2T}\)
Therefore, the correct answer is (B).
A disk of radius \(0.10\,\mathrm{m}\), initially at rest, undergoes a constant angular acceleration of \(2.0\,\mathrm{rad\,s^{-2}}\). If the disk only rotates, find the total distance traveled by a point on the rim of the disk in \(4.0\,\mathrm{s}\).
Correct Answer: \(\boxed{\mathrm{D}}\)
For constant angular acceleration,
\(\Delta\theta=\omega_0t+\frac{1}{2}\alpha t^2\)
Since the disk starts from rest, \(\omega_0=0\).
\(\Delta\theta=\frac{1}{2}(2.0)(4.0)^2=16\,\mathrm{rad}\)
The distance traveled by a point on the rim is the arc length,
\(s=r\Delta\theta\)
\(s=(0.10)(16)=1.6\,\mathrm{m}\)
Alternatively, the final angular speed is
\(\omega=\omega_0+\alpha t=0+(2.0)(4.0)=8.0\,\mathrm{rad\,s^{-1}}\)
The average angular speed is
\(\omega_{\mathrm{avg}}=\frac{0+8.0}{2}=4.0\,\mathrm{rad\,s^{-1}}\)
Thus,
\(\Delta\theta=\omega_{\mathrm{avg}}t=(4.0)(4.0)=16\,\mathrm{rad}\), giving the same result.
Therefore, the correct answer is (D).
