Unit 5: Torque and Rotational Dynamics
Weightage : 10-15%
A mechanical wheel initially at rest on the floor begins rolling forward with an angular acceleration of \(2\,\mathrm{rad\,s^{-2}}\). If the radius of the wheel is \(0.5\,\mathrm{m}\), what is the linear velocity of the wheel after \(5\,\mathrm{s}\)?
Correct Answer: \(\boxed{\mathrm{C}}\)
The wheel starts from rest, so \(\omega_0=0\).
Using the rotational kinematics equation,
\(\omega=\omega_0+\alpha t\)
\(\omega=0+(2)(5)=10\,\mathrm{rad\,s^{-1}}\)
Since the wheel rolls without slipping,
\(v=r\omega\)
\(v=(0.5)(10)=5\,\mathrm{m\,s^{-1}}\)
Thus, after \(5\,\mathrm{s}\), the center of the wheel moves with a linear velocity of
\(\boxed{5\,\mathrm{m\,s^{-1}}}\)
Therefore, the correct answer is (C).
A solid cylinder of mass \(m\) and radius \(R\) has a string wound around it. A person holding the string pulls it vertically upward, as shown above, such that the cylinder is suspended in midair for a brief time interval \(\Delta t\) and its center of mass does not move. The tension in the string is \(T\), and the rotational inertia of the cylinder about its axis is \(\frac{1}{2}mR^{2}\).
The linear acceleration of the person’s hand during the time interval \(\Delta t\) is
Correct Answer: \(\boxed{\mathrm{B}}\)
Since the cylinder’s center of mass remains stationary,
\(\sum F=0\)
Therefore,
\(T=mg\)
Applying Newton’s second law for rotation,
\(\sum\tau=I\alpha\)
\(TR=\frac{1}{2}mR^{2}\alpha\)
Substituting \(T=mg\),
\(mgR=\frac{1}{2}mR^{2}\alpha\)
\(\alpha=\frac{2g}{R}\)
Because the string does not slip on the cylinder,
\(a=\alpha R\)
\(a=\frac{2g}{R}\times R=2g\)
The person’s hand moves with the same linear acceleration as the string, so its acceleration is \(2g\) upward.
Therefore, the correct answer is (B).
A particle is moving in a circle of radius \(2\,\mathrm{m}\) according to the relation \( \theta = 3t^{2}+2t \), where \( \theta \) is measured in radians and \( t \) in seconds.
The speed of the particle at \(t=4\,\mathrm{s}\) is
Correct Answer: \( \boxed{\mathrm{D}} \)
The angular velocity is the time derivative of angular position:
\( \omega=\frac{d\theta}{dt}=6t+2 \)
At \(t=4\,\mathrm{s}\),
\( \omega=6(4)+2=26\,\mathrm{rad/s} \)
The linear speed is related to angular velocity by
\( v=r\omega \)
Therefore,
\( v=(2)(26)=52\,\mathrm{m/s} \)
Therefore, the correct answer is (D).
