Unit 5: Torque and Rotational Dynamics
Weightage : 10-15%
Two blocks of masses \(M\) and \(2M\) are connected by a light string. The string passes over a pulley, as shown above. The pulley has radius \(R\) and moment of inertia \(I\) about its center. The tensions in the string on either side of the pulley are \(T_1\) and \(T_2\), and the blocks accelerate with magnitude \(a\).
Which of the following equations best describes the pulley’s rotational motion while the blocks accelerate?
Correct Answer: \( \boxed{\mathrm{C}} \)
The pulley rotates because the tensions on the two sides of the string are different, producing a net torque about its center.
Applying Newton’s second law for rotation,
\(\tau_{\mathrm{net}}=I\alpha\)
The net torque on the pulley is
\(\tau_{\mathrm{net}}=T_2R-T_1R=(T_2-T_1)R\)
Since the string does not slip on the pulley, the linear acceleration of the blocks and the angular acceleration of the pulley are related by
\(\alpha=\dfrac{a}{R}\)
Substituting into the rotational equation gives
\((T_2-T_1)R=I\alpha=I\dfrac{a}{R}\)
This equation correctly relates the net torque on the pulley to its angular acceleration.
Therefore, the correct answer is (C).
Three equal-mass objects (A, B, and C) are each initially at rest horizontally on a pivot, as shown in the figure.
Object A is a \(40\,\mathrm{cm}\) long, uniform rod, pivoted \(10\,\mathrm{cm}\) from its left edge.
Object B consists of two heavy blocks connected by a very light rod. It is also \(40\,\mathrm{cm}\) long and pivoted \(10\,\mathrm{cm}\) from its left edge.
Object C consists of two heavy blocks connected by a very light rod that is \(50\,\mathrm{cm}\) long and pivoted \(20\,\mathrm{cm}\) from its left edge.
Which of the following correctly ranks the objects’ angular acceleration about the pivot point when they are released?
Correct Answer: \(\boxed{\mathrm{D}}\)
The angular acceleration is determined by
\(\alpha=\frac{\tau_{\mathrm{net}}}{I}\)
where \(\tau_{\mathrm{net}}\) is the net gravitational torque and \(I\) is the moment of inertia about the pivot.
Objects A and B: Both have the same net gravitational torque because their centers of mass are located the same perpendicular distance from the pivot.
For Object A, the weight acts at its center of mass, producing a torque of
\(\tau_A=W(10\,\mathrm{cm})\)
For Object B, the torques from the two masses are
\((10\,\mathrm{N})(30\,\mathrm{cm})-(10\,\mathrm{N})(10\,\mathrm{cm})=200\,\mathrm{N\cdot cm}\)
Thus, \(\tau_A=\tau_B\). However, Object B has a larger moment of inertia because its mass is concentrated farther from the pivot. Therefore,
\(\alpha_A>\alpha_B\)
Object C: The net torque is smaller,
\((10\,\mathrm{N})(30\,\mathrm{cm})-(10\,\mathrm{N})(20\,\mathrm{cm})=100\,\mathrm{N\cdot cm}\)
and its moment of inertia is even larger than that of Object B because the masses are farther from the pivot.
Therefore,
\(\alpha_B>\alpha_C\)
Combining the results,
\(\alpha_A>\alpha_B>\alpha_C\)
Therefore, the correct answer is (D).
A light rigid rod with masses attached to its ends is pivoted about a horizontal axis as shown above. When released from rest in a horizontal orientation, the rod begins to rotate with an angular acceleration of magnitude
Correct Answer: \(\boxed{\mathrm{A}}\)
Apply Newton’s second law for rotation:
\(\sum\tau=I\alpha\)
The clockwise and counterclockwise torques due to gravity are
\(\sum\tau=(3M_0g)(l)-(M_0g)(2l)=M_0gl\)
The total moment of inertia about the pivot is
\(I=(3M_0)l^2+(M_0)(2l)^2=3M_0l^2+4M_0l^2=7M_0l^2\)
Therefore,
\(\alpha=\frac{\sum\tau}{I}=\frac{M_0gl}{7M_0l^2}=\frac{g}{7l}\)
The initial angular acceleration depends on the net gravitational torque and the rotational inertia of the system about the pivot.
Therefore, the correct answer is (A).
