Unit 5: Torque and Rotational Dynamics
Weightage : 10-15%
A wheel of radius \(R\) and negligible mass is mounted on a horizontal frictionless axle so that the wheel is in a vertical plane. Three small objects having masses \(m\), \(M\), and \(2M\), respectively, are mounted on the rim of the wheel, as shown. If the system is in static equilibrium, what is the value of \(m\) in terms of \(M\)?
Correct Answer: \( \boxed{\mathrm{C}} \)
Since the wheel is in static equilibrium, the net torque about the axle must be zero:
\( \sum \tau = 0 \)
Take counterclockwise torques as positive.
The mass \(m\) is located at the leftmost point of the wheel, producing a counterclockwise torque of
\( \tau_m = mgR \)
The mass \(M\) is located \(60^\circ\) above the horizontal. Its lever arm is the horizontal distance from the center:
\( R\cos 60^\circ \)
Therefore,
\( \tau_M = Mg(R\cos 60^\circ) \)
The mass \(2M\) at the right side produces a clockwise torque:
\( \tau_{2M} = 2MgR \)
Applying rotational equilibrium,
\( mgR + Mg(R\cos 60^\circ) – 2MgR = 0 \)
Using \( \cos 60^\circ = \frac{1}{2} \),
\( mgR + \frac{1}{2}MgR – 2MgR = 0 \)
Dividing by \(gR\),
\( m + \frac{1}{2}M – 2M = 0 \)
\( m = \frac{3}{2}M \)
Therefore,
\( \boxed{m=\frac{3M}{2}} \)
Hence, the correct answer is (C).
A friend is balancing a fork on one finger. Which of the following are correct explanations of how he accomplishes this? Select two answers.
Correct Answer: \( \boxed{\mathrm{C\ and\ D}} \)
For an object to balance, it must be in rotational equilibrium.
The condition for rotational equilibrium is
\( \sum \tau = 0 \)
This means the clockwise torque must be equal in magnitude to the counterclockwise torque.
Therefore, choice (D) is correct.
An object balances when its center of mass lies directly above the support point (the finger). In that position, the weight acts through the pivot, producing no net torque.
Thus, choice (C) is also correct.
Choice (A) is not the reason the fork balances. Energy conservation does not determine whether an object is in rotational equilibrium.
Choice (B) is incorrect because the fork has a nonzero moment of inertia. Any extended object with mass distributed away from an axis has rotational inertia.
Since the center of mass is above the finger and the net torque is zero, the fork remains balanced.
Hence, the correct answers are (C) and (D).
A \(5\)-meter uniform plank of mass \(100\ \mathrm{kg}\) rests on the top of a building with \(2\ \mathrm{m}\) extended over the edge as shown. How far can a \(50\ \mathrm{kg}\) person venture past the edge of the building on the plank before the plank just begins to tip?
Correct Answer: \( \boxed{\mathrm{B}} \)
When the plank is just about to tip, it is in rotational equilibrium about the edge of the building. Therefore, the net torque about the edge must be zero.
Choose the edge of the building as the pivot point.
The plank is \(5\ \mathrm{m}\) long, so its center of mass is at the midpoint, \(2.5\ \mathrm{m}\) from either end.
Since \(2\ \mathrm{m}\) of the plank extends beyond the edge, the center of mass of the plank is
\( 2.5 – 2.0 = 0.5\ \mathrm{m} \)
to the left of the pivot.
Applying rotational equilibrium:
\( \sum \tau = 0 \)
Counterclockwise torque from the plank’s weight equals clockwise torque from the person’s weight:
\( (100\,\mathrm{kg})g(0.5\,\mathrm{m}) = (50\,\mathrm{kg})g(r) \)
Canceling \(g\),
\( (100)(0.5) = 50r \)
\( 50 = 50r \)
\( r = 1.0\ \mathrm{m} \)
Therefore, the person can move \(1.0\ \mathrm{m}\) past the edge before the plank begins to tip.
Hence, the correct answer is (B).
