A uniformly thick wheel with moment of inertia I and radius R is free to rotate about its centre of

Q: Why do we need to consider the moment of inertia of the pulley?

In ideal problems, pulleys are often assumed to be massless. However, in this problem, the pulley has a specified moment of inertia $I$. This means it has mass and requires energy to rotate. We must account for its rotational kinetic energy $\frac{1}{2}I\omega^2$ in our energy conservation equation.

Q: Why is the net loss in potential energy $(m_1 - m_2)gh$?

As the heavier mass $m_1$ moves down by a distance $h$, it loses potential energy $m_1gh$. Because they are connected by a string, the lighter mass $m_2$ must move up by the same distance $h$, gaining potential energy $m_2gh$. The net change is the loss minus the gain, which is $(m_1 - m_2)gh$.

Q: What does the 'no-slip' condition mean?

The 'no-slip' condition means that the string moves exactly with the rim of the pulley without sliding over it. This allows us to directly relate the linear speed of the string (and the blocks) $v$ to the angular speed of the pulley $\omega$ using the equation $v = \omega R$.

Q: Could we solve this using Newton's laws instead of energy conservation?

Yes, absolutely! You could set up force equations for both blocks ($m_1g - T_1 = m_1a$ and $T_2 - m_2g = m_2a$) and a torque equation for the pulley ($(T_1 - T_2)R = I\alpha$). Then, using $a = \alpha R$, you can solve for the acceleration $a$, and finally use kinematics ($v^2 = u^2 + 2ah$) to find the velocity. However, the energy method is often faster and less prone to algebraic errors.

Q: What if the pulley was a uniform solid disc?

If the pulley was explicitly stated to be a uniform solid disc of mass $M$, its moment of inertia would be $I = \frac{1}{2}MR^2$. You would simply substitute this value into the final expression.

Question

A uniformly thick wheel with moment of inertia $I$ and radius $R$ is free to rotate about its centre of mass (see figure). A massless string is wrapped over its rim and two blocks of masses $m_1$ and $m_2(m_1 > m_2)$ are attached to the ends of the string. The system is released from rest. The angular speed of the wheel when $m_1$ descents by a distance $h$ is

Options:

(A) $\left[ \frac{2(m_1 + m_2)gh}{(m_1 + m_2)R^2 + I} \right]^{1/2}$

(B) $\left[ \frac{m_1 + m_2}{(m_1 + m_2)R^2 + I} gh \right]^{1/2}$

(C) $\left[ \frac{m_1 - m_2}{(m_1 + m_2)R^2 + I} gh \right]^{1/2}$

(D) $\left[ \frac{2(m_1 - m_2)gh}{(m_1 + m_2)R^2 + I} \right]^{1/2}$

Sigma Percentile · Accepted Answer

### The Atwood Machine with a Massive Pulley

Imagine a classic Atwood machine, but with a twist: the pulley isn't just a massless, frictionless idealization. It's a real, physical wheel with a moment of inertia $I$ and a radius $R$. When we release the system from rest, the heavier mass $m_1$ accelerates downwards, pulling the lighter mass $m_2$ upwards and causing the massive pulley to spin.

### The Energy Perspective

To find the angular speed of the wheel after $m_1$ has descended by a distance $h$, the most elegant approach is to use the **Principle of Conservation of Mechanical Energy**.

As the system moves, it loses gravitational potential energy and gains kinetic energy. Let's break this down:

1. **Loss in Potential Energy:** The mass $m_1$ moves down by $h$, losing potential energy $m_1gh$. Simultaneously, $m_2$ moves up by $h$, gaining potential energy $m_2gh$. The net loss in potential energy is the difference between the two:
   $$\Delta U = m_1gh - m_2gh = (m_1 - m_2)gh$$

2. **Gain in Kinetic Energy:** This lost potential energy doesn't just vanish; it transforms into the kinetic energy of the moving parts. The two blocks gain translational kinetic energy, and the pulley gains rotational kinetic energy:
   $$\Delta K = \frac{1}{2}m_1v^2 + \frac{1}{2}m_2v^2 + \frac{1}{2}I\omega^2$$

### The No-Slip Condition

Here's the crucial link: because the string doesn't slip on the pulley, the linear speed $v$ of the string (and thus the blocks) is directly tied to the angular speed $\omega$ of the pulley's rim. This relationship is given by:
$$v = \omega R$$

By substituting this into our kinetic energy equation, we can express the entire kinetic energy in terms of the angular speed $\omega$:
$$\Delta K = \frac{1}{2}m_1(\omega R)^2 + \frac{1}{2}m_2(\omega R)^2 + \frac{1}{2}I\omega^2$$

Factoring out $\frac{1}{2}\omega^2$, we get:
$$\Delta K = \frac{1}{2}\omega^2 \left[ (m_1 + m_2)R^2 + I \right]$$

### The Final Calculation

Now, we simply equate the net loss in potential energy to the total gain in kinetic energy:
$$(m_1 - m_2)gh = \frac{1}{2}\omega^2 \left[ (m_1 + m_2)R^2 + I \right]$$

To find the angular speed $\omega$, we isolate it on one side of the equation:
$$\omega^2 = \frac{2(m_1 - m_2)gh}{(m_1 + m_2)R^2 + I}$$

Taking the square root yields our final, beautiful result:
$$\omega = \left[ \frac{2(m_1 - m_2)gh}{(m_1 + m_2)R^2 + I} \right]^{1/2}$$

This equation perfectly encapsulates the physics of the system: the driving force (gravity acting on the mass difference) is in the numerator, while the total inertia (the mass of the blocks plus the rotational inertia of the pulley) resists the motion in the denominator.

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Visualized Solution

$Visualizing the System$

$Conservation of Mechanical Energy$

$Loss in Potential Energy$

$Gain in Kinetic Energy$

$No-Slip Condition$

$Substituting v = ω R$

$Equating Energies$

$Final Answer$

The Sigma Insight: Work and Energy in Rotational Motion

The Atwood Machine with a Massive Pulley

The Energy Perspective

The No-Slip Condition

The Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the System

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conservation of Mechanical Energy

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Loss in Potential Energy

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Gain in Kinetic Energy

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } No-Slip Condition

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting v=ωR

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equating Energies

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer