A cord is wound round the circumference of wheel of radius r. The axis of the wheel is horizontal an

Q: Why don't we consider the work done by the tension in the cord?

Tension is an internal force within the wheel-mass system. The positive work done by tension on the wheel is exactly canceled by the negative work done by tension on the falling mass. Thus, it doesn't affect the total mechanical energy.

Q: Why is the initial kinetic energy taken as zero?

The problem states that the weight 'falls from rest'. This implies that both the initial linear velocity of the mass and the initial angular velocity of the wheel are zero.

Q: Can we solve this using Newton's laws and torque instead of energy?

Absolutely! You can write $mg - T = ma$ for the mass and $Tr = I\alpha$ for the wheel. Using $a = r\alpha$, you can find the acceleration $a$, and then use kinematics ($v^2 = u^2 + 2ah$) to find the final velocity. However, the energy method is much faster and less prone to algebraic errors.

Q: What does the no-slip condition $v = r\omega$ physically mean here?

It means the cord unwinds from the wheel without sliding over its surface. For every meter the mass falls, exactly one meter of cord unwinds from the circumference of the wheel.

Q: Does the moment of inertia $I$ depend on the mass of the wheel?

Yes, the moment of inertia $I$ is a property that depends on both the mass of the wheel and how that mass is distributed relative to the axis of rotation. The problem gives it as a general variable $I$ to keep the result universally applicable.

Question

A cord is wound round the circumference of wheel of radius $r$. The axis of the wheel is horizontal and the moment of inertia about it is $I$. A weight $mg$ is attached to the cord at the end. The weight falls from rest. After falling through a distance $h$, the square of angular velocity of wheel will be

Options:

(A) $\frac{2mgh}{I + 2mr^2}$

(B) $\frac{2mgh}{I + mr^2}$

(C) $2gh$

(D) $\frac{2gh}{I + mr^2}$

Sigma Percentile · Accepted Answer

### Understanding the Setup

Imagine a classic physics scenario: a wheel of radius $r$ and moment of inertia $I$ is mounted on a horizontal, frictionless axis. A cord is tightly wound around its circumference, and a mass $m$ hangs from the free end of this cord.

When the system is released from rest, gravity pulls the mass downwards. As the mass falls, it pulls on the cord, causing the wheel to spin. Our goal is to find the square of the angular velocity ($\omega^2$) of the wheel after the mass has fallen through a vertical distance $h$.

### The Power of Energy Conservation

While we could solve this problem using Newton's laws of motion and torque equations, the principle of conservation of mechanical energy offers a much more elegant and direct path.

Since there are no non-conservative forces doing work on the system (tension is an internal force), the total mechanical energy is conserved. The loss in gravitational potential energy of the falling mass is entirely converted into the kinetic energy of the system.

$$\Delta PE = \Delta KE_{translational} + \Delta KE_{rotational}$$

The loss in potential energy as the mass falls a distance $h$ is $mgh$. This energy is distributed into two forms:
1. The translational kinetic energy of the falling mass: $\frac{1}{2}mv^2$
2. The rotational kinetic energy of the spinning wheel: $\frac{1}{2}I\omega^2$

Equating the energy loss to the energy gain, we get our master equation:

$$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$$

### Connecting Linear and Rotational Motion

Our master equation contains two unknowns: the linear velocity $v$ and the angular velocity $\omega$. To solve for $\omega$, we need to express $v$ in terms of $\omega$.

Because the cord is wound around the wheel and does not slip, the linear speed of the cord (and thus the falling mass) must equal the tangential speed of the rim of the wheel. This gives us the crucial kinematic constraint:

$$v = r\omega$$

### The Final Algebraic Sprint

Now, we substitute this constraint back into our energy equation:

$$mgh = \frac{1}{2}m(r\omega)^2 + \frac{1}{2}I\omega^2$$

Expanding the squared term, we get:

$$mgh = \frac{1}{2}mr^2\omega^2 + \frac{1}{2}I\omega^2$$

Notice that both terms on the right side share a common factor of $\frac{1}{2}\omega^2$. Let's factor it out to isolate $\omega$:

$$mgh = \frac{1}{2}(I + mr^2)\omega^2$$

Finally, we rearrange the equation to solve for $\omega^2$. We multiply both sides by 2 and divide by the bracketed term $(I + mr^2)$:

$$\omega^2 = \frac{2mgh}{I + mr^2}$$

And there we have it! The square of the angular velocity is elegantly expressed in terms of the given parameters.

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Visualized Solution

$S y s t e m S e t u p$

$C o n ser v a t i o n o f E n er g y$

$E n er g y E q u a t i o n$

$K in e ma t i c C o n s t r ain t$

$S u b s t i t u t i o n$

$F a c t or in g$

$F ina l E x p r ess i o n$

$T h e W a y F or w a r d$

The Sigma Insight: Work and Energy in Rotational Motion

Understanding the Setup

The Power of Energy Conservation

Connecting Linear and Rotational Motion

The Final Algebraic Sprint

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

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