A string is wrapped around the rim of a wheel of moment of inertia… | JEE Main 2024 (9 Apr, Shift 1)

Q24JEE Main 2024 · 9 APR · SHIFT 1Medium

Rotational Kinetic Energy, Work & Power

A string is wrapped around the rim of a wheel of moment of inertia $0.40\text{ kg m}^{2}$ and radius $10\text{ cm}$ . The wheel is free to rotate about its axis. Initially the wheel is at rest. The string is now pulled by a force of $40\text{ N}$ . The angular velocity of the wheel after $10\text{ s}$ is $x\text{ rad/s}$ , where $x$ is __________.

Numerical answer type. Work it out, then open the answer.

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Answer: 100

Torque $\tau = FR = 40 \times 0.1 = 4$ N m
$\alpha = \frac{\tau}{I} = \frac{4}{0.40} = 10$ rad/s $^2$
From rest: $\omega = \alpha t = 10 \times 10 = 100$ rad/s
So $x = 100$ rad/s.

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