A string is wrapped along the rim of a wheel of the moment of inertia $0.10 ext{ kg-m}^2$ and radius $10 e

Question

A string is wrapped along the rim of a wheel of the moment of inertia $0.10 	ext{ kg-m}^2$ and radius $10 	ext{ cm}$. If the string is now pulled by a force of $10 	ext{ N}$, then the wheel starts to rotate about its axis from rest. The angular velocity of the wheel after $2 	ext{ s}$ will be:

Accepted Answer

Given:
Moment of inertia, $I = 0.10 	ext{ kg-m}^2$
Radius, $R = 10 	ext{ cm} = 0.1 	ext{ m}$
Force, $F = 10 	ext{ N}$
Time, $t = 2 	ext{ s}$
Initial angular velocity, $\omega_0 = 0$ (since it starts from rest)

The torque acting on the wheel is given by:
$	au = F 	imes R = 10 	imes 0.1 = 1 	ext{ N m}$

Using Newton's second law for rotational motion, $	au = I\alpha$, we can find the angular acceleration $\alpha$:
$\alpha = \frac{	au}{I} = \frac{1}{0.10} = 10 	ext{ rad/s}^2$

Now, using the first equation of rotational kinematics to find the final angular velocity $\omega$:
$\omega = \omega_0 + \alpha t$
$\omega = 0 + 10 	imes 2 = 20 	ext{ rad/s}$

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