Question 402: Centre of Mass and Rotational Motion

Question

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

Accepted Answer

To find the angular velocity ($\omega$) of the wheel after $10$ seconds, we first need to understand the relationship between the force applied through the string, the torque produced by this force, and how this torque affects the wheel's angular acceleration ($\alpha$).
The torque ($	au$) produced by the force ($F$) is given by the product of the force and the radius ($r$) of the wheel through which the force is applied:
$	au = F \cdot r$
Given that $F = 40 \, \mathrm{N}$ and $r = 10 \, \mathrm{cm} = 0.1 \, \mathrm{m}$, the torque can be calculated as:
$	au = 40 \cdot 0.1 = 4 \, \mathrm{Nm}$
The torque is related to the angular acceleration ($\alpha$) and the moment of inertia ($I$) of the wheel by the equation:
$	au = I \cdot \alpha$
Given that $I = 0.40 \, \mathrm{kg \cdot m}^2$, we can rearrange the above formula to solve for $\alpha$:
$\alpha = \frac{	au}{I} = \frac{4}{0.40} = 10 \, \mathrm{rad/s}^2$
With the angular acceleration ($\alpha$), we can calculate the angular velocity ($\omega$) after a given time ($t$) using the formula:
$\omega = \omega_0 + \alpha \cdot t$
Where $\omega_0$ is the initial angular velocity. Since the wheel starts from rest, $\omega_0 = 0$. Thus, for $t = 10 \, \mathrm{s}$:
$\omega = 0 + 10 \cdot 10 = 100 \, \mathrm{rad/s}$
Therefore, the angular velocity ($\omega$) of the wheel after $10$ seconds is $100 \, \mathrm{rad/s}$, so $x = 100$.

Centre of Mass and Rotational Motion

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