A wheel is subjected to uniform angular acceleration about its axis. Initially, its angular velocity is zero.

Question

A wheel is subjected to uniform angular acceleration about its axis. Initially, its angular velocity is zero. In the first $2	ext{ s}$, it rotates through an angle $	heta_1$. In the next $2	ext{ s}$, it rotates through an additional angle $	heta_2$. The ratio of $\frac{	heta_2}{	heta_1}$ is:

Accepted Answer

Let the uniform angular acceleration be $\alpha$.
Given initial angular velocity, $\omega_0 = 0$.
Angle rotated in the first $2	ext{ s}$ is $	heta_1 = \omega_0 t + \frac{1}{2}\alpha t^2 = 0 + \frac{1}{2}\alpha (2)^2 = 2\alpha$.
Angle rotated in the first $4	ext{ s}$ (i.e., $2	ext{ s} + 2	ext{ s}$) is $	heta_{total} = \omega_0 t' + \frac{1}{2}\alpha t'^2 = 0 + \frac{1}{2}\alpha (4)^2 = 8\alpha$.
The additional angle rotated in the next $2	ext{ s}$ is $	heta_2 = 	heta_{total} - 	heta_1 = 8\alpha - 2\alpha = 6\alpha$.
Therefore, the ratio $\frac{	heta_2}{	heta_1} = \frac{6\alpha}{2\alpha} = 3$.

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