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# A constant torque is acting on a wheel. If starting from rest, the wheel makes a rotation in t seconds, show that the angular acceleration is given by $\alpha = \dfrac{{4\pi n}}{{{t^2}}}$ rad ${s^{ - 2}}$.

Last updated date: 20th Jul 2024
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Hint: To answer this question, we first need to know the rotational motion.The motion of an object along a circular path in a fixed orbit is known as rotational motion. The dynamics of rotational motion are somewhat similar to those of linear or translational motion. Many of the equations for rotating object dynamics are similar to the linear motion equations.

and number of rotations completed in t seconds = $n$
therefore, total angular displacement in t seconds = $2πn$
Total angular displacement up to time t = ${w_0}t + \dfrac{1}{2}\alpha {t^2}$.....(here ${w_0}$ is the initial rotational velocity and $\alpha$ is the angular acceleration)
Here ${w_0}$=0 and angular displacement= $2\pi n$
$2\pi n$= $\dfrac{1}{2}\alpha {t^2}$
$\therefore \alpha = \dfrac{{4\pi n}}{{{t^2}}}$