Question

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 ${s^{ - 2}}$.

Answer 1

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.

Complete step by step answer:
As per given in the question initial angular velocity = 0
and number of rotations completed in t seconds = $n$
therefore, total angular displacement in t seconds = $2πn$
Therefore from the 3rd equation of motion in the rotational system.
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= $2\pi n$
Substituting the values, now
$2\pi n$= $\dfrac{1}{2}\alpha {t^2}$
$\therefore \alpha = \dfrac{{4\pi n}}{{{t^2}}}$

Note: The motion of an object along a circular path in a fixed orbit is known as rotational motion. It can also be described as a body's motion around a fixed point in which all of its particles travel in a circular motion with the same angular velocity—for example, Earth's rotation around its axis.