Question

A wheel accelerates from rest to 34.7 rad/s at a rate of 47.0 rad/s^2. Through what angle (in radians) did the wheel turn while accelerating?

Answer 1

12.8 rad

Explanation:

The angular displacement $\theta$ through which the wheel turned can be determined from the equation below:

$\omega^2 = \omega_0^2 + 2\alpha\theta$ (1)

where

$\omega_0 = 0$

$\omega = 34.7\:\text{rad/s}$

$\alpha = 47.0\:\text{rad/s}^2$

Using these values, we can solve for $\theta$ from Eqn(1) as follows:

$2\alpha\theta = \omega^2 - \omega_0^2$

or

$\theta = \dfrac{\omega^2 - \omega_0^2}{2\alpha}$

$\:\:\:\:= \dfrac{(34.7\:\text{rad/s})^2 - 0}{2(47.0\:\text{rad/s}^2)}$

$\:\:\:\:= 12.8\:\text{rad}$