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

Step-by-step 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 = (\omega^2 - \omega_0^2)/(2\alpha)`

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

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