Question

At the end of a 2.62 s time interval, a wheel has rotated 9 revolutions and has an angularvelocity of 29 rad•s^-1A) Calculate the constant angular acceleration (in rad•s^-2) of the wheel.

Answer 1

Given data

*The given time interval is t = 2.62 s

*The given angular velocity is

`\omega=29\text{ rad/s}`

*The given number of revolutions is 9 revolutions.

The formula for the constant angular acceleration is calculated by the rotational equation of motion as

`\theta=\omega t-(1)/(2)\alpha t^2^{}`

Substitute the known values in the above expression as

`\begin{gathered} 9*2\pi=29*2.62-(1)/(2)(\alpha)(2.62)^2 \\ \alpha=5.66rad/s^2 \end{gathered}`

Hence, the constant angular acceleration is 5.66 rad/s^2