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

$\theta=\omega t-\frac{1}{2}\alpha t^2^{}$

Substitute the known values in the above expression as

$\begin{gathered} 9\times2\pi=29\times2.62-\frac{1}{2}(\alpha)(2.62)^2 \\ \alpha=5.66rad/s^2 \end{gathered}$

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