Question

The time it takes for a disk to rotate 45 revolutions is 5.82 s. The angular velocity at the end of the 5.82 s time interval is 68 rad⋅s−1.a) Calculate the constant angular acceleration (in rad⋅s−2) of the disk.

Answer 1

Given data

*The given time is t = 5.82 s

*The given angular velocity is

$\omega=68\text{ rad/s}$

*The number of the revolution is n = 45 revolutions

*The angular distance traveled is

$\theta=2\pi n=2\pi(45)=90\pi\text{ rad}$

(a)

The formula for the constant angular acceleration is given 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

$\begin{gathered} 90\pi=(68)(5.82)-\frac{1}{2}\alpha(5.82)^2 \\ \alpha=6.68rad.s^{-2} \end{gathered}$

Hence, the constant angular acceleration of the disk is 6.68 rad/s^2.