Question

A rotating wheel requires 7.00 s to rotate 30.0 revolutions. Its angular velocity at the end of the 7.00 s interval is 96.0 rad/s. What is the constant angular acceleration of the wheel? (Do not assume that the wheel starts at rest.)

Answer 1

The angular displacement of the wheel is, 0 = 30 rev 27 rad | lrev =188.49 rad From equation of angular motion the initial angular velocity of the wheel is, 0=0, + at 0 =0)-at =96rad/s - a(7.0s)

Again from equation of angular motion, the angular acceleration of the wheel is, O=0t + at 188.49 rad = (96rad/s-a( 7.0s))(75)+a( 7s)? 188.49 rad = 672rad - ( 49sº)a +(24.5sº)a (49sº)a-(24.5sº)a=672rad – 188.49rad (24.5sº)a = 483.51rad =483.51rad 24.5s? = 19.735rad/s2