Question

Problem 3A solid uniform sphere of mass 120 kg and radius 1.7 m starts from rest and rolls without slipping down an inclined plane of vertical height 5.3 m. What is the angular speed of the sphere at the bottom of the inclined plane

Answer 1

Answer:

5.1 rad/s

Explanation:

Mechanical energy of the system is conserved since no external work is done on the sphere.

$mgh = mv^2/2 + I\omega^2/2$

Substituting v = ωr and I = 2 m r^2/5, we get,

=> $mgh=m(\omega r)^2/2 + (2\omega r^2/5)\omega^2/2$

=> $mgh = m\omega^2r^2/2 + m\omega^2r^2/5$

=> $gh =\omega^2r^2/2+\omega^2r^2/5$

=>  $gh = 7\omega^2 r^2/10$

=>  $\omega r = (10gh/7)^{1/2}$

=> $\omega = (1/r)(10gh/7)^{1/2} = (1 / 1.7)(10\times 9.8\times 5.3 / 7)^{1/2}$ = 5.1 rad/s