A 40 lb sphere is released from rest in a circular trough At

A 40 lb sphere is released from rest in a circular trough. At the time of release, the sphere is 6 ft from the bottom of the trough. The sphere and trough radii are 1 ft and 10 ft, respectively. Assuming the sphere does not slip, what is the maximum angular velocity? 19.5 rad/s 11.5 rad/s 17.9 rad/s 15.2 rad/s 19.6 rad/s

Solution

Solution:

Given problem can be solved using engery conservation theorem which states that energy is neither created nor destroyed, it just get converted from one form to another.

At 6 ft from the bottom of trough

sphere is having potiential energy equal to mgh

at bottom all this energy get converted into KE & sphere will possess max KE & velocity

mgh = 0.5mv^2

v^2 = 2 g h

= 2 x 32.2 x 6

v=19.6 ft/s

v = angular velocity x radius

here radius is given as 1 ft

therefore max angular velocity possible is 19.6 rad/s