The smooth bar rotates in the horizontal plane with constant

The smooth bar rotates in the horizontal plane with constant angular velocity omega _0 = 60 rpm (revolutions per minute). If the 2-1b collar A is released at r = 1 ft with no radial velocity, what is the magnitude of its velocity when it reaches the end of the bar?

Solution

calculate the angular velocitty.

w = 60*2*pie/60

w = 6.28rad/s

calculate the magnitude of the velocity.
a = v*dv/dr = (w)^2*r
V^2/2 = (w)^2*r^2/2
V^2 = 6.28^2 *(4-1)
V=10.88ft/s