loss of potential energy = m g h = 3*9.81* 7 = 206 Joules
= translational Ke + rotational Ke
= (1/2) m v^2 + (1/2)I omega^2
Note
I = (2/5) m r^2
v = omega r
so
206=(1/2)m v^2 + (1/2)(2/5)m r^2 v^2
412 = 1.4 *3 v^2
v = 9.9 m/s
then calculate (1/2) (2/5)(3) v^2
A solid ball of mass 3 kg, rolls down a hill that is 7 meters high. What is the rotational KE at the bottom of the hill?
1 answer
1 answer