mgh=rotational energy + translational energy
I assume the 20 is an angle
mg(1.90 sin20)=1/2 I w^2+1/2 mv^2
but v=wr so
= 1/2 I w^2+1/2 m w^2 r^2
so calculate w.
then you have the two terms, and the total energy, solve for the fraction.
An 9.00-cm-diameter, 400 sphere is released from rest at the top of a 1.90-m-long, 20.0 incline. It rolls, without slipping, to the bottom.
A) What is the sphere's angular velocity at the bottom of the incline?
B) WWhat fraction of its kinetic energy is rotational?
1 answer