Question

A solid disk with a radius of 33.3 cm is moving so that its center of mass is initially moving at 4 m/s while also rolling without slipping at 12rad/s along a horizontal surface. It rolls up an incline, coming to rest as shown before rolling back down (drawing is not to scale). Consider the disk at the bottom of the hill. What is the ratio of the rotational kinetic energy to the total kinetic energy of the disk, Krot/Ktotal?A. 1/2B. 1/3C. 1D. 1/4

Answer 1

Given

solid disc with mass m and radius is R ,whose moment of inertia about the center of mass is I = m*R^2/2

and rotational k.e = 0.5*I*W^2

and translational k.e = 0.5*m*v^2

relation between v,W ,r in rolling without slipping is V = r*W

so the total k.e = k.e_r + k.e_t

k.e = 0.5*I*W^2+ 0.5*m*V^2

k.e = 0.5*0.5*m*R^2*v^2/R^2 + 0.5*m*v^2 = (3/4)m*v^2

the rotational k.e =0.5*0.5*m*R^2*v^2/R^2 = 1/4*m*v^2

now the ratio is (1/4*m*v^2) /(3/4)m*v^2 = 1/3

so the answer is option C 1/3