Question

Starting from rest, a basketball rolls from the top to the bottom of a hill, reaching a translational speed of 5.4 m/s. Ignore frictional losses. (a) What is the height of the hill? (b) Released from rest at the same height, a can of frozen juice rolls to the bottom of the same hill. What is the translational speed of the frozen juice can when it reaches the bottom?

Answer 1

(a)

kinetic energy of basket ball

K = (1/2)*I*w^2

I = moment of inertia = (2/3)*M*r^2

w(omega) angular speed = v/r

kinetic energy K = (1/2)*(2/3)*M*r^2*(v/r)^2

K = (1/3)*M*v^2

from energy conservation

total energy at he top = KE at the bottom

PE = KE

M*g*h = (1/3)*M*v^2

h = V^2/(3g)

h = 5.4^2/(3*9.81)

h = 1 .0 m

=================

(b)

potential energy E at the top = M*g*h

KE at the bottom KE = (1/2)*I*w^2

I = (1/2)*M*r^2

w = (v/r)

KE = (1/2)*(1/2)*M*r^2(v/r)^2 = (1/4)*M*v^2

from energy conservation

KE = PE

(1/4)*M*V^2 = M*g*h

V = sqrt(4*g*h)

V = sqrt(4*9.81*1)

V = 6.26 m/s