Energy of a rolling sphere?

Question

A solid sphere of mass 0.6 kg rolls without slipping along a horizontal surface with a translational speed of 5.22 m/s. It comes to an incline that makes an angle of 33 with the horizontal surface. Neglecting energy losses due to friction,

a) what is the total energy of the rolling sphere?

(b) to what vertical height above the horizontal surface does the sphere rise on the incline?

Answer 1

θ = 33º
M = 0.6 kg
v = 5.22 m/s

Since the sphere is rolling, it has rotational energy. To calculate any energy equations, we need to know the moment of inertia for a sphere.

I = 2/5 M R²


The initial energy in the sphere is purely kinetic energy.

KE = ½ I ω² + ½ m v²


This problem would be undoable without the radius, except...

ω = v / R


KE = ½ I ω² + ½ m v²
KE = ½ I (v / R)² + ½ m v²
KE = ½ (2/5 m R²) v² / R² + ½ m v²
KE = 1/5 m v² + ½ m v²
KE = 7/10 m v²
KE = 7/10 (0.6) (5.22)²
KE = 11.44 J


The height the sphere reaches can be determined by energy calculations. We already know the initial energy is 11.44 J. The final energy is m g h. So..

11.44 = m g h
11.44 = 0.6 (9.8) h

h = 1.95 m

Answer 2

(a)
two components to the energy:

**rolling energy: Kr = .5 * I * w^2
I = moment of inertia
w = omega = angular speed

for a solid sphere, I = 2/5 * m * r^2
and w = v^2 / r^2
where v = translational velocity

**translational energy: Kt = 1/2 * m * v^2

add Kr + Kt = E_total since there' no potential
i guess you need the sphere's radius for that though. do you know the radius? i cant think of a way to do it without the radius (EDIT: good call Mr. Boozer see below)

(b)
potential energy due to gravity Ug = mgh

set mgh = E_total and solve for h to get the vertical height