Rotational Energy and Momentum

Question

A spherical shell of mass M and radius R is released from rest and rolls without slipping down an inclined plane with a height of 3 m.

I am very confused on how to solve this when I'm only given the height

a) Calculate the speed of the shell's center of mass at the bottom of the incline.

b) Calculate the ratio of center of mass kinetic energy to rotational kinetic energy at the bottom of the incline.

Answer 1

a)

At the bottom of the incline all of the potential energy is converted into kinetic energy.

Conservation of Energy

Potential energy (PE) = Mgh

Kinetic energy (KE) comes in two forms

Center of mass kinetic energy (KE_cm): KE_cm = ½Mv² (also called translational kinetic energy)

Rotational kinetic energy: KE_rot = ½Iω²

So ½Mv² + ½Iω² = Mgh

Velocity Equation

v = ωR which means in the equation above we can substitute ω² = v²/R² to obtain

½Mv² + ½Iv²/R² = ½v²(M + I/R²) = Mgh which means v² = 2Mgh/(M + I/R²)

Which gives us our velocity equation

Moment of Inertia

The moment of inertia of a spherical shell is I = ⅔MR²

So I/R² = ⅔M in our velocity equation and the denominator becomes

(1 + ⅔)M = (5/3)M giving

6gh/5 = 6(9.8)(3)/5 = 35.28, so

v = √(35.28) = 5.9 m/s

b)

KE_cm = ½Mv²

and KE_rot = ½Iω²

substituting ω² = v²/R² and I = ⅔MR² we get

KE_rot = ½(⅔MR²)( v²/R²) = ⅓Mv²

So the ratio of center of mass kinetic energy to

rotational kinetic energy at the bottom of the incline

is given by

KE_cm/KE_rot = (½Mv²)/(⅓Mv²) = (½)/(⅓) = 3/2