Question

Rolling Sphere in a Hemisphere

Must answer both parts

A small sphere, with radius 2.9 cm and mass 6.0 kg, rolls without slipping on the inside of a large fixed hemisphere with radius 1.00 m and a vertical axis of symmetry. It starts at the top from rest. What is its kinetic energy of the sphere at the bottom?

What fraction of its kinetic energy at the bottom (in percent) is associated with rotation about an axis through its center of mass?

Answer 1

A) We know that, Total energy of the sphere initially= potential energy,U of sphere
So, U= mgh m: mass of sphere, h= Radius of hemisphere
Now , Kinetic energy at bottom= Potential energy at the top=mgh
So,Kinetic energy at bottom = 6*9.81*1= 58.86J

B) The kinetic energy associated with rotation is given as:
KE1 = (1/2) I w², where w : rotational velocity.
Now,since the sphere rolls without slipping,
So,the translational velocity of its cener of mass is simply:
v = w*r
Substituting this into the expression for the rotational kinetic energy:
KE1 = (1/2) (I / r²) v²

Also,the expression for the translational kinetic energy is:
KE2 = (1/2) m v²
Thus, the total kinetic energy of the ball at the base of the hemisphere is;
KE = KE1 + KE2 = (v²/2)( I/r² + m)
The fraction of the rotational kinetic energy over the total is:

f = KE1 / KE = (v²/2)( I/r²) / (v²/2)( I/r² + m)

f = ( I/r²)/( I/r² + m)
The moment of inertia, I, of a sphere is:
I = (2/5)mr²
Substituting this into the expression for f:
f = (2/5)m/( (2/5)m + m)
So,f = (2/5)/( (2/5) + 1)
So,f = 2/7
Hense, this is the required fraction.