Question

A cylinder of radius 2.59 cm and a spherical shell of radius 6.72 cm are rolling without slipping along the same floor. The two objects have the same mass. If they are to have the same total kinetic energy, what should the ratio of the cylinder\'s angular speed to the spherical shell\'s angular speed be?

Answer 1

Kinetic energy of the cylinder = (1/2)mV² + (1/2)I_Cw²
Where m is the mass of cylinder , V is the linear velocity of cylinder, I_C is the moment of inertia of cylinder and w is the angular speed.
I_c = (mr² /2 ) where r is the radius of the cylinder
We also know that V = wr ,
w = V/r
Kinetic energy of the cylinder = (1/2)mv² + (1/2)(mr²/2)(V/r)² = (1/2)mv² + (1/2)(m/2)(V²) = mV²(1/2 + 1/4)
= (3/4)mV² ----(1)
Now for the Spherical shell
Kinetic energy of the spherical shell = (1/2)mv² + (1/2)I_Sw_s²  
where v is the linear velocity of the spherical shell , w_s is the angular speed of the spherical shell
Moment of inertia of the spherical shell (I_S) = (2/3)mr_s²
Kinetic energy of the spherical shell = (1/2)mv² + (1/2)(2/3)mr_s²(v/r_s)² = mv²(1/2 + 1/3) = (5/6)mv² ---------(2)
It is given that the kinetic energy of both is same
(5/6)mv² = (3/4)mV² , hence
(5/6)m(w_sr_s)² = (3/4)m(wr)²
(5/6)m(w_s*6.72)² = (3/4)m(w*2.59)²
37.632w_s² = 5.03w²
(w/w_s)² = (37.632/5.03)
w/w_s = 2.735