Question

A cylinder of radius 4.09 cm and a spherical shell of radius 7.47 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

The expression for the total kinetic energy of each body

T_sphere = 1/2 Is ws^2 + 1/2 m Vs^2

T_cylinder = 1/2 Ic wc^2 + 1/2 m Vc^2

Here, ws,c are the angular velocities of the sphere,cylinder and Vs,c are the velocities of the center of mass of sphere,cylinder, and Is,c is the moment of inertia of each body. Masses are the same for both.

Now, the moments of inertia of each body is given as -  

Is = (2/3) m Rs^2

Ic = (1/2) m Rc^2

For the motion, rolling without slipping -

Rw = V.

Making Ts=Tc and substituting the moments of inertia and the expression for rolling motion we get

1/2 ( 2/3 m Rs^2) ws^2 + 1/2 m (Rs ws)^2 = 1/2 (1/2 m Rc^2) wc^2 + 1/2 m (Rc wc) ^2

the equation above reduces to

(1/3+ 1/2) (Rs ws)^2 = (1/4 + 1/2) (Rc wc) ^2

thus, the ratio ws/wc is just

ws/wc = Rc/Rs sqrt ((1/4 + 1/2)/(1/3+1/2)) = (Rc / Rs) * sqrt((0.25+0.50) / (0.33+0.50))

= (Rc / Rs)* sqrt(0.75 / 0.83)

= (4.09 / 7.47) * 0.95 = 0.52

Therefore -

Cylinder's angular speed / Spherical shell's angular speed = 0.52 (Answer)