Question

A solid cylinder and a solid sphere both roll without slipping and both have identical masses (6 kg) and identical radii (0.2 m). I start them at the top of a 30 m high hill and roll them down the hill. What is the velocity of each at the bottom of the hill? Which one wins?

Answer 1

for solid cyllinder

let w is the angular speed and v is the linear speed of the solid cyllinder at the bottom.

Apply conservation of energy

final kinetic energy = initial potential energy

(1/2)*m*v^2 + (1/2)*I*w^2 = m*g*h

(1/2)*m*v^2 + (1/2)*(1/2)*m*R^2*w^2 = m*g*h

(1/2)*m*v^2 + (1/4)*m*(R**w)^2 = m*g*h

(1/2)*m*v^2 + (1/4)*m*v^2 = m*g*h

(3/4)*m*v^2 = m*g*h

v_cyiilinder = sqrt(4*g*h/3)

= sqrt(4*9.8*30/3)

= 19.8 m/s

for solid sphere

let w is the angular speed and v is the linear speed of the solid sphere at the bottom.

Apply conservation of energy

final kinetic energy = initial potential energy

(1/2)*m*v^2 + (1/2)*I*w^2 = m*g*h

(1/2)*m*v^2 + (1/2)*(2/5)*m*R^2*w^2 = m*g*h

(1/2)*m*v^2 + (1/5)*m*(R**w)^2 = m*g*h

(1/2)*m*v^2 + (1/5)*m*v^2 = m*g*h

(7/10)*m*v^2 = m*g*h

v_sphere = sqrt(10*g*h/7)

= sqrt(10*9.8*30/7)

= 20.5 m/s

clearly, v_sphere > v_cyllinder

so, spheres wins