Question

Determine the order in which the following objects released from the same height, initially at rest and rolling down an incline without slipping, will reach the bottom: A hollow sphere, a solid sphere, a hollow pipe, and a solid cylinder.

Please explain in simplest terms and give an example using units given.

Answer 1

solid sphere will reach first.

for solid sphere,

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

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

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

for hollow sphere,

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)*(2/3)m*r^2*w^2 = m*g*h

(1/2)*m*v^2 + (1/3)*m*(r*w)^2 = m*g*h

(1/2)*m*v^2 + (1/3)*m*v^2 = m*g*h

(5/6)m*v^2 = m*g*h

v_hollowsphere = sqrt(6*g*h/5)

for hollow pipe,

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)*m*r^2*w^2 = m*g*h

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

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

m*v^2 = m*g*h

v_hollowpipe = sqrt(g*h)

for solid cyllinder,

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)*m*r^2*w^2 = m*g*h

(1/2)*m*v^2 + (1/2)*(1/2)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_solidcyllinder = sqrt(4*g*h/3)

clearly,

v_hollowpipe < v_hollowsphere < v_solidcyllinder < v_soildsphere

so, solid sphere will reach first.