Question

Two objects roll down a hill: a hoop and a solid cylinder. The hill has an elevation change of 1.4-m and each object has the same diameter (0.55-m) and mass. Calculate the velocity of each object at the bottom of the hill and rank them according to their speeds.

[Hint: When an object is rolling, the angular speed and the velocity of the center of mass are related by , where is the radius of the object.]

Veyi= 4.3 m/s Vhoop= 3.7 m/s

Answer 1

here,

for a solid cyclinder

let the speed at bottom be v0

using conservation of energy

0.5 * m * v0^2 + 0.5 * I * w0^2 = m * g * h

0.5 * m * v0^2 + 0.5 * (0.5 * m * r^2) * (v0/r)^2 = m * g * h

0.75 * v0^2 = g * h

v0 = sqrt(9.81 * 1.4 /0.75) = 4.3 m/s

height , h = 1.4 m

for a Hoop

let the speed at bottom be v

using conservation of energy

0.5 * m * v^2 + 0.5 * I * w^2 = m * g * h

0.5 * m * v^2 + 0.5 * ( m * r^2) * (v/r)^2 = m * g * h

v^2 = g * h

v = sqrt(9.81 * 1.4) = 3.75 m/s