Question

Two objects of equal mass m are at rest at the top of a hill of height h. Object 1 is a circular hoop of radius r, and object 2 is a solid disc, also of radius r. The object are released from rest and roll without slipping.

A) Provide expressions for the LINEAR VELOCITY of each object once it reaches the bottom of the hill. Careful - you should provide two answers!

B) Considering your results from part A, which object reaches the bottom of the hill first? Provide a BRIEF justification in English for your answer.

C) For EACH object, provide an expression for what percentage of its total energy is rotational after it reaches the bottom of the hill. Careful - you should provide two answers!

Answer 1

let m is the mass of each objec and h is the hight of the hill.

A) for hoop,

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_hoop = sqrt(g*h)

for disk,

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

B) disk recahes the bottom first.

because, v_disk > v_hoop

C) for hoop = KE_rotationa/KE_total

= (1/2)*m*v^2/(m*v^2)

= 50%

for disk = KE_rotationa/KE_total

= (1/4)*m*v^2/((3/4)m*v^2)

= 33.3%