Question

A string is attached to the rim of a small hoop of radius r= 8.00×10⁻² m and mass m = 0.180 kg and then wrapped several times around the rim. If the free end of the string is held in place and the hoop is released from rest and allowed to drop, as shown in the figure (Figure 1) , calculate the angular speed and the translational speed of the rotating hoop after it has descended h = 0.750 m . Use g = 9.80 m/s2 for the acceleration due to gravity.

What is the translational speed of the rotating hoop after it has descended a distance 0.750 m ?

Express your answer in meters per second

Answer 1

using law of conservation of energy

initial energy = final energy

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

I is the moment of inertia of hoop = m*R^2

w is the angular speed

and v is the translational speed

then

m*g*h = (0.5*m*R^2*w^2)+(0.5*R^2*w^2)

0.18*9.8*0.75 = (0.5*0.18*0.08^2*w^2)+(0.5*0.18*0.08^2*w^2)

w = 33.88 rad/s

v = R*w = 0.08*33.88 = 2.71 m/s