Consider a playground merry-go-round with mass 250 kg and radius 1.46 m. Approxi
ID: 1457384 • Letter: C
Question
Consider a playground merry-go-round with mass 250 kg and radius 1.46 m. Approximate the merry-go-round as a solid, flat cylinder rotating about its axis of symmetry; you can look up the moment of inertia for this geometry in any college physics textbook. (E.g., the OpenStax College Physics figure captioned "Some rotational inertias".)
(a) Calculate the rotational kinetic energy in the merry-go-round when it has an angular velocity of 16.5 rpm. In J? (b) Find the number of revolutions you would have to push the merry-go-round at the edge with force 16.5 N to achieve this angular velocity starting from rest.
(c) Calculate the force you would need to exert to stop the merry-go-round in 2 revolutions.
Explanation / Answer
a) w = 16.5 rpm = 16.5*2*pi/60 = 1.727 rad/s
moment of inertia of merry-go-round, I = 0.5*M*R^2
= 0.5*250*1.46^2
= 266.45 kg.m^2
so, KE = 0.5*I*w^2
= 0.5*266.45*1.727^2
= 397.3 J
b) Apply, kinetic energy gained = workdone
= T*theta
397.3 = 16.5*1.46*theta
==> theta = 397.3/(16.5*1.46)
= 16.5 rad
= 2.63 revolutions
c) Apply, KE = F*R*theta
F = KE/(R*theta)
= 397.3/(1.46*2*2*pi)
= 21.65 N
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.