A horizontal 800.0 N merry-go-round with a radius
of 1.5 m is started from rest by a constant horizontal
force of 50.0 N applied tangentially to the merry-go-
round. Find the kinetic energy of the merry-go-round
after 3.0 s. Assume it is a solid cylinder.
KE = (1/2)(Iw^2)
The Attempt at a Solution
Answer in the back is 280 J
My answer, however, turns out to be 40ish.
Here's how I did it:
I found moment of inertia I of combined objects (I treated the forces as objects here):
I = I_(mgr) + I_(object_at_edge)
I = (.5MR^2) + (MR^2)
I = (.5*80*1.5^2) + (5*1.5^2)
I = 101.25 kg.m^2
Then, I chose the 50 N constant force as the force that maintains circular motion, so
50 = (mv^2)/r
50 = ((I/r^2)*v^2)/r
50 = ((101.25/1.5^2)*v^2)/1.5
v^2 = 1.667
v = 1.29
Now, KE = (.5)(Iw^2)
= 43.5 J
Haha, awfully messy.