- #1

- 12

- 0

## Homework Statement

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.

## Homework Equations

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)

= (.5)(101.25*(0.86)^2)

= 43.5 J

Haha, awfully messy.

Last edited: