Rotational Kinetic Energy of ride

[mathmuncher]

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.

 

[mjsd]

note you have not used the fact that it is asking you what happen after 3.0s. it said a constant force is applied tangentially so can't equate 50 = mv^2/r either... my guess from the wording of the question is that the merry go round is under constant acceleration.. you should work out the angular acceleration and torque... $$\tau= I\alpha$$ where, I, here is moment of inertial of object.

 

[robb_]

Why do you state "combined objects?"
The 50 N is applied tangentially.

 

[mathmuncher]

Thanks for the help. I have made a fundamental error here and I should have known better. I have reached the solution by using the net torque identity, which equals to I*angular acceleration.

t=Ia

Then since t=Fd

Fd = Ia
(50)(1.5) = (.5*80*1.5^2)(a)

a = 0.83 rad/s^2

Since a = (w_f - w_i)/t

0.83 = w_f/3

w_f = 2.49

Now we can easily conclude KE_f:

KE_f = (.5)(90)(2.49)^2
= 280 J