Moment of Inertia of a Changing System

[jcd2012]

Homework Statement

A person grabs on to an already spinning merry-go-round. The person is initially at rest and has a mass of 29.5 kg. They grab and cling to a bar that is 1.70 m from the center of the merry-go-round, causing the angular velocity of the merry-go-round to abruptly drop from 45.0 rpm to 19.0 rpm. What is the moment of inertia of the merry-go-round with respect to its central axis?

Homework Equations

Angular Momentum = I * ω

Momentum is conserved, so L(initial) = L(final), and so, I(initial) * ω(initial) = I(final) * ω(final)

Inertia = MR^2

The Attempt at a Solution

Inertia of the system changes when the person grabs on. So I(final) = I(merry) + I(person)

The inertia of the person is given by MR^2 = 29.5 kg * 1.7^2 = 85.255.

Using the conservation of momentum, here is what I put together:

L(final) = L(person, final) + L(mgr, final) = I(merry) * ω(final) + I(person) * ω(final)
L(final) = I(merry) * 19.0rpm + (82.255) * (19.0)

Solved for L to get L = 60.1094. If L is conserved then that should be the inertia of the merry-go-round. So I solved 60.1094 = I * 45.0rpm for I, which came out to be I = 1.33576. kgm^2. I am not sure where I could have gone wrong with this.

 

[TSny]

Using the conservation of momentum, here is what I put together:

L(final) = L(person, final) + L(mgr, final) = I(merry) * ω(final) + I(person) * ω(final)
L(final) = I(merry) * 19.0rpm + (82.255) * (19.0)

Solved for L to get L = 60.1094.

Hello, jcd2012. Can you show how you solved for L?

 

[jcd2012]

Hello, jcd2012. Can you show how you solved for L?

Forgot that . Here:

If L is conserved, then L(initial) = L(final), so L(final) = I(merry) * ω(initial).
So

I(merry) * ω(initial) = I(merry) * ω(final) + I(person) * ω(final). Solving for I(merry), shortened to I:

I * 45rpm = I * 19.0 + 82.255 * 19.0

45I = 19I + 1562.845
27I = 1562.845
I = 60.1094

 

[TSny]

That looks good.

 

[jcd2012]

That looks good.

It says it is not the right answer though. That is why I am confused.

 

[TSny]

Oh, it looks like you used 82.255 instead of 85.255 when doing the calculation.

Also, note that rpm is not the SI unit for angular velocity. However, in this case it's ok to use rpm since each term in the angular momentum conservation equation has a factor of ω and the conversion factor from rpm to rad/s will cancel out. But be careful in other problems where you might need to convert to rad/s.