# Momentum Problem

A girl weighing 60 KG is at the rim of a merry-go-round. The radius of the merry-go-round is 2 M. Her speed on the rim is 2 radians a second. The merry-go-round weighs 100KG.

I of student=MR2 (thats MR squared)
I of merry-go-round=1/2MR2 (thats MR squared also)

What is her speed if she moves to be .5 M from the center of the merry-go-round. I know the answer is 4.09 radian, but I cannot figure out how to get that, thanks a lot for the help

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BobG
Homework Helper
You know how to calculate the moment of inertia. The angular velocity is given. If you know moment of inertia and angular velocity, you can calculate the angular momentum.

Angular momentum stays constant, regardless of where she is on the merry-go-round. If the moment of inertia decreases, the angular velocity has to increase.

It's a question to do with the conservation of angular momentum.

You work out the total angular momentum for the girl + roundabout in the starting situation described.

Now you need a formula that links angular momentum with rotational speed with the girl at her new location.

You then just have to work out the rotational speed for the whole system that gives the same angular momentum as it was at the start.

The situation here is a bit like an ice skater pulling her arms in while pirouetting - as the mass moves closer to the centre, the rotational speed has to increase to maintain the same angular momentum.

Original:
L=1/2(100)(r^2)+(60)(2^2)(2)

After
L=1/2(100)(r^2)+60(.5^2)(w)
Is that correct?

Im partly confused on what goes in for R for the merry go round part

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BobG
Homework Helper
Jacob87411 said:
Original:
L=1/2(100)(r^2)+(60)(2^2)(2)

After
L=1/2(100)(r^2)+60(.5^2)(w)
Is that correct?

Im partly confused on what goes in for R for the merry go round part
Modify it just slightly.

L=[1/2(100)(r^2)+(60)(2^2)](2)

After
L=[1/2(100)(r^2)+60(.5^2)](w)

You sum your moments of inertia to get the system's overall moment of inertia. Angular momentum is the system's moment of inertia times angular velocity.

And the merry-go-round's 'r' is the merry-go-round's radius (2 meters).

1/2(100)(2^2)+(60)(2^2)](2) so that equals its moment of inertia initially and

1/2(100)(r^2)+60(.5^2)](w) is equal to the moment of inertia after?

I get confused again after that, I was thinking setting the initial angular momentum to the final momentum (IW Initial = IW Final), but what would go in for W there?

Jacob87411 said:
Original:
L=1/2(100)(r^2)+(60)(2^2)(2)

After
L=1/2(100)(r^2)+60(.5^2)(w)
Is that correct?

Im partly confused on what goes in for R for the merry go round part
I think you meant for the original to be L = I*w = (1/2(100)(r^2)+(60)(2^2))*(2). Both objects have the same angular velocity.
Also, you will find it more illuminating (and less wasteful when the problems get more complicated) to work in symbolic form for almost all problems, plugging in numbers only when you have the solution in symbolic form. In that respect, let $$I_g$$ be the angular momentum of the girl and let $$I_m$$ be the angular momentum of the merry-go-round. Then we have :
$$L_{before} = (I_m + I_{g before})\omega_{before}$$
which should be equivalent by conservation of angular momentum to
$$L_{after} = (I_m + I_{g after})\omega_{after}$$
The only unknown is $$\omega_{after}$$. Although in this case, nothing cancels out, you do save some ambiguity and paper. :) You can also see that the the moment of inertia of the merry-go-round doesn't change (R is 2 m for both). Note that the angular velocity of the girl is the same as that of the merry-go-round.

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hypermorphism said:
I think you meant for the original to be L = I*w = (1/2(100)(r^2)+(60)(2^2))*(2). Both objects have the same angular velocity.
Also, you will find it more illuminating (and less wasteful when the problems get more complicated) to work in symbolic form for almost all problems, plugging in numbers only when you have the solution in symbolic form. In that respect, let $$I_g$$ be the angular momentum of the girl and let $$I_m$$ be the angular momentum of the merry-go-round. Then we have :
$$L_before = (I_m + I_{g before})\omega_before$$
which should be equivalent by conservation of angular momentum to
$$L_after = (I_m + I_{g after})\omega_after$$
The only unknown is $$\omega_after$$. Although in this case, nothing cancels out, you do save some ambiguity and paper. :) You can also see that the the moment of inertia of the merry-go-round doesn't change (R is 2 m for both).
So does
$$L_before = (I_m + I_{g before})\omega_before$$ = $$L_after = (I_m + I_{g after})\omega_after$$

Jacob87411 said:
So does
$$L_{before} = (I_m + I_{g before})\omega_{before}$$ = $$L_{after} = (I_m + I_{g after})\omega_{after}$$
By conservation of angular momentum in the absence of external torque. :)

Thanks a lot, really appreciate it