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Conservation of Angular Momentum on a Merry-Go-Round

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mrshappy0
#1
Nov16-11, 12:25 PM
P: 98
1. The problem statement, all variables and given/known data
A merry-go-round with a a radius of R = 1.94 m and moment of inertia I = 184 kg-m2 is spinning with an initial angular speed of ω = 1.66 rad/s. A person with mass m = 76 kg and velocity v = 4.1 runs on a path tangent to the merry-go-round. Once at the merry-go-round they jump on and hold on to the rim of the ride.
1)What is the magnitude of the initial angular momentum of the merry-go-round? 2)What is the magnitude of the initial angular momentum of the person 2 meters before they jump on the merry-go-round? 3)What is the magnitude of the initial angular momentum of the person just before they jump on to the merry-go-round? 4)What is the angular speed of the merry-go-round after the person jumps on? 5)Once the merry-go-round travels at this new angular speed, what force does the person need to hold on? 6)Once the person gets half way around, they decide to simply let go of the merry-go-round to exit the ride.
What is the linear velocity of the person right as they leave the merry-go-round?

2. Relevant equations

L=Iω
L=mvR

3. The attempt at a solution

So I was easily able to answer all the questions up to questions 5, and 6. I don't even know where to start with 5. It seems like it requires more than just angular momentum. #6 doesn't seem bad. My thought is that it would be the same speed regardless of where the person lets go of the ride. So I would just calculate the linear velocity from the angular velocity that i got from the earlier parts. This would be the speed he would exit the merry-go-round.
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Delphi51
#2
Nov16-11, 12:51 PM
HW Helper
P: 3,394
I agree with all you say. In #5, I think you are asked for the centripetal force that must be overcome by hanging on.


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