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Conservation of Angular Momentum on a MerryGoRound 
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#1
Nov1611, 12:25 PM

P: 98

1. The problem statement, all variables and given/known data
A merrygoround with a a radius of R = 1.94 m and moment of inertia I = 184 kgm2 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 merrygoround. Once at the merrygoround they jump on and hold on to the rim of the ride. 1)What is the magnitude of the initial angular momentum of the merrygoround? 2)What is the magnitude of the initial angular momentum of the person 2 meters before they jump on the merrygoround? 3)What is the magnitude of the initial angular momentum of the person just before they jump on to the merrygoround? 4)What is the angular speed of the merrygoround after the person jumps on? 5)Once the merrygoround 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 merrygoround to exit the ride. What is the linear velocity of the person right as they leave the merrygoround? 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 merrygoround. 


#2
Nov1611, 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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