A merry-go-round with a a radius of R = 1.91 m and moment of inertia I = 210 kg-
ID: 1620785 • Letter: A
Question
A merry-go-round with a a radius of R = 1.91 m and moment of inertia I = 210 kg-m2 is spinning with an initial angular speed of = 1.65 rad/s in the counter clockwise direection when viewed from above. A person with mass m = 74 kg and velocity v = 4.9 m/s runs on a path tangent to the merry-go-round. Once at the merry-go-round the person jumps on and holds on to the rim of the merry-go-round.
1)
What is the magnitude of the initial angular momentum of the merry-go-round?
kg-m2/s
2)
What is the magnitude of the angular momentum of the person 2 meters before she jumps on the merry-go-round?
kg-m2/s
3)
What is the magnitude of the angular momentum of the person just before she jumps on to the merry-go-round?
kg-m2/s
4)
What is the angular speed of the merry-go-round after the person jumps on?
rad/s
5)
Once the merry-go-round travels at this new angular speed, with what force does the person need to hold on?
N
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 magnitude of the linear velocity of the person right as they leave the merry-go-round?
m/s
7)
What is the angular speed of the merry-go-round after the person lets go?
rad/s
Explanation / Answer
1)
Merry-go-round has moment of inertia I = 210 kg-m2
its initial angular speed is = 1.65 rad/s,
So initial angula momentum is,
L = I = (210 kg-m2)(1.65 rad/s)
or, L = 346.5 kg-m/s2
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2)
The magnitude of the angular momentum of the person is,
L = mvR = (74 kg)(4.9 m/s)(1.91 m)
or, L = 692.566 kg-m/s2
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3)
The magnitude of angular momentum before the person jumps on the merry-go-round will be same as the magnitude of the angular momentum of the person when she is 2 meters away from the point when she jumps on the merry-go-round.
As long as the person remains on the tangent line, her angular momentum about the center of the merry-go-round will remain the same, as her perpendicular distance from the center will remain equal to R as long as she remains on the tangent line.
So, the magnitude of angular momentum before the person jumps on the merry-go-round will be,
L = 692.566 kg-m/s2
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4)
Initial angular momentum of the merry-go-round+person system is,
Li = 346.5 kg-m/s2 + 692.566 kg-m/s2
or, Li = 1039.066 kg-m/s2
Final moment of inertia of the merry-go-round+person system is,
I = 210 kg-m2 + (74 kg)(1.91 m)2
or, I = 479.9594 kg-m2
Let the final angular speed after the person jumps on the merr-go-round be
then final angular momentum of merry-go-round+person system is,
Lf = I = (479.9594 kg-m2)
From principle of conservation of angular momentum,
Lf = Li
or, (479.9594 kg-m2) = 1039.066 kg-m/s2
or, = 2.16 rad/s
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5)
Force needed is, F = m2R = (74 kg)(2.16 rad/s)2(1.91 m)
or, F = 662.43 N
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6)
magnitude of the linear velocity of the person is,
v = R = (2.16 rad/s)(1.91 m)
or, v = 4.13 m/s
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7)
The angular speed of the merry-go-round after the person lets go will reamain same as before at = 2.16 rad/s.
This is because the angular momentum of person+merry-go-round system remains the same before and after the person leaves the meery-go-round.
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This concludes the answers. Check the answer and let me know if it's correct. If you need any more clarification, modification or correction, feel free to ask.....