A merry-go-round with a a radius of R = 1.91 m and moment of inertia I = 210 kg-

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.....

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