Once the person gets half way around, they decide to simply Let go of the merry-

Question

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 A merry-go-round with a a radius of R = 1.93 m and moment of inertia I = 191 kg-m2 is spinning with an initial angular speed of omega = 1.69 rad/s in the counter clockwise direction when viewed from above. A person with mass m = 62 kg and velocity y = 4.6 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-m^2/s 2) What is the magnitude of the angular momentum of the person 2 meters before she jumps on the merry-go-round? ___________ kg-m^2/s Submit 3) What is the magnitude of the angular momentum of the person just before she jumps on to the merry-go-round? kg-m^2/s Submit 4) What is the angular speed of the merry-go-round after the person jumps on? rad/s Submit 5) Once the merry-go-round travels at this new angular speed, with what force does the person need to hold on? _________ N Submit

Explanation / Answer

1)

intial angular momentum L1 = I1w1 = 191*1.69 = 322.79 kg-m^2/s

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2)

0 kg-m^2/s

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3)

before jumping

L' = mvr = 62*4.6*1.93 = 550.4 kg-m^2/s

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4)
after jumps on to rim moment of inertai I2 = 322.79 + (62*1.93^2) = 553.73 kg-m^2

from conservation of angular momentum

I1w1 = I2w2

322.79 = 553.73 w2

w2 = 0.583 rad/s
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5)

Force , N = mrw2^2 + mg = 62*1.93*0.583^2 + 62*9.8 = 648.3 N
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6)

mv^2/r = N-mg = 40.66

v = 1.125 m/s

7)

angular speed w = v/r = 1.125/1.93 = 0.583 rad/s
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