Given: A woman of mass 75 kg running across the ground with a velocity of 8 m/s

Question

Given: A woman of mass 75 kg running across the ground with a velocity of 8 m/s tangential to the edge of a stationary merry-go-round jumps on and sets the merry-go-round into motion. The merry-go-round has a radius of 2 m and a moment of inertia about its axis of I = 1200 kg m2. The axle about which the merry-go-round rotates is frictionless. (Treat the woman as a point mass.)

Question: What is the angular velocity of the system, in radians per second, after the woman has jumped on? (For this question, after the woman lands, she is stationary with respect to the merry-go-round.)

Given: A woman of mass 75 kg running across the ground with a velocity of 8 m/s tangential to the edge of a stationary merry-go-round jumps on and sets the merry-go-round into motion. The merry-go-round has a radius of 2 m and a moment of inertia about its axis of I = 1200 kg m2. The axle about which the merry-go-round rotates is frictionless. (Treat the woman as a point mass.) Question: What is the angular velocity of the system, in radians per second, after the woman has jumped on? (For this question, after the woman lands, she is stationary with respect to the merry-go-round.)

Explanation / Answer

conserving angular momentum,

Original angular momentum = m . v . r = 75 . 8 . 2 = 1200 kgm^2/s

I of the roundabout = 1200 + woman's I (mr^2)
So final momentum = 1200 = I . w = (1200 + (75 . 2^2)) w
w = 1200 / 1500 = 0.8 rad/s

when the woman reaches the centre then also conserving angular momnetum,

so , Iw=I2w2

I2=1200

w2=1

K.E.=0.5Iw2 = 600J

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