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: The woman now walks to the middle of the merry-go-round. What is the kinetic energy of the system after she gets to the center of 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: The woman now walks to the middle of the merry-go-round. What is the kinetic energy of the system after she gets to the center of 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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