A merry-go-round with moment of inertia 400 kg-m^2 and radius 2.0m is rotating w

Question

A merry-go-round with moment of inertia 400 kg-m^2 and radius 2.0m is rotating with angular speed 0.50 rad/s in the clockwise direction about a fixed axis. A child of mass 40 kg runs tangentially to the merry-go-round with speed 3.0m/s and grabs onto the outside edge of the merry-go-round.

a. What is the final angular velocity of the system (merry-go-round plus child)/ What is the final tangential speed of the child?

b. What is the change in kinetic energy?

c. If instead the merry-go-round is initially rotating with angular speed 3.0 rad/s in the counterclockwise direction what would the final angular velocity be/ What is the final tangential speed of the child?

Explanation / Answer

Given

moment of inertia of the merry go round is I1 = 400 kg m2

radius r = 2m

initial angular speed is W1 = 0.5 rad/s

when child of mass m = 40 kg runs tangentially and at the rim

a)
here the conservation of angular momentum

L = I*W

L1 = L2

I1W1 = I2W2

W2 = I1W1/I2

I2 = I1+mr^2 = 400+40*2^2 = 560 kg m2

now W2 = 400*0.5/560 rad/s = 0.357143 rad/s
the tangential speed of the child is v = r*W2 = 2*0.357143 m/s = 0.714286 m/s

b) change in kinetic energy is

rotational k.e = 0.5*I*W^2

change in k.e =0.5(I2*W2^2 - I1*W1^2)

= 0.5( 560*0.357143^2-400*0.5^2)

= -14.28568571428 J

= 14.286 J

c) if W1 = 3 rad/s

I2 = I1+mr^2 = 400+40*2^2 = 560 kg m2

now W2 = 400*3/560 rad/s = 2.142857 rad/s

the tangential speed of the child is v = r*W2 = 2*2.142857 m/s = 4.285714 m/s

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