A. A child is spinning on a chair with her arms outstretched. She is initially s

Question

A. A child is spinning on a chair with her arms outstretched. She is initially spinning with an angular speed of 0.850 rev/s. She quickly brings her arms in. What will happen to the child? Explain. b. The child's initial rotational inertia was 4.50 kgm^2 and it changes to a final value of 1.20 kgm^2. What is her speed just after bringing her arms in? c. Going from her initial speed to her final speed, how much rotational kinetic energy was lost or gained? d. Where does the extra kinetic energy come from (if it increases) or go to (if it decreases)? e. Once her arms are in, the girl slows to a stop in a time of 22.8 s. Assuming a constant torque was acting on her as she slowed down, how large was it?

Explanation / Answer

Q5. a. A child is spinning on a chair with her arms outstretched.

When she quickly brings her arms in her inertia will be decreased. As no external torque was present and angular momentum is conserved, so her angular speed will increase.

b. Initial moment of inertia = I1= 4.5 kg.m2

Initial angular velocity = w1 = 0.85 rev/s =0.85 x 2 x 3.141 =5.34 rad/s

Final moment of inertia=I2= 1.2 kg.m2

final angular velocity=w2=?

As angular momentum is conserved,

I1w1=I2w2

so, 4.5 x 5.34 = 1.2 x w2

so, w2= her speed just after bringing her arms in = 20.02 rad/s

c. Initial rotational kinetic energy= 1/2 I1w12 = 1/2 x 4.5 x 5.342 =64.16 J

final rotational kinetic energy = 1/2 x I2 x w22 =1/2 x 1.2 x 20.022 = 240.48 J

rotational kinetic was gained = 240.48 - 64.16 = 176.32 j

d. This exta energy came from the child itself.

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