Part A A figure skater rotating at 5.00 rad/s with arms extended has a moment of

Question

Part A

A figure skater rotating at 5.00 rad/s with arms extended has a moment of inertia of 2.25 kg·m2. If the arms are pulled in so the moment of inertia decreases to 1.80 kg·m2, what is the final angular speed?

Part B

Which of the following statement correctly describes the change in the rotational kinetic energy of the figure skater?

KErot remained the same KErot decreased by a factor of 1.56 KErot  increased by a factor of 1.25 KErot decreased by a factor of 1.25 KErot increased by a factor of 1.56

Explanation / Answer

I1 = 2.25 kg-m^2 ; I2 = 1.8 kg m^2

w1 = 5 rad/s ; w2 = ?

A)from the conservation of angular momentum

Li = Lf

I1 w1 = I2 w2

w2 = (I1/I2)w1

w2 = (2.25/1.8)5 = 6.25 rad/s

Hence, w2 = 6.25 rad/s

B)KEi = 1/2 I1 w1^2 = 1/2 x 2.25 x 5^2 = 28.13 J

KEf = 1/2 I2 w2^2 = 1/2 x 1.8 x 6.25^2 = 35.16 J

KEf/Kei = 35.16/28.13 = 1.25

KEf = 1.25 KEi

KErot  increased by a factor of 1.25

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