(7%) Problem 13: An ice skater is spinning at 6.8 rev/s and has a moment of iner

Question

(7%) Problem 13: An ice skater is spinning at 6.8 rev/s and has a moment of inertia of 0.56 kg-m2. 33% Part(a) Calculate the angular momentum in kilogram meters squared per second, of the ice skater spinning at 6.8 rev/s. Deductions Potential sin0 cotan0 asinacos0 atan0 acotan sinh0 coshO Submission Attempts rer (5% per atte detailed vie 0 DegreesRadians I give up! Feedback: 3%deduction per feedback. Hints: 2% deduction per hint. Hints remaining:- 4 33% Part b He reduces his rate of rotation by extending his arms and increasing his moment of inertia. Find the value of his r inertia (in kilogram meters squared) if his rate of rotation decreases to 1.5 rev/s. 33% Part (c) Suppose instead he keeps his arms in and allows friction of the ice to slow him to 325 revs what is the magnitu average torque that was exerted, in N·m, if this takes 11 s? TA, LLC

Explanation / Answer

A.

Angular momentum is given by:

Li = I*w

w = 6.8 rev/sec = 42.72 rad/sec

Li = 0.56*42.72 = 23.92 kg-m^2/sec

B.

Using momentum conservation

L1 = L2

I1*w1 = I2*w2

I2 = I1*w1/w2

I2 = 0.56*6.8/1.5

I2 = 2.54 kg-m^2

C.

Torque = I*alpha

we know that

wf = wi + alpha*t

alpha = (wf - wi)/t

alpha = 2*pi*(6.8 - 3.25)/11 = 2.027 rad/sec^2

So,

torque = 0.56*2.027 = 1.135 N-m

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