An ice skater is spinning at 6.2 rev/s and has a moment of inertia of .48 kgm^2.

Question

An ice skater is spinning at 6.2 rev/s and has a moment of inertia of .48 kgm^2.

f1=6.2 rev/s f2=1.25 rev/s 3.75 rev/s I=.48kgm^2 t=17s

Calculate the angular momentum, in kilogram meters squared per second, of the ice skater spining at 6.2 rev/s

He reduces his rate of rotation by extending his arms and increasing his moment of inertia. Find the value of his moment of inertia (in kilogram meters squared) if his rate of rotation decreases to 1.25 rev/s.

Suppose instead he keeps his arms in and allows friction of the ice to slow him to 3.75 rev/s. What is the magnitude of the average torque that was exerted, in N*m, if this takes 17 s?

Explanation / Answer

Given,

moment of inertia of .48 kgm^2

I=0.48kgm^2

f1= 6.2 rev/s ,

f2= 1.25 rev/s ,

t=17s

a)

we know,

L= I *w =(I)*(2*3.14*f1)

=(0 .48)*(2*3.14*6.2)

L   = 18.68928 kg.m2 /sec

b)

I1 * f1=I2 * f2

I2 =(I1 *f1) / f2

I2 = ( 0.48*6.2 ) / (1.25)

I2= 2.3808kg.m2

c) let

f1= 6.2 ,

f2=3.75 rev/s ,

t=17s

2*pi* f2 = 2 * pi * f1 -Et

E= 2*pi* (f1-f2) / (t)

= 2 * 3.14 *(6.2 - 3.75 )/( 17s ) = 0.9050 rad /sec2

Torque = I *E

= 0.48 * 0.9050

= 0.4344 N-m

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