Question

An ice skater with a moment of inertia of 0.390 kg*m² is spinning at 6.00 rev/s

a. What is the angular velocity of the ice skater in rad/s?

b. What is the angular momentum of the ice skater at this angular velocity

c. He reduces his rate of spin (angular velocity) by extending his arms and increasing his moment of inertia. Find the value of his moment of inertia if his angular velocity drops to 1.70 rev/s

d. Suppose instead he keeps his arms in and allows friction with the ice to slow him from 6.00 rev/s to 3.00 rev/s. What average torque was exerted if this takes 14.0 seconds?

e. Assuming the torque from friction is constant, clearly sketch the angular velocity and angular acceleration versus time as the ice skater slows from 6.00 rev/s to 3.00 rev/s.

Answer 1

(a)

angular velocity ,

w = 2*pi*f

w = 2*3.14*6

w = 37.68 rad/s

(b)

The angular momentum of the ice skater can be found from

L = I*w

L = 0.390*37.68

L = 14.69 kg*m²/s

(c)

Angular momentum remains constant ,Hence moment of inertia

I = L/w

w = 2*pi*f = 2*3.14*1.7

w = 10.67 rad/s

I = 14.69/10.67

I = 1.37 kg*m²

(d)

Average torque was exerted

= I*

As we know that

angular acceleration = w_final - w_initial / t

= 2*pi*f_final - 2*pi*f_initial /t

= (2*pi/t)*(f_final - f_initial)

= (2*3.14/14)*(3 - 6)

= -1.34 rad/s²

So,

= 0.390 * (-1.34)

= 0.524 Nm