(a) Calculate the angular momentum of an ice skater spinning at 6.00 rev/s given

Question

(a) Calculate the angular momentum of an ice skater spinning at 6.00 rev/s given his moment of inertia is 0.490 kg·m2.
_____________ kg · m2/s
(b) He reduces his rate of spin (his 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 2.05 rev/s.
__________ kg·m2
(c) Suppose instead he keeps his arms in and allows friction with the ice to slow him to 3.00 rev/s. What average torque was exerted if this takes 12.0 seconds?
__________ N·m

Explanation / Answer

a)

moment of inertia , I = 0.490 kg.m^2

angular speed , w = 6 rev/s

angulat momentum , L = I*w

L = 0.490 * 6 *2pi

L = 18.47 Kg.m/s

the angular momentum is 18.47 kg.m/s

b)

as angular momentum is conserved,

I * 2.05 = 6 * 0.490

I = 1.434 kg.m^2

the moment of inertia is 1.434 kg.m^2

c)

let the torque is T

using first equation of motion

3 * 2pi - 6 *2pi = 12 * a

a = 1.571 rad/s^2

Now, torque = I * a

Torque = 1.571 * 0.490

Torque = 0.77 N.m

the torque of fritcion is 0.77 N.m

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