Question

What is the angular momentum of a figure skater spinning at 2.3 rev/s with arms in close to her body, assuming her to be a uniform cylinder with a height of 1.5 m, a radius of 16 cm , and a mass of 49 kg ?

How much torque(in magnitude) is required to slow her to a stop in 4.8 s , assuming she does not move her arms?

Answer 1

Given

initial angular velocity is w = 2.3 rev/s = 2.3*2pi rad/s

figure skater is like a solid cylinder with m = 49 kg , height h = 1.5 m, radius R = 16 cm

time t = 4.8 s

we know the relation between the torque and angular momentum is  

T = dL/dt

and L = I*W

I is moment of inertia of the solid cylinder = M*R^2 /2

L = W* M*R^2/2 = 2.3*2pi*49*0.16^2 /2 = 9.06387 kg m^2/s

now torque is  

T = (M*R^2/2)(dW/dt)

T = (49*0.16^2 /2)(2.3*2pi/4.8) N.m

T = 1.8883066243177 N.m

T = 1.88831 N.m