Question

Starting from rest a field player gives himself a constant angular acceleration of 0.350 rad/s² for exactly 6 complete rotations with his arms held horizontally outward. Soon afterwards, he pulls his arms into his torso and spins quicker. The player's arms each have a mass of 7.50 kg and his total mass in 75.0 kg.

A. What is the player's angular speed after the initial 6 rotations?

B. What is the expressions for and calculate the player's moment of inertia before and after he pulls his arms in. What are the assumptions made to arrive at the expressions.

C. What was the magnitude of the torque the player had applied to himself during the initial angular acceleration (the 6 revolutions)?

D. What is the final angular speed of the player after he pulls his arms in?

* I didn't use any geometric parameters in the question, please make an assumption and justify it.

Answer 1

a] By third equation of rotation,

w_f = sqrt(2*alpha*theta)

= sqrt(2*0.35*6*2pi)

= 5.137 rad/s

b] Assuming his body is taken as solid cylinder with 0.15 m radius, hands are taken as rods of length 0.6 m.

Before he pulls his arms in, Moment of inertia = moment of inertia of a cylinder + moment of inertia of two rods

= 0.5m1r^2 + 2*m2/2 *L^2/3

= 0.5*(75-7.5)*0.15^2 + 7.5*0.6^2/3

= 1.659 kg m^2

After he pulls his arms in, Moment of inertia = moment of inertia of a cylinder

= 0.5m1r^2

= 0.5*75*0.15^2

= 0.844

c] Torque = I alpha = 1.659*0.35 = 0.581 Nm

d] angular momentum will be conserved

i1w1 = i2w2

w2 = i1w1/i2

= 1.659*5.137/0.844

= 10.1 rad/s