someone spinns with an angular velocity of 4.55rad/s with arms held out. his rotational inertia is 2.7kgm^2. whn the he pulls his arms in his angular velocity increases to 7.60rad/s. what is his rotary ek?

Answer 1

Use conservation of angular momentum to find his final moment of inertia, I2.
I1*w1 = I2*w2

I2 = I1*(4.55/7.60) = 1.616 kg m^2

The final rotational KE is

(1/2) I2*(w2)^2 = (1/2)*1.616*(7.60)^2
(joules)

Answer 2

thanks

Answer 3

i got 2.790 is that correct