Question

A professor sits on a rotating stool that spins at 10.0rpm while she holds a 1.00-kg weight in each of her hands. Her outstretched arms are 0.75m from the axis of rotation which passes through her head into the center of the stool. When she draws the weights in toward her body, her angular speed increases to 20.0 rpm. Neglecting the mass of her arms, how far are the weights from the rotational axis at the increased speed?

Answer 1

here conservation of angular momentum

that is

L_i = L_f       initial angular momentum = final angular momentum

we know that angular momentum L = Iw    I is sum of moment of inertia of two weights of mass 1 kr each

L_i = (mr1^2 +mr1^2)w_i= 2mr1^2*w_i

L_f = (mr2^2 +mr2^2)w_f= 2mr2^2*w_f

Li = Lf === > 2mr1^2*wi = 2mr2^2*w_f

                 r2= r1(sqrt (w_i / w_f))

                      = 0.75 sqrt(10*2*pi / 20*2*pi)

                      = 0.75 sqrt(1/2)

                   r2 = 0.53033 m

the new position of the weights from the rotational axis is 0.53033 m