Answer
(a) $0.344m$
(b) $1.18KJ;1.42KJ$
Work Step by Step
(a) We can find the required distance as follows:
$r_f=\sqrt{\frac{(1_s+2mr_i^2)\frac{\omega_i}{\omega_f}-I_s}{2m}}$
We plug in the known values to obtain:
$r_f=\sqrt{\frac{(5.43Kg.m^2+2(1.25Kg)(0.759m)^2)(\frac{2.95rev/s}{3.54rev/s})-(5.43Kgm^2)}{2(1.25Kg)}}$
$r_f=0.344m$
(b) We can calculate the initial and final kinetic energy as follows:
$K.E_i=\frac{1}{2}I_i\omega_i^2$
$K.E_i=\frac{1}{2}(I_s+2mr_i^2)\omega_i^2$
We plug in the known values to obtain:
$K.E_i=\frac{1}{2}(5.43Kg.m^2+2(1.25Kg)(0.759m)^2)[(2.95rev/s)(\frac{2\pi rad}{rev})]^2$
$K.E_i=1.18KJ$
The final kinetic energy is given as
$K.E_f=\frac{1}{2}[I_s+2mr_f^2]\omega_f^2$
We plug in the known values to obtain:
$K.E_f=\frac{1}{2}[5.43Kgm^2+2(1.25Kg)(0.344m)^2][(3.54rev/s)(\frac{2\pi rad}{rev})]^2$
$K.E_f=1.42KJ$
