Question

A 70 kg Person sitting on a frictionless stool Is hulking to 5 kg weight a distance of 0.8 m from their center. Assume the person's moment of inertia is that of a uniform cylinder 0.2m in radians if the person. Is initially rotating at omega = 3 red/s and then pulls the two weights to their body (0.2 m from their centre) find there are angular velocity omega the change in their kinetic energy delta KE
Please explain the process.

Answer 1

The initial angular momentum is equal to final angular momentum so that angular momentum is conserved.

$I_1 \omega_1 =I_2\omega_2$

$(\frac{1}{2}mr^2 + 2mr_1^2) \omega_1 =(\frac{1}{2}mr^2 + 2mr_2^2)\omega_2$

$(\frac{1}{2}(70)(0.2)^2 + 2(70)(0.8)^2) (3) =(\frac{1}{2}(70)(0.2)^2 + 2(70)(0.2)^2)\omega_2$

$\omega_2= 39 rad/s$

b)

The change in rotation kinetic energy is

$\Delta K = \frac{1}{2}I_2\omega_f^2 - \frac{1}{2}I_1\omega_i^2$

$\Delta KE = \frac{1}{2}(\frac{1}{2}(70)(0.2)^2 + 2(70)(0.2)^2)(39)^2 -\frac{1}{2} (\frac{1}{2}(70)(0.2)^2 + 2(70)(0.8)^2) (3)^2$

$\Delta KE = 4914J$