A horizontal platform in the shape of a circular disk rotates on a frictionless bearing about a vertical axle through the centre of the disk. The platform has a mass of $\mathbf{150}\mathbf{}\mathbf{kg}$, a radius of$\mathbf{2.0}\mathbf{}\mathbf{m}$, and a rotational inertia of$\mathbf{300}\mathbf{}\mathbf{kg}\mathbf{.}\mathbf{}{\mathbf{m}}^{\mathbf{2}}$about the axis of rotation. A$\mathbf{60}\mathbf{}\mathbf{kg}$student walks slowly from the rim of the platform toward the centre. If the angular speed of the system is$\mathbf{1}\mathbf{.5}\mathbf{}\mathbf{rad}\mathbf{/}\mathbf{s}$when the student starts at the rim, what is the angular speed when she is$\mathbf{0.50}\mathbf{}\mathbf{m}$from the centre?

1. Mass of the platform,$\mathrm{m}=15\text{0kg}$
2. Radius of platform,$\mathrm{r}=2.0\text{m}$
3. Initial angular speed of the system${\mathrm{\omega }}_{\mathrm{i}}=1.5\text{rad/s}$
4. Distance of the student from the center,$\mathrm{R}=0.\text{5m}$
5. ${\mathrm{I}}_{\mathrm{disk}}=300{\text{kg.m}}^{\text{2}}$
   

Calculate initial and final rotational inertias. Then, apply law of conservation of angular momentum to find final angular speed. According tothe conservation of momentum, momentum of a system is constant if no external forces are acting on the system.

Formula are as follow:

${\mathbf{I}}_{\mathbf{student}}\mathbf{=}{\mathbf{M}}_{\mathbf{student}}\mathbf{\times }{\mathbf{R}}^{\mathbf{2}}$

$\mathbf{Initial}\mathbf{}\mathbf{angular}\mathbf{}\mathbf{momentum}\mathbf{=}\mathbf{Final}\mathbf{}\mathbf{angular}\mathbf{}\mathbf{momentum}$

Where,${\mathbf{I}}_{\mathbf{student}}$ is moment of inertia of student, ${\mathbf{M}}_{\mathbf{student}}$is mass of student and R is radius.

The initial rotational inertia of the system,

$\mathrm{I}={\mathrm{I}}_{\mathrm{disk}}+{\mathrm{I}}_{\mathrm{student}}$

$\mathrm{I}=300+60\times 2^2=540{\text{kg.m}}^{\text{2}}$

The final rotational inertia when she reaches$\text{0.5m}$from the center,

$\begin{array}{l}{\mathrm{I}}_{\mathrm{f}}={\mathrm{I}}_{\mathrm{disk}}+{\mathrm{I}}_{\mathrm{student}}\\ {\mathrm{I}}_{\mathrm{f}}=300+60\times {0.5}^2=315{\text{kg.m}}^{\text{2}}\end{array}$

According to law of conservation of angular momentum:

$\begin{array}{c}{\mathrm{L}}_{\mathrm{i}}={\mathrm{L}}_{\mathrm{f}}\\ {\mathrm{I}}_{\mathrm{i}}{\mathrm{\omega }}_{\mathrm{i}}={\mathrm{I}}_{\mathrm{f}}{\mathrm{\omega }}_{\mathrm{f}}\\ 540\times 1.5=315\times {\mathrm{\omega }}_{\mathrm{f}}\end{array}$

Hence,

${\mathrm{\omega }}_{\mathrm{f}}=2.6\frac{\text{rad}}{{\text{s}}^{\text{2}}}$

Hence,angular speed of student when she is at$\text{0.5m}$from the center of a circular disk is${\mathrm{\omega }}_{\mathrm{f}}=2.6\frac{\text{rad}}{{\text{s}}^{\text{2}}}.$

Therefore, using law of conservation of angular momentum, the final angular velocity of the student at the given distance can be found.