Question

A person of $80kg$ mass is standing on the rim of a circular platform of mass $200kg$ rotating about its axis at $5$ revolutions per minute (rpm). The person now starts moving towards the center of the platform. What will be the rotational speed (in rpm) of the platform when the person reaches its center _____?

Answer 1

Step 1. Given data

Mass of the circular platform, $M=200kg$

Mass of the person, $m=80kg$

Revolutions per minute, ${\omega }_1$$=5$

Radius of sphere $=R$

Step 2. Finding the angular velocity when the person reaches the center, ${\omega }_2$

Applying the conservation of angular momentum,

Initial angular momentum, $A_i$$=$Final angular momentum, $A_f$

$I_1{\omega }_1=I_2{\omega }_2$ [$eq\left(1\right)$]

The initial moment of inertia, $I_1=$Moment of inertia about the center $+$ Moment of inertia of the person

$I_1=\frac{M{R}^2}{2}+m{R}^2$

The final moment of inertia, $I_2=\frac{M{R}^2}{2}$

Now, from $eq\left(1\right)$, we get

$\left(\frac{M{R}^2}{2}+m{R}^2\right){\omega }_1=\left(\frac{M{R}^2}{2}\right){\omega }_2$

${\omega }_2=\left(1+\frac{2m}{M}\right){\omega }_1$

${\omega }_2=\left(1+\frac{2\times 80}{200}\right){\omega }_1$

${\omega }_2=1.8\times 5$

${\omega }_2=9rpm$

Therefore, the angular speed of the platform is $\mathbf{9}\mathbf{}\mathit{r}\mathit{p}\mathit{m}$.