Question

A horizontal square platform of mass $m$ and side $a$ is free to rotate about a vertical axis passing through its centre $O$. The platform is stationary and a person of the same mass $\left(m\right)$ as the platform is standing on it at point $A$. The person now starts walking along the edge from $A$ to $B$ ( see figure). The speed $v$ of the person with respect to the platform is constant. Taking $v=5\text{m/s and a = 1 m}$,

• A. Angular velocity of platform when the person is at $A$ is $6$ rad/s.
• B. Angular velocity of platform increases first and then decreases again as person reaches $B$.
• C. Angular velocity of platform decreases first and then increases again as person reaches $B$.
• D. Angular velocity of platform remains constant as the person goes from $A$ to $B$.

Answer 1

Answer: (A), (D)

The correct options are
A Angular velocity of platform when the person is at $A$ is $6$ rad/s.
D Angular velocity of platform remains constant as the person goes from $A$ to $B$.



$v_m=v_{mp}+v_p$
$=\sqrt{v^2+{\omega }^2{x}^2+2v\omega xcos(90+\theta )}$
$=\sqrt{v^2+{\omega }^2{x}^2-2v\omega xsin\theta }$

By conservation of angular momentum,
$L_i=0=L_f$
$\Rightarrow I\omega +m(v-v_{P}cos(90-\theta )\frac{a}{2}=0$
$\Rightarrow -\frac{m}{12}(a^2+a^2)\omega +m(v-x\omega \sin \theta )\frac{a}{2}=0$
$\Rightarrow -\frac{m}{6}{a}^2\omega -m\frac{a^2}{4}\omega +mv\frac{a}{2}=0$
($\because xsin\theta =\frac{a}{2}$)

$\Rightarrow \frac{a^2}{6}\omega +\frac{a^2}{4}\omega =\frac{va}{2}$
$\Rightarrow \omega =\frac{6v}{5a}$

Therefore, angular velocity of platform remains constant (D)

and $\omega =\frac{6\times 5}{5\times 1}=6\text{rad/s}$ (A)