Question

A man of $80\, kg$ mass is standing on the rim of a circular platform of mass $200\, kg$ rotating about its axis. The mass of the platform with the man on it rotates at $12.0\, rpm$. If the man now moves to centre of the platform, the rotational speed would become

• A. 16.5 rpm
• B. 25.7 rpm
• C. 32.3 rpm
• D. 31.2 rpm

Answer 1

Answer: (D)

Here, mass of man, $m=80\, kg$
Mass of platform, $M=200\, kg$
Let $R$ be the radius of platform.
When man is standing on the rim,
$I_{1} =M(R / 2)^{2}+m R^{2}$
$=\left(\frac{R}{2}\right)^{2}(M+4\, m)$
When man reaches the centre of platform,
$I_{2}=M(R / 2)^{2}+m \times 0=m(R / 2)^{2}$
As angular momentum is conserved,
$\frac{I_{1}}{I_{2}}-\frac{\omega_{2}}{\omega_{1}}-\frac{2 \pi n_{2}}{2 \pi n_{1}}-\frac{n_{2}}{n_{1}}$
$n_{2} =\frac{I_{1}}{I_{2}} \times n_{2}$
$=\frac{(M+4 m)(R / 2)^{2}}{M(R / 2)^{2}} \times 12$
$=\frac{(200+4 \times 80)}{200} \times 12$
$=\frac{520 \times 12}{200}$
$n_{1} =31.2$ rpm