Question

A man of mass $M=50\text{kg}$ standing on the edge of a platform (moment of inertia of the platform is $I$ and radius $R=1\text{m}$) is rotating in anticlockwise direction at an angular speed of $20\text{rad/s}$. The man starts walking along the rim with a speed $1\text{m/s}$ relative to the platform, also in the anticlockwise direction. The new angular speed of the platform in $\left(\text{rad/s}\right)$ is (take $(I=0.02{\text{kg-m}}^2)$)

Answer 1

Given,
Moment of inertia of platform (disc) $I=0.02{\text{kg-m}}^2$
Radius of platform (disc) $R=1\text{m}$
Speed of man w.r.t. platform $v_{mp}=1\text{m/s}$
Mass of man $M=50\text{kg}$
Angular speed initially, $\omega =20\text{rad/s}$
Initially:


Initial angular momentum about central axis
$L_i=(I+M{R}^2)\omega$
Finally:


Final angular momentum about central axis
$L_f=I{\omega }^{\prime }+MvR$
$L_f=(I+M{R}^2){\omega }^{\prime }+MvR$
Here,
$(I+M{R}^2)$ is the MOI of the system (man+platform)
${\omega }^{\prime }=$ angular velocity when man starts moving with velocity $v$ w.r.t platform
Here, disc and man both are moving in anti-clockwise direction. Hence according to right hand thumb rule, direction of angular momentum for both is same (upward axial direction)

As we can see, there is no external torque on the system, so the angular momentum of the system about the rotating axis remains conserved or same.
Hence $L_i=L_f$
$(I+M{R}^2)\omega =(I+M{R}^2){\omega }^{\prime }+MvR$
${\omega }^{\prime }=(\omega -\frac{MvR}{I+M{R}^2})$
${\omega }^{\prime }=\left[20\right(\text{rad/s})-\frac{\left(50\text{kg}\right)\left(1\text{m/s}\right)\left(1\text{m}\right)}{0.02\left({\text{kgm}}^2\right)+50\left(\text{kg}\right)\times (1\text{m}{)}^2}]$
${\omega }^{\prime }=[20-\frac{50}{50.02}]$
${\omega }^{\prime }=19\text{rad/s}$