Question

A $60kg$ woman stands at the western rim of a horizontal turntable having a moment of inertia of $500kg{m}^2$ and a radius of $2m$. The turntable is initially at rest and is free to rotate about a frictionless, vertical axle through its center. The woman then starts walking around the rim clockwise (as viewed from above the system) at a constant speed of $1.5m/s$ relative to the Earth. The final angular velocity of the woman and the turntable system is

• A. $0.36rad/s\text{(counter clockwise)}$
• B. $3.6rad/s\text{(clockwise)}$
• C. $0.36rad/s\text{(clockwise)}$
• D. $1.8rad/s\text{(counter clockwise)}$

Answer 1

The correct option is A $0.36rad/s\text{(counter clockwise)}$

Given : $m=60kg,I_{table}=500kg{m}^2,r=2m,v_{women}=1.5m/s$

Initially turntable was at rest hence it had zero angular momentum

Now as women starting walking along the circumference turntable starting rotating in opposite direction to maintain angular momentum of system as zero

From conservation of angular momentum for the system of the woman and the turntable, we have

$L_f=L_i$=0

$L_f=I_{woman}{\omega }_{woman}+I_{table}{\omega }_{table}=0$

${\omega }_{table}$ = (-$\frac{I_{women}}{I_{table}})$${\omega }_{woman}$

(=-$\frac{I_{women}{r}^2}{I_{table}})$ $(\frac{V_{women}}{r}$)

${\omega }_{woman}=-60\times \frac{2}{500}\times 1.5=-0.36rad/s$ (counterclockwise)

FINAL ANSWER: ($a$)