Question

Suppose a 65 kg person stands at the edge of a 6.5 m diameter merry-go-round turntable that is mounted on frictionless bearings and has a moment of inertia of 1850 kg m2. The turntable is at rest initially, but when the person begins running at a speed of 3.8 m/s (with respect to the turntable) around its edge, the turntable begins to rotate in the opposite direction. Calculate the angular velocity of the turntable.

Answer 1

Answer:0.316 rad/s

Explanation:

Given

mass of Person $m=65 kg$

velocity of person $v=3.8 m/s$

diameter of turntable $d=6.5 m$

moment of Inertia of the table $I_0=1850 kg-m^2$

Moment of inertia of Person I

$I=mr^2=65\times (\frac{6.5}{2})^2$

$I=686.56 kg-m^2$

initial angular velocity $\omega _1=\frac{v}{r}$

$\omega _1=\frac{3.8}{3.25}=1.17 rad/s$

Conserving Angular momentum

$I\omega _1=(I+I_0)\omega _2$ , where $\omega _2=final\ angular\ velocity$

$686.56\times 1.17=(686.56+1850)\times \omega _2$

$\omega _2=\frac{686.56}{2536.56}\times 1.17$

$\omega _2=0.316 rad/s$