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 kgm2. 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. (Hint: use what you know about relative velocity to help solve the problem

Answer 1

Answer:

$0.3165\ \text{rad/s}$

Explanation:

m = Mass of person = 65 kg

d = Diameter of round table = 6.5 m

r = Radius = $\dfrac{d}{2}=3.25\ \text{m}$

v = Velocity of person running = 3.8 m/s

$I_t$ = Moment of inertia of turntable = $1850\ \text{kg m}^2$

Moment of inertia of the system is

$I=I_t+mr^2\\\Rightarrow I=1850+65\times 3.25^2\\\Rightarrow I=2536.5625\ \text{kg m}^2$

As the angular momentum of the system is conserved we have

$L_i=L_f\\\Rightarrow mvr=I\omega_f\\\Rightarrow \omega_f=\dfrac{mvr}{I}\\\Rightarrow \omega_f=\dfrac{65\times 3.8\times 3.25}{2536.5625}\\\Rightarrow \omega_f=0.3165\ \text{rad/s}$

The angular velocity of the turntable is $0.3165\ \text{rad/s}$.