A boy with mass m is at the rim of a horizontal turntable which possesses a rota
ID: 1481679 • Letter: A
Question
A boy with mass m is at the rim of a horizontal turntable which possesses a rotational inertia I and radius R. Both the boy and the turntable are initially at rest and the system is free to rotate without friction about the vertical axis of the turntable which is at the center of it. The boy begins to walk counterclockwise at a speed v relative to the ground. What is the angular speed of the turntable and in which direction rotates?
The answer is supposed to be mRv/I, clockwise but I don't know how to get there.
Explanation / Answer
let
mass of boy m, moment of inertia of turnable is I,
initially both are at rest so the angular momentum is zero,
when the boy begins to move in anticlock wise direction wity v speed
gains angular momentum , turnable having moment of inertia so the turnable will move
in the opposite direction to the boy.
L1 =L2 conservation of momentum
L1= 0 , L2 = I2W2
I2 = (Ib+I_turnable)W2
now conservation of angular momentum
Ib*W2 = - I_turnable *W2
mr^2*v/r = I_turnable *W2
W2 = - m*v*r / I_turnable
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