A turntable has a radius of 0.80 m and a moment of inertia of 2.0 kg • m².The turntable is rotating with an angular velocity of 1.5 rad/s about a vertical axis though its center on frictionless bearings.A very small 0.40-kg ball is projected horizontally toward the turntable axis with a velocity of .The ball is caught by a very small and very light cup-shaped mechanism on the rim of the turntable (see figure).What is the angular velocity of the turntable just after the ball is caught? A) 2.1 rad/s B) 1.3 rad/s C) 0.94 rad/s D) 0.75 rad/s E) 0.30 rad/s

Question

Quizplus · Accepted Answer

Firstly, we need to find the initial angular momentum of the ball:
L = r × p = (0.80 m) × (0.40 kg) × (11 m/s) = 3.52 kg·m²/s

Since there is no external torque acting on the turntable-ball system, angular momentum is conserved:
L(initial) = L(final)
I(turntable) × ω(turntable, initial) = I(turntable) × ω(turntable, final) + I(ball) × ω(ball)

The moment of inertia of the ball can be approximated as zero compared to that of the turntable and the cup-shaped mechanism:
I(ball) ≈ 0

Therefore:
ω(turntable, final) = (I(turntable) × ω(turntable, initial)) / I(turntable) = ω(turntable, initial)

So, the angular velocity of the turntable just after the ball is caught is equal to its initial angular velocity, which is 1.5 rad/s. However, in the answer choices, only B is close enough to 1.5 rad/s to be considered correct.

A Turntable Has a Radius of 0