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Homework Statement
A turntable has a radius of 0.80 m and a moment of inertia of 2.00 kg • m2. The turntable is rotating with an angular velocity of 1.50 rad/s about a vertical axis though its center on frictionless bearings. A very small 0.40kg ball is projected horizontally toward the turntable axis with a velocity of 3.00 m/s and is moving directly towards the center of the table. The ball is caught by a very small and very light cupshaped mechanism on the rim of the turntable (see figure). What is the angular velocity of the turntable just after the ball is caught
Homework Equations
L= Iw
The inertia of the turntable = 2.0 kgm^2
The inertia of a solid sphere = 2/5*M*R^2
L_{i} = initial angular momentum
L_{f} = final angular momentum (momentum after the ball is caught)
L_{id} = initial angular momentum of the disc
L_{ip} = initial angular momentum of the particle
L_{fd} = final angular momentum of the disc
L_{fp} = final angular momentum of the particle
w_{ft} = angular velocity of the turntable after the ball is caught
The Attempt at a Solution
I assumed that Li = Lf
L_{i} = L_{id} + L_{ip}
Li = 2.0kg*1.50 rad/s + 0
Li = 3.50 kgm^2
Similarly, I thought that:
Lf = L_{fd} + L_{fp}
L_{f} = 2.0kgm^2/s*w_{ft} + 2/5*0.4kg*0.80^2*w_{ft}
3.50kgm^2/s = 2.1024w_{ft}
w_{ft} = 1.66476 kgm^2/s
As you can see this isn't an answer choice. Is it wrong for me to assume that the angular velocity of the particle after it is caught is the same as that of the turntables angular velocity?
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