Question

A 2.5 kg , 20-cm-diameter turntable rotates at 50 rpm on frictionless bearings. Two 470 gblocks fall from above, hit the turntable simultaneously at opposite ends of a diameter, and stick.

What is the turntable's angular velocity, in rpm, just after this event?

Answer 1

Using angular momentum conservation

L1 = L2
I1w1 = I2w2
I1 = Initial Moment of inertia of turntable = 0.5*M*R^2
I2 = final Moment of inertia of turntable + blocks = 0.5*M*R^2 + 2*m*R^2

M = mass of turntable = 2.5 kg
m = mass of blocks = 470 gm = 0.47 kg
R = radius of table = 20/2 cm. = 0.1 m
So,

I1 = 0.5*2.5*(0.1)^2 = 0.0125

I2 = 0.5*2.5*(0.1)^2 + 2*0.47*(0.1)^2 = 0.0219
w1 = Initial Angular Speed = 50 rpm
w2 = final angular speed = ?
So,
w2 = w1*(I1/I2)
Using above values:
w2 = 50*(0.0125/0.0219)
w2 = 28.54 rpm = new angular speed of turntable

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