A 2.1 kg , 20-cm-diameter turntable rotates at 70 rpm on frictionless bearings. Two 520 g blocks 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

Explanation:

There is change in moment of inertia of turntable due to which its angular velocity changes . We shall apply law of conservation of angular momentum

initial moment of inertia = 1 / 2 M r² , M is mass and r is radius of table

I₁ = 1/2 x 2.1 x .10²

= .0105 kg m²

final moment of inertia

I₂ = I₁ + 2 x m r²  , m is mass of fallen blocks on the table

= .0105 + 2 x .52 x .1²

= .0105 + .0104

= .0209 kg m²

initial angular velocity ω₁ = 2π n , n is revolution per second

= 2π  x 70 / 60

= 2.3333 π rad /s

Final angular velocity ω₂ = ?

Applying law of conservation of angular momentum

I₁ ω₁ = I₂ ω₂

.0105 x 2.3333 π =  .0209 x ω₂

ω₂ = 1.172  π

= .586 x 2 π rad / s

no of revolution per sec = .586

no of revolution per minute = .586  x 60

= 35.16 rpm .

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