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

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.