Question

A turntable, with a mass of 1.5 kg and diameter of 20 cm, rotates at 70 rpm on frictionless bearings. Two 540 g blocks fall from above, hit the turntable simultaneously at opposite ends of a diameter, and stick to it.
Required:
What is the turntable's angular speed, in rpm, just after this event?

Answer 1

Answer:

The turntable's angular speed after the event is 28.687 revolutions per minute.

Explanation:

The system formed by the turntable and the two block are not under the effect of any external force, so we can apply the Principle of Conservation of Angular Momentum, which states that:

$I_{T}\cdot \omega_{o} = (2\cdot r^{2}\cdot m +I_{T})\cdot \omega_{f}$ (1)

Where:

$I_{T}$ - Moment of inertia of the turntable, in kilogram-square meters.

$r$ - Distance of the block regarding the center of the turntable, in meters.

$m$ - Mass of the object, in kilograms.

$\omega_{o}$ - Initial angular speed of the turntable, in radians per second.

$\omega_{f}$ - Final angular speed of the turntable-objects system, in radians per second.

In addition, the momentum of inertia of the turntable is determined by following formula:

$I_{T} = \frac{1}{2}\cdot M\cdot r^{2}$ (2)

Where $M$ is the mass of the turntable, in kilograms.

If we know that $\omega_{o} \approx 7.330\,\frac{rad}{s}$, $M = 1.5\,kg$, $m = 0.54\,kg$ and $r = 0.1\,m$, then the angular speed of the turntable after the event is:

$I_{T} = \frac{1}{2}\cdot M\cdot r^{2}$

$I_{T} = 7.5\times 10^{-3}\,kg\cdot m^{2}$

$I_{T}\cdot \omega_{o} = (2\cdot r^{2}\cdot m +I_{T})\cdot \omega_{f}$

$\omega_{f} = \frac{I_{T}\cdot \omega_{o}}{2\cdot r^{2}\cdot m +I_{T}}$

$\omega_{T} = 3.004\,\frac{rad}{s}$ ($28.687\,\frac{rev}{min}$)

The turntable's angular speed after the event is 28.687 revolutions per minute.