Question

A grinding stone with a mass of 50 kg and a radius of .75 m is rotating with an angular velocity of 30 rev/s. A second, smaller stone with a mass of 20 kg and a radius of .5 m is dropped onto the first without slipping. What is the new angular velocity of the two stones?

Answer 1

`\dot n = 25.471\,(rev)/(s)`

Step-by-step explanation:

The situation is described reasonably by the Principle of Angular Conservation:

`(1)/(2)\cdot (50\,kg)\cdot (0.75\,m)^(2)\cdot \left(30\,(rev)/(s) \right) = (1)/(2)\cdot \left[(50\,kg)\cdot (0.75\,m)^(2)+ (20\,kg)\cdot (0.5\,m)^(2) \right] \cdot \dot n`

The final angular velocity is:

`\dot n = 25.471\,(rev)/(s)`

Answer 2

The new angular velocity of the two stones = 160.064 rad/s

Step-by-step explanation:

This is a case of conservation of angular momentum.

For initial case:

Mass = 50 kg

Radius = 0.75 m

Angular velocity N = 30 rev/s

We must convert to rad/s w

w = 2¶N = 2 x 3.142 x 30 = 188.52 rad/s

Moment of inertia I = m x r^2

I = 50 x 0.75^2 = 28.125 kgm2

Angular momentum = I x w

= 28.125 x 188.52 = 5302.125 kgm2-rad/s

For second case smaller stone has

m = 20 kg

Radius = 0.5 m

I = m x r^2 = 20 x 0.5^2 = 5 kgm2

Therefore,

Total moment of inertia of new system is

I = 28.125 + 5 = 33.125 kgm2

Final angular momentum = I x Wf

Where Wf = final angular speed of the system.

= 33.125 x Wf = 33.125Wf

Equating the two angular moment, we have,

5302.125 = 33.125Wf

Wf = 5302.125/33.125 = 160.064 rad/s