Answer:
The new angular velocity of the two stones = 160.064 rad/s
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
Answer:
[tex]\dot n = 25.471\,\frac{rev}{s}[/tex]
Explanation:
The situation is described reasonably by the Principle of Angular Conservation:
[tex]\frac{1}{2}\cdot (50\,kg)\cdot (0.75\,m)^{2}\cdot \left(30\,\frac{rev}{s} \right) = \frac{1}{2}\cdot \left[(50\,kg)\cdot (0.75\,m)^{2}+ (20\,kg)\cdot (0.5\,m)^{2} \right] \cdot \dot n[/tex]
The final angular velocity is:
[tex]\dot n = 25.471\,\frac{rev}{s}[/tex]
