Respuesta :
The given question is incomplete. The complete question is as follows.
A merry-go-round rotates at the rate of 0.14 rev/s with an 84 kg man standing at a point 2.4 m from the axis of rotation. What is the new angular speed when the man walks to a point 0.6 m from the center? Assume that the merry-go-round is a solid 25-kg cylinder of radius 2.0 m.
Explanation:
The given data is as follows.
Initial angular velocity ([tex]\omega_{1}[/tex]) = 0.14 rev/s
[tex]\omega_{1} = (0.14 \times 2 \pi) rad/s[/tex]
= 0.87 rad/s
Mass of man (M) = 84 kg, Mass of cylinder (m) = 25 kg
Radius of the cylinder (r) = 2.4 m
Let [tex]r_{1}[/tex] be the initial distance of man from the axis of rotation = 2.0 m
Final distance of man from the axis of rotation ([tex]r_{2}[/tex]) = 0.6 m
Formula to calculate the moment of inertia of cylinder is as follows.
I = [tex]\frac{Mr^{2}}{2}[/tex]
And, moment of inertia of man = [tex]mr^{2}[/tex]
Therefore, initial moment of inertia of the system is as follows.
[tex]I_{1} = \frac{Mr^{2}}{2} + mr^{2}_{1}[/tex]
= [tex]\frac{84 \times (2.4 m)^{2}}{2} + 25 \times (2)^{2}[/tex]
= 241.92 + 100
= 341.92 [tex]kg m^{2}[/tex]
Now, final moment of inertia of the system will be calculated as follows.
[tex]I_{2} = \frac{Mr^{2}}{2} + mr^{2}_{2}[/tex]
= [tex]\frac{84 \times (2.4)^{2}}{2} + 25 \times (0.6)^{2}[/tex]
= 241.92 + 9
= 250.92
Now, according to the law of conservation of angular momentum is as follows.
Initial angular momentum = Final angular momentum
[tex]I_{1} \omega_{1} = I_{2} \omega_{2}[/tex]
or, [tex]\omega_{2} = \frac{I_{1}\omega_{1}}{I_{2}}[/tex]
= [tex]\frac{341.92 \times 0.87}{250.92}[/tex]
= 1.18 rad/s
Thus, we can conclude that the new angular speed is 1.18 rad/s.
