Answer:
ωf = 2.95 rad/sec
Explanation:
- Assuming no external torques acting while the person jumps on, total angular momentum must be conserved.
- Angular momentum for a rotating rigid body can be expressed as follows:
[tex]L = I * \omega (1)[/tex]
- where I = moment of inertia regarding the rotating axis, and ω= angular velocity.
- Since total angular momentum must be conserved, this means that the following equality must be satisfied:
[tex]L_{o} = L_{f} (2)[/tex]
- The initial angular momentum, taking into account that the Merry-Go-Round can be modeled as solid disk, can be expressed as follows:
[tex]L_{o} = I_{o} * \omega_{o} = \frac{1}{2}* M* R^{2}* \omega_{o} =\\ \frac{1}{2} * 77.0 kg* (2.70m)^{2}* 7.40 rad/sec = 2076.92 kg*m2*rad/sec (3)[/tex]
- The final angular momentum, is just the product of the new moment of inertia times the final angular velocity.
- The new moment of inertia, is just the sum of the original moment of inertia I₀ and the moment of inertia due to the person that jumps on.
- Assuming that we can treat him as a point mass, his moment of inertia is just the product of his mass times to the distance to the axis of rotation (the radius of the Merry-Go-Round) squared.
- So, we can write the new moment of inertia If as follows:
[tex]I_{f} = I_{o} +( m_{p} * R^{2}) = (\frac{1}{2} * M* R^{2}) + ( m_{p} * R^{2}) =\\ (\frac{1}{2} * 77.0 kg* (2.70m)^{2}) +( 58.0 kg * (2.70m)^{2}) = \\ 280.67 kg*m2 + 422.82 kg*m2 = \\ 703.49 kg*m2 (4)[/tex]
- The final angular momentum can be written as follows:
[tex]L_{f} = I_{f} * \omega_{f} (5)[/tex]
- Since (3) and (5) must be equal each other, replacing If by its value from (4) in (5), we can solve for ωf, as follows:
[tex]\omega_{f} = \frac{L_{o} }{I_{f}} = \frac{2076.92kg*m2*rad/sec}{703.49kg*m2} = 2.95 rad/sec (6)[/tex]
