Question

An example of conservation of angular momentum is jumping on a Merry-Go-Round. Watch this video (it starts part way through but the only thing you miss is the people pushing the Merry-Go-Round) to see someone jumping on a Merry-Gr-Round in motion like this problem. You can model the Merry-Go-Round as a solid disk with a radius of 2.70 m and a mass of 77.0 kg. Initially the Merry-Go-Round has an angular velocity 7.40 radians / second. Then the person jumps on and change the Moment of Inertia of the system. The person lands on the outer edge of the Merry-Go-Round and has a mass of 58.0 kg. What is the final angular velocity of the system after the person jumps on

Answer 1

ωf = 2.95 rad/sec

Step-by-step 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:

`L = I * \omega (1)`

• 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:

`L_(o) = L_(f) (2)`

• The initial angular momentum, taking into account that the Merry-Go-Round can be modeled as solid disk, can be expressed as follows:

`L_(o) = I_(o) * \omega_(o) = (1)/(2)* M* R^(2)* \omega_(o) =\\ (1)/(2) * 77.0 kg* (2.70m)^(2)* 7.40 rad/sec = 2076.92 kg*m2*rad/sec (3)`

• 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:

`I_(f) = I_(o) +( m_(p) * R^(2)) = ((1)/(2) * M* R^(2)) + ( m_(p) * R^(2)) =\\ ((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)`

• The final angular momentum can be written as follows:

`L_(f) = I_(f) * \omega_(f) (5)`

• 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:

`\omega_(f) = (L_(o) )/(I_(f)) = (2076.92kg*m2*rad/sec)/(703.49kg*m2) = 2.95 rad/sec (6)`