A 39.0kg child runs with a speed of 2.70m/s tangential to the rim of a stationary merry-go-round . The merry-go-round has a moment of inertia of 506 kg m^2 and a radius of 2.51m . When the child jumps onto the merry-go-round, the entire system begins to rotate.
Calculate the final kinetic energy of the system.
Angular momentum is conserved. The boy has an initial angular momentum about the merry-go-round's axis of
m V R. V is the boy's initial velocity
Let the final angular velocity of boy and merry-go-round be w.
m V R = (I + m R^2) w
w = mVR/(I + m R^2)
Final KE = (1/2)(I + mR^2)w^2
= (1/2) m^2 V^2 R^2/(I + mR^2)
= (1/2) m V^2 /{[I/(mR^2)] + 1]}
Note that KE must be lost since the denominator [I/(mR^2) + 1]
is greater than 1
To calculate the final kinetic energy of the system, we need to consider the initial kinetic energy of the child and the final kinetic energy of the system after the child jumps onto the merry-go-round.
To calculate the initial kinetic energy of the child, we can use the formula:
Kinetic energy = (1/2) * mass * (speed)^2
Plugging in the values, we have:
Kinetic energy of the child = (1/2) * 39.0 kg * (2.70 m/s)^2
= 113.14 J
The angular momentum of the child can be calculated using the formula:
Angular momentum = moment of inertia * angular velocity
Since the merry-go-round is initially stationary, the angular velocity is also zero. Therefore, the angular momentum of the child is zero.
When the child jumps onto the merry-go-round, the angular momentum must be conserved. Assuming no external torques act on the system, the final angular momentum of the system (child + merry-go-round) is equal to the initial angular momentum of the child.
We can calculate the final angular velocity of the system using the equation:
Angular momentum of the system = moment of inertia of the system * angular velocity of the system
Since the child is initially running tangentially to the rim of the merry-go-round, we can treat this as the final state of the system. The moment of inertia of the system is given by the sum of the moment of inertia of the child and the moment of inertia of the merry-go-round.
Moment of inertia of the system = moment of inertia of the child + moment of inertia of the merry-go-round
= 506 kg m^2 + 39.0 kg * (2.51 m)^2
= 506 kg m^2 + 992.79 kg m^2
= 1498.79 kg m^2
To calculate the final angular velocity of the system, we rearrange the equation:
Angular velocity of the system = Angular momentum of the system / Moment of inertia of the system
Plugging in the values, we have:
Angular velocity of the system = 0 / 1498.79 kg m^2
= 0 rad/s
Since the angular velocity of the system is zero, the final kinetic energy of the system is also zero.
Therefore, the final kinetic energy of the system is zero.