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A playground merry-go-round of radius 3 m has a moment of inertia 200 kg.m^2 and is rotating at 10 rev/min about a frictionless vertical axle. Facing the axle, a 40 kg child hops onto the merry-go-round, and manages to sit down on the edge.

The child initially is not on the merry-go-round. I think when the child hops onto the merry-go-round, he will generate an external force acting on the merry-go-round. Why is the net angular momentum of the system remains constant?

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closed as off-topic by Steeven, John Rennie, Yashas, ZeroTheHero, Kyle Kanos May 6 '17 at 11:17

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  • $\begingroup$ Can you be more specific about what you want to know? The total momentum of a system is always conserved when there are no external forces acting on it. $\endgroup$ – BowlOfRed May 6 '17 at 3:12
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    $\begingroup$ Momentum is not conserved. Angular momentum of the merry-go-round plus child is conserved. This is because the external forces exert no net torque about the merry-go-round axis; the only torque on the merry-go-round is from the child, and the only torque on the child is from the merry-go-round. $\endgroup$ – pwf May 6 '17 at 3:54
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Considering the child and the merry-go-round as the system then if there are no external torques acting about the axis of rotation of the merry-go-round then the angular momentum of the system must be conserved.

What happens when the child hits the merry-go-round is that internal frictional forces start to act.

If there is no radial slipping then the radial components of those frictional forces do no work and also the torque about the axis of rotation of those radial frictional forces is zero.

A Newton third law pair of tangential frictional forces act on the child and the merry-go-round which are equal in magnitude, opposite in direction and act for exactly the same time.
This means that the impulsive frictional torque about the axis of rotation on the child is equal in magnitude and opposite in direction to the impulsive frictional torque on the merry-go-round.
This means that the angular momentum lost by the merry-go-round in its original direction of rotation is equal to the angular momentum gained by the child in that direction.
Thus the angular momentum of the child and merry-go-round is unchanged.
However because there is relative motion between the child and the merry-go-round work is done by the tangential frictional forces so heat is generated and the rotational kinetic energy of the system is decreased.

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We define the angular momentum as $$l=r \times P$$ then the time rate of change of the angular momentum is $$\dot l=({\dot r \times p}) + (r \times \dot p)$$ but we also know that $$p=m\dot r$$ $$\dot p=F_{ext}$$ Thus substituting the above equations into the equation of the time rate of change of angular momentum we get, $$\dot l=({\dot r \times m\dot r }) + (r \times F_{ext} )$$ Since we know the cross product of any two vectors in the same direction is zero $${\dot r \times m\dot r }=0$$ $$\dot l=r \times F_{ext} $$

Moreover we know that $$τ=r \times F_{ext} $$

where τ is the net external torque on the system

$$\dot l=τ$$

In your question we notice that there is no net torque on the system in the direction of the vertical axle lets call it the z axis thus $$\dot l_z=0$$ Which implies that the z component of the angular momentum of the system is conserved

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The child initially is not on the merry-go-round. Why is the momentum conserved?

If you are asking about the momentum that the child carries with the "hop", which it got by giving the earth an equal and opposite momentum, one has to look at the boundary conditions of the problem.

Take a motionless merry go round and the hop on a diameter of the circle of the merry go round. The circle will rock, part of the kinetic energy will turn to friction and the circle will rock up and down ( first up, and then due to the mechanics reflected down), and the earth will finally pick up the the momentum, of smaller magnitude due to friction, through the axis. Friction carries off momentum as heat/radiation.

If there is a perpendicular to the diameter component of the momentum it will start the rotation, as the other answers state.

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