Converting tangential momentum to angular momentum

AI Thread Summary
When a person jumps onto a frictionless merry-go-round, linear momentum is conserved overall, but not within the isolated system of the person and the merry-go-round. The person's linear momentum remains constant in magnitude, but its direction changes as they move in a circular path. To convert their linear momentum into angular momentum, the person's speed and distance from the center of the merry-go-round can be used to calculate angular speed (omega). The angular momentum can then be determined by multiplying the moment of inertia (I) of the person as a point mass by the angular speed (W). This approach allows for the analysis of momentum transfer in a simplified system.
jforce93
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"converting" tangential momentum to angular momentum

If someone is running and jumps onto a merry-go-round, momentum is still conserved, correct? (ignoring friction). So, would the momentum of the person while they were running be the same as the angular momentum of the merry-go-round after they jumped onto it? If not, how would I "convert" (for lack of a better term) their momentum into angular momentum? I know about

L = IW
and all, but I am really confused.

Thanks,

Jordan
 
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For simplicity's sake assume that the merry-go-round has no mass and therefore possesses no momentum of its own and the person's speed is the same before he or she jumps on (when they are running) and after (when they are sitting on the merry-go-round). Also assume no friction and that the person is a "point" mass (again just for simplicity).

Linear momentum is conserved, but in this case the person's momentum is transferred to the Earth through the merry-go-round as the merry-go-round exerts a force on the person towards its center. In other words, linear momentum is covered overall (if you include the Earth) but if you only include the person and the merry-go-round in your system then linear momentum is NOT conserved. Note the magnitude of the person's linear momentum stays the same since the person continues to travel at the same speed, but the momentum changes since momentum is a vector quantity and the person's direction is continually changing as they go around the merry-go-round.

To convert, just get the person's speed before the jump and combine that with the distance from the center to come up with the omega (the W, angular speed) value. So for example if the person is going 15 ft per second and is 5 feet from the center of the merry-go-round, they are doing speed (ft/sec) / circumference (ft/revolution) = 15/(2*5*pi) revolutions/sec = .477 rev/sec = 3 radians/sec (I'm not sure if my math is right, but you get the idea).

To get I:http://en.wikipedia.org/wiki/List_of_moments_of_inertia"

use the one for point mass (m*r*r) so you get mass of person * 5 * 5

multiply I and W together to get angular momentum! Note that this problem might be doable if you are not given the mass of the person, set it up and see if the m's cancel. Good luck!
 
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