As shown in the figure, a merry-go-round (MGR) is free to spin about a frictionl
ID: 1769799 • Letter: A
Question
As shown in the figure, a merry-go-round (MGR) is free to spin about a frictionless vertical axle (marked O); the figure is drawn looking down on the MGR. The MGR is initially not spinning. A small cannon (not drawn in the figure) is mounted on the western rim of the MGR; the cannon can be fired in any of the four horizontal directions which are drawn in the figure and labeled as A, B, C, and D (direction A is horizontally South and direction D is horizontally West). Consider the system of cannon, cannonball, and MGR, and consider the angular momentum of this system about the axle O. Initially, all parts of the system are at rest, and "just after firing" the cannonball has just left the barrel of the cannon and is traveling in one of the four labeled directions, but has not yet traveled far enough to have experienced any significant effects from the force of air resistance. For which of the four labeled firing directions will the angular momentum of the system, about the axle O, be constant during the described action?
For all four directions. For all directions except D. For only direction D. For none of the four directions. For only directions B and C. For only direction A. For all directions except B and C. For all directions except A. N A C' B AExplanation / Answer
For all the directions except B and C
In D angular momentum about O is zero so marry go round will not rotate at all.
In A angular momentum of canon ball will be mvr and that will be equal to Iw. And that will be conserved.
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