The small mass m sliding without friction along the looped track shown in the fi

Question

The small mass m sliding without friction along the looped track shown in the figure is to remain on the track at all times, even at the very top of the loop of radius r. Part A In terms of the given quantities, determine the minimum release height, h. h = ___ r Part B If the actual release height is 2h, calculate the normal force exerted by the track at the bottom of the loop. FN = _____ mg Part C If the actual release height is 3h, calculate the normal force exerted by the track at the top of the loop. FN = _____ mg Part D If the actual release height is 2h, calculate the normal force exerted by the track after the block exits the loop onto the flat section. FN = _____ mg I need explanations for this please, as this will be on my exam soon and I want to understand how to do it, not just know the answers. Thank you!

Explanation / Answer

Part A According to the law of conservation of energymgh=1/2mv2 v=v2gh and this should be equal to v=v5gr for thecircular motion in the bottom of the circle. Equating both we get h=2.5r Part B mg5h=1/2mv2 v=v10gh At the bottom of the circle the normal reaction force is thecentripetal force at that point. F=mv2/r=m10gh/r=25mg Part C At the top v should be v=vgr mg2h=1/2mv2 v=2vgh As this is at the bottom v=v5gr Hence v5gr=2vgh vgr=2/v5*vgh Hence the centripetal force at the top ismv2/r=m*2/5*gh=mgh Part D When the block comes at the bottom its velocityv=v14gh=v5gr(as h=7h) So it exits with this velocity. But we are not bothered about the velocity with which itsgoing as it is independent of normal reaction force. Its normal reaction force is just its weight W=mg.

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