Confusing Physics Please Help Dr. A or someone else!!?
The small mass sliding without friction along the looped track shown in the figure, http://session.masteringphysics.com/problemAsset/1057653/6/GIANCOLI.ch08.p090.jpg. (it's not a Roller coaster and my picture has the end being flat) is to remain on the track at all times, even at the very top of the loop of radius r.
a) In terms of the given quantities, determine the minimum release height h.
b) If the actual release height is 5h, calculate the normal force exerted by the track at the bottom of the loop.
c) If the actual release height is 3h, calculate the normal force exerted by the track at the top of the loop.
d) If the actual release height is 3h, calculate the normal force exerted by the track after the block exits the loop onto the flat section.
The answer for a) Please answer with units. The answer for b) wants it in terms of mg. PLEASE EXPLAIN HOW YOU GOT IT!!
actechkarizma2012-11-24T14:30:54Z
Favorite Answer
a)
At the top of the loop, the centripetal force mv^2/r is acting downwards and the gravitational force (mg) is also acting downwards..
-mg =- mv^2/r
where v is the velocity of the mass and r is the radius of the loop....
To find speed at the top of the loop where the mass is still attached
mg = mv^2/r gr = v^2............(1)
To find minimum release height (h), use conservation of energy
Energy at the release point = Energy at the top of the loop
KE(rel) + PE (rel) = KE (rel) + PE (rel).......KE-->kinetic energy and PE -->potential energy
0 + mgh = ½ mv^2 + mg(2r).............where 2r is the height of the object where it is at the top of the loop
gh = ½ gr + 2gr............from (1) gr=v^2
h = ½r+ 2r = 2½ r
b) actual release height = 5h
Force exerted by the track at the bottom of the loop..
now similar principle to the question a)
now we have three forces, force exerted by the track on the mass (Fb) acting upwards, gravitational force (mg) acting downwards and centripetal force (mv^2/r) acting upwards.. lets put them all together..