Confusing Physics Please Help Dr. A or someone else!!?

Question

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!!

actechkarizma · Accepted 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..


Fb = mv^2/r + mg..........................................(2)

Conservation of energy:

Energy at the point of release = Energy at the bottom of the loop

KE(rel) + PE (rel) = KE (bottom) + PE (Bottom)

0 + mg(5h) = ½ mv^2 + 0

v^2 = 10gh

we know that h= 2.5r

v^2 = 10g(2.5r) = 25gr------------------(3)

substituting (3) in (2)

Fb = m (25gr)/r + mg = 25mg + mg = 26mg


c) Release height = 3h

Force exerted by the track at the top of the loop (Ft)

in addition to the gravitational force (mg), the centripetal force (mv^2/r) has to counter the force exerced by the track (Ft)

It means that  Ft is acting downwards, mg is acting downwards, mv^2/r is acting downwards


-Ft-mg=-mv^2/r

Ft = mv^2/r - mg -----------------------------(4)

Conservation of energy.. similar to a) and b)


0 + mg(3h) = ½ mv^2 + mg(2r)

v^2 = 6gh - 4gr =  6g(2.5r)- 4gr =  15gr - 4gr = 9gr  ----------------------(5)

substituting (5) in (4)


Ft = m (9gr)/r - mg  =  9mg-mg = 8mg



d)

To find the force at the flat area of the track,

simply we have to equate 

Ff - mg = 0

Ff = mg

Hope it helps :)

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Confusing Physics Please Help Dr. A or someone else!!?

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