For the following answers, use only the symbols given. (a) Determine the minimum

Question

For the following answers, use only the symbols given.

(a) Determine the minimum release height h in terms of the radius r.

Now assume the actual release height is 3h. Give the following forces in terms of given parameters m, r, and g.
(b) Calculate the magnitude of the normal force exerted by the track at the bottom of the loop.
(c) Calculate the magnitude of the normal force exerted by the track at the top of the loop.
(d) Calculate the magnitude of the normal force exerted by the track after the block exits the loop onto the flat section.

Explanation / Answer

a)To remain on the track the mass remain in contect with the surface at the top.Let it,s velocity at top be v

Initialy the mass has only potential energy mgh Nm

From law conservation of energy

mgh=1/2mv2+2mgr .....................(1)

where 1/2mv2 is mechanical energy of the mass at top of loop and 2mgr is potential enrgy at the top of loop

For the mass to be in contect of loop at top

mg=mv2 /r

=>v=gr...................(2)

From equation (1) and ( 2) we get

mgh=1/2mgr+2mgr

=>h=5r/2

Now according to the problem the actual hight H=3h=15r/2

b) At the bottom of the loop the mass has only mechenical energy

So from law of conservation of energy

mgx15r/2=1/2mv2

=>v=15rg

Now the Normal reaction , force due to gravity and center pital force are in eulibrium

So N=mg+mv2/r

=>N=mg+mX15rg/r

=>N=16mg

c)At the top of the loop

From law of conservation of energy

15rmg/2=2rmg+1/2mv2

=>v=11gr

For equlibirium at top

N+mg=mv2 /r

=>N=11mg-mg

=>N=10mg

d)After the block exits the loop it does not have centeripital force.

So, the normal force exerted by the track is mg

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