A marble of mass m and radius r rolls along the looped rough track of the figure

Question

A marble of mass m and radius r rolls along the looped rough track of the figure.(Figure 1) Ignore frictional losses.

Part A

What is the minimum value of the vertical height h that the marble must drop if it is to reach the highest point of the loop without leaving the track? Assume r?R.

Express your answer in terms of R.

Part B

What is the minimum value of the vertical height h that the marble must drop if it is to reach the highest point of the loop without leaving the track? Do not make assumption that r?R.

Express your answer in terms of R and r.

Explanation / Answer

Part a)
Using Conservation of Energy -
PEi = PEf + KEf + Krot-f
mgh = mg(2R) + 1/2mv^2 + 1/2(I)(w^2)
Where,  
I (moment of Inertia of sphere) = 2/5mr^2
v=r*w
For the marble not to leave the track at the highest point of the loop,
m*V²/R = m*g
V² = R*g.

Substituing Values -

mgh = mg(2R) + 1/2m(gR) + 1/2(2/5mr^2)(v/r)^2
mgh = mg(2R) + 1/2m(gR) + 1/2(2/5mr^2)([gR]/r^2)
h = 2R + R/2 + 1/5 R
h = 2.7R

Part b)
Using Energy Conservation -
Conservation of Energy: PEi = PEf + KEf + Krot-f
mg(h+r) = mg(2R-r) + 1/2mv^2 + 1/2(I)(w^2)

Where,  
I (moment of Inertia of sphere) = 2/5mr^2
v=r*w
For the marble not to leave the track at the highest point of the loop,
m*V²/(R-r) = m*g
V² = (R-r)*g.
Substituing Values -

mg(h+r) = mg(2R-r) + 1/2mg(R-r) + 1/2(2/5mr^2)(v/r)^2
h +r = 2R - 2r +1/2R - 1/2r + 1/5*R - 1/5r
h= 2.7R - 2.7r

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