Question

A spherical marble that has a mass of 50.0 g and a radius of 0.500 cm rolls without slipping down a loop-the-loop track that has a radius of 25.0 cm. The marble starts from rest and just barely clears the loop to emerge on the other side of the track.

1)What is the minimum height that the marble must start from to make it around the loop?

Answer 1

VERTICAL LOOP

at the top of the loop

Fnet = mg + Ntop

Ntop = normal force on the marble

from newtons second law

Fnet = m*a = m*vtop^2/R

Ntop + mg = m*vtop^2/R

the minimum speed for a marble to clear the top

for vtop to be minimm Ntop = 0

m*g = m*vtop^2/R

vtop = sqrt(g*R)

total energy at the top

Etop = m*g*2R + (1/2)*m*vtop^2

total energy at the minimum height E1 = m*g*h

from energy conservation

m*g*h = m*g*2R + (1/2)*m*vtop^2 + (1/2)*I*w^2

I = (2/5)*m*vtop^2

w = vtop/R

m*g*h = m*g*2R + (1/2)*m*vtop^2 + (1/2)*(2/5)*m*vtop^2

g*h = g*2R + (1/2)*g*R + (1/5)*g*R

h = 2.7*R

h = 0.675 m <<<===========answer