The small mass m sliding without friction along the looped track shown in the fi

Question

The small mass m sliding without friction along the looped track shown in the figure(Figure 1) is to remain on the track at all times, even at the very top of the loop of radius r.

If the actual release height is 5 h, calculate the normal force exerted by the track at the bottom of the loop.

THE ANSWER CANNOT CONTAIN mg

If the actual release height is 2 h, calculate the normal force exerted by the track at the top of the loop.

If the actual release height is 6 h, calculate the normal force exerted by the track after the block exits the loop onto the flat section.

Note: It says the answer does not depend on mg so I'm not clear how to figure out the answer then. Thanks for the help.

Explanation / Answer

According to the given problem,

As for the mass to remain on the track at the top of the loop, the centrifugal force there must at least equal the weight of the mass

Centrifugal Force = m*v2/r

Weight = m*g

Condition to stay on track C.F > Weight; m*v2/r = m*g; v > [r*g]

The kinetic energy at the top of the loop is then 1/2*m*v2= 1/2*m*r*g

The potential energy change from the release point to the top of the loop PE = m*g*y

y = h - 2*r; PE = m*g*(h - 2*r)

Equate KE to PE to get 1/2*m*r*g = m*g*(h - 2*r)

1/2*r = h - 2*r

h > (½*r + 2*r) = 2.5*r

I.)Given H = 5h = 12.5*r

At the bottom of the loop

PE = m*g*H = m*g*12.5*r
KE = 1/2*m*v2
v2 = 25*g*r
C.F = m*v2/r = 25*m*g
Weight = m*g

Total force against track at bottom = 26*m*g

II.) Given H = 2h = 5*r

At the top of the loop

PE = m*g*(H - 2*r) = m*g*(5*r - 2*r) = m*g*3*r

KE = 1/2*m*v2

m*g*3*r = 1/2*m*v2
v2 = 6*g*r
CF = m*v2/r = 6.0*m*g

normal force = CF - m*g = 5.0*m*g

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