Aroller coaster car crosses the top of a circular loop-the-loop at twice the critical speed.  what is the ratio of the normal force to the gravitational force?

R = radius of the loop

m = mass of the roller coaster car

for looping the loop critical speed at the top is given as

v₀ = critical speed at the top = sqrt(2gR)

given that :

v = speed of crossing at the top = 2 v₀ = 2  sqrt(2gR)

F = normal force at the top in downward direction

W = weight in downward direction = mg

force equation at the top is given as

F + W = m v²/R

F + mg = m (2  sqrt(2gR))²/R

F + mg = m (4 x 2 gR)/R

F + mg = m (8gR)/R

F + mg = 8 mg

F = 7 mg

F/(mg) = 7

so the ratio comes out to be 7:1

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg