height at top of loop = 2 R
so A is (3.5 - 2) = 1.5 R below drop point
so
(1/2) m v^2 = m g (1.5 R)
v^2 = 3 g R
F =m (v^2/R-g) = m (3g-g) = 2 m g
m = .005
g = 9.81
A bead slides without friction around a loop-the-loop. The bead is released from a
height h = 3.5R. A is at the highest point of the loop and R is the radius of the loop
(a) What is its speed at point A?
(b) How large is the normal force on it if its mass is 5g?
3 answers
on a hot summer night
Solution and explain