Question
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?
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?
Answers
Damon
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
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
alex
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Wahabu kakooza
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