Asked by ATT

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?

Answers

Answered by 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
Answered by alex
on a hot summer night
Answered by Wahabu kakooza
Solution and explain
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