In the figure, a small block of mass m = 0.030 kg can slide along the frictionle

Question

In the figure, a small block of mass m = 0.030 kg can slide along the frictionless loop-the-loop, with loop radius R = 15 cm. The block is released from rest at point P, at height h = 5R above the bottom of the loop. What are the magnitudes of (a) the horizontal component and (b) the vertical component of the net force acting on the block at point Q? (c) At what height h should the block be released from rest so that it is on the verge of losing contact with the track at the top of the loop? (On the verge of losing contact means that the normal force on the block from the track has just then become zero).

Explanation / Answer

Here ,

height , h = 5R

h = 5 * 0.15 = 0.75 m

let the speed at the bottom is v

v = sqrt(2 * g * h)

v = sqrt(2 * 9.8 * 0.75)

v = 3.834 m/s

Now , at the bottom of the loop

a)

as there is no acceleration in the horizontal direction at the bottom of the loop

the net force in horizontal direction is zero

b)
Vertical component of net force on block = m * v^2/R

Vertical component of net force on block = 0.030 * 3.834^2/.15

Vertical component of net force on block = 2.94 N

the Vertical component of net force on block = 2.94 N

c)

let the height is h

for just losing contact

m * v^2/r = m * g

(2 * g * (h - 2R)) = 9.8

2 * (h - 2 R) = 1

h = 0.80

the minium height needed is 0.80 m

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