A bead slides without friction around a loop-the-loop (see figure below). a) The

Question

A bead slides without friction around a loop-the-loop (see figure below). a) The bead is released from rest at a height h = 3.90R. What is its speed at point ? Use the following as necessary: the acceleration due to gravity g, and R.) b) How large is the normal force on the bead at point if its mass is 4.20 kg? c) The bead is released from rest at a height h = 2.10R. What is its speed at point ? (Use the following as necessary: the acceleration due to gravity g, and R.) (d) How large is the normal force on the bead at point if its mass is 5.60 kg?

Explanation / Answer

(a)

energy at h = energy at A

m*g*h = m*g*2R + (1/2)*m*vA^2

g*3.9*R = (g*2R) + (1/2)*VA^2

vA^2/2 = 1.9*g*R

vA = sqrt(3.8*g*R)

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(b)

N = m*vA^2/R - mg

N = 3.8*m*g - mg

N = 2.8*m*g

N = 2.8*4.2*9.8 = 115.25 N

downwards

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(c)

m*g*h = m*g*2R + (1/2)*m*vA^2

g*2.1*R = g*2R + vA^2/2

vA^2/2 = 0.1*g*R

vA = sqrt(0.2*g*R)

(d)

N = mg - m*vA^2/R

N = mg - 0.8*m*g

N = 0.8*5.6*9.8 = 44 N

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