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

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 = 3.10 R. What is its speed at point A? (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 A if its mass is 5.80 kg? (c) The bead is released from rest at a height h = 2.10R. What is its speed at point A? (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 A if its mass is 4.10 kg?

Explanation / Answer

a) Apply conservation of mechanical energy

mecahincal energy at point A = initial mechanical energy

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

g*2*R + (1/2)*v^2 = g*3.1*R

(1/2)*v^2 = 3.1*g*R - 2*g*R

v^2 = 2*1.1*g*R

v = sqrt(2.2*g*R )

b) use, Fnety = m*g + N

m*a_rad = m*g + N

m*v^2/R = m*g + N

==> N = m*v^2/R - mg

= m*(2.2*g*R)/R - m*g

= 1.2*m*g

= 1.2*5.8*9.8

= 68.2 N

direction : downward

c)

Apply conservation of mechanical energy

mecahincal energy at point A = initial mechanical energy

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

g*2*R + (1/2)*v^2 = g*2.1*R

(1/2)*v^2 = 2.1*g*R - 2*g*R

v^2 = 2*0.1*g*R

v = sqrt(0.2*g*R )

d) use, Fnety = m*g + N

m*a_rad = m*g + N

m*v^2/R = m*g + N

==> N = m*v^2/R - mg

= m*(0.2*g*R)/R - m*g

= 0.8*m*g

= 0.8*4.1*9.8

= 32.1 N

direction : upward

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