1. An 80.0-kg actor swing from a 4.00-m long rope that is at 80.0 degree angle w

Question

1. An 80.0-kg actor swing from a 4.00-m long rope that is at 80.0 degree angle with respect to the vertical when he starts. At the bottom, he picks up a 50.0-kg costar. a) If his initial velocity is 3.00 m/s, and assuming that there is no friction in the system, what is the maximum height that the two actors can reach on their upward swing? b) Repeat part a) but in this case assume that there is friction in the system and that some of the mechanical energy gets dissipated because of it. Specifically, assume the following: - During the downswing, which involves only the 80.0-kg actor, 10% of his initial mechanical energy is dissipated through friction. - During the upswing, which involves the two partners moving together from the bottom of the swing until they reach the max height, 10% of their initial energy, i.e., the energy they possess right after they embraced each other, dissipates through friction. (Hints: The air friction plays no role during the embrace of the two actors. Also, be cautions how the energy that dissipates is to be used in the equations. Is it positive or negative?)

Explanation / Answer

at the 80 degree position

mechanical energy E2 = M*g*L*(1-cos80) + (1/2)*M*v1^2

Ei = M*g*L*(1-cos80) + (1/2)*M*v^2

at the lowest point

Mechanical energy E2 = (1/2)*M*v2^2

E2 = E1

V2 = sqrt(2*g*L*(1-cos80) + v1^2)

v2 = sqrt(2*9.8*4*(1-cos80) + 3^2)

v2 = 8.60 m/s

after picking up co star , let v be the common speed

v = Mv2/(M + m)

v = (80*8.6)/(80 + 50)

v = 5.3 m/s

let the height reached after collision be h

mechanical energy immediately after collision E3 = (1/2)*(M+m)*v^2

mechanical energy at h , E4 = (M+m)*g*h

E4 = E3

h = v^2/(2g)

h = 5.3^2/(2*9.8)

h = 1.433 m

=====================

(b)

E2 = E1 - 10% of E1

E2 = 0.9*E1

V2 = sqrt(0.9*(2*g*L*(1-cos80) + v1^2))

v2 = sqrt(0.9*(2*9.8*4*(1-cos80) + 3^2))

v2 = 8.15 m/s

after picking up co star , let v be the common speed

v = Mv2/(M + m)

v = (80*8.15)/(80 + 50)

v = 5.02 m/s

let the height reached after collision be h

mechanical energy immediately after collision E3 = (1/2)*(M+m)*v^2

mechanical energy at h , E4 = (M+m)*g*h

E4 = E3 - 10%E3 = 0.9*E3

h = 0.9*v^2/(2g)

h = 0.8*5.02^2/(2*9.8)

h = 1.03 m

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