Question

In a circus act, a 84.5 kg trapeze artist starts from rest with the 4.65 m trapeze rope horizontal. What is the tension in the rope when it is vertical?

Answer 1

For this equation we need to use centripetal force, because the trapeze artist starts horizontal, then moves radially under the center point of rotation.


there are three forces at work here in the y-direction
F_w: weight of the trapeze artist
F_t: tension in the rope
F_c: centripetal force

F_t = F_c +F_w
F_t = (mv^2)/r+mg
where we know:
m = 84.5 kg
g = 9.81 m/s^2
r = 4.65 m
v = ? [solve for]

thus we will use the conservation of energy to solve for velocity
PE = KE
mgh = (1/2) iw^2
where i is inertia and w is angular velocity
i = m*r^2 [inertia for a point mass]
i = 84.5*4.65^2
i = 1827.1 kg m^2
w = ? [solve for]
84.5*9.81*4.65 = .5*1827.1*w^2
w = 2.05 rad/sec

convert angular velocity into tangential velocity
v_tangetial = r*w
v_t = 4.65*2.05
v_t = 9.5 m/s
v_t = v [for notation use into former equation]

return to the summed forces in the y-direction
F_t = ( m v^2 ) / r + m g
F_t = (84.5*9.5^2) / 4.65 + 84.5*9.81
F_t = 1640 + 828.9
F_t = 2469 newtons