A circus trapeze consists of a bar suspended by two parallel ropes, each of leng

Question

A circus trapeze consists of a bar suspended by two parallel ropes, each of length , allowing performers to swing in a vertical circular arc. Suppose a performer with mass m holds the bar and steps off an elevated platform, starting from rest with the ropes at an angle i with respect to the vertical. Suppose the size of the performer's body is small compared to the length , that she does not pump the trapeze to swing higher, and that air resistance is negligible.

- Determine the angle i for which the force needed to hang on at the bottom of the swing is 1.55 times the performer's weight.
___ °

Explanation / Answer

m = mass
g = gravity
h = height
v = velocity
L = length of rope

Potential (at the top) = m g h
Kinetic (at the bottom) = 1/2 m v²

m g h = 1/2 m v²
g h = 1/2 v²
v² = 2 g h

But we don't know height. We can, however, write height as a trig component of the angle i and the rope length. Draw a triangle to make sure you've got the correct function and understand why we use (L - h).
Cos[i] = (L - h)/L
L Cos[i] = L - h
h = L - L Cos[i]

Now go back to our energy equation.
v² = 2 g h
v² = 2 g (L - L Cos[i])

Here's where the wording can be tricky. At the bottom we need F = 1.55 m g. With the trap at rest, you've already got 1 m g. So let's just minus that out as given and only deal with centripetal force.

F(needed) = 1.55 m g - 1 m g = 0.55 m g

Centripetal force is given as F = m v² / r, where r = L in this problem. Just plug and chug... it's nice that we only had to solve for v² earlier rather than taking a root.
F = m v² / L
0.55 m g = m v² / L
0.55 g = v² / L
0.55 g = (2 g (L - L Cos[i])) / L
0.55 = 2 - 2 Cos[i]
1.45 = 2 Cos[i]
0.725 = Cos[i]
i = cos^-1[0.725]
i = 43.53°

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.