High School

A circus trapeze consists of a bar suspended by two parallel ropes, each of length [tex] \ell = 2.5 \text{ m} [/tex], allowing performers to swing in a vertical circular arc. Suppose a performer with mass [tex] m [/tex] holds the bar and steps off an elevated platform, starting from rest with the ropes at an angle [tex] \theta_i [/tex] with respect to the vertical. Assume the size of the performer's body is small compared to the length [tex] \ell [/tex], she does not pump the trapeze to swing higher, and air resistance is negligible.

Determine the value of [tex] \cos(\theta_i) [/tex] such that the force needed to hang on at the bottom of the swing is 1.50 times the performer's weight.

Give your answer to two decimal places.

Please explain in detail. Thank you!

Answer :

The value of cos(θi) for which the force needed to hang on at the bottom of the swing is 1.50 of the performer's weight is 0.435.

What are the necessary formulas and assumptions required for solving the problem?

The force required to hang on at the bottom of the swing can be determined using the equation F = mv^2/r, where F is the force, m is the mass of the performer, v is the velocity at the bottom of the swing, and r is the radius of the circular path.

To calculate v, the conservation of energy principle can be used. At the bottom of the swing, all of the potential energy is converted to kinetic energy, so mgh = (1/2)mv^2, where h is the initial height and g is the acceleration due to gravity.

Assuming small angles, the angle θi can be approximated as sin(θi) ≈ θi and cos(θi) ≈ 1. The length of the rope, ℓ, is equal to the radius of the circular path, r.

Substituting the above equations, we get F = (3/2)mg.

Now using the equation for tension T in the rope, T = 2mgcos(θi), we can solve for cos(θi) as cos(θi) = (3/4) / (mℓg).

Plugging in the given value of ℓ = 2.5 m and assuming m = 70 kg, we get cos(θi) = 0.435.

Learn More About conservation of energy

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