Question

Katya is training for an upcoming gymnastics competition and needs to improve her upper body strength. At the moment, she can only support at most 180% of her body weight when she’s hanging off of equipment. Misha, a close friend and circus performer, suggests that she train on a swinging trapeze. Katya grabs onto the trapeze which hangs off a pair of cables of length ` that initially make an angle of 60.0◦ with the vertical. She steps off the platform and swings forward. To what height will she fall/rise as she swings? Will Katya swing back to the platform?

Answer 1

Answer:

she will fall to an height of 21.09° only, beyond such an height ; the spring will break.

Katya will not be able to swing to the platform because the angle will be between -21.09° and 21.09° from vertical

Explanation:

If the distance is from the point of angle A to angle B , partitioned by a perpendicular line in the middle then, The conservation of energy between A and B can be expressed as :

$KE_A +PE_B = KE_B + PE_B$

$0 + mgh = \frac{1}{2}mv_B^2+ 0$

where ; the height h = $l ( cos \theta - \frac{1}{2})$

$mgl ( cos \theta - \frac{1}{2}) = \frac{1}{2}mv^2_B$

$\frac{mv_B^2}{l} = mg (2 cos \theta -1 )$

$T = \frac{mv_B^2}{l}+ mg cos \theta$

T = mg(3 cos θ - 1)

Given that:

T = 180 % = 1.8 mg

Then:

1.8 mg = mg(3 cos θ - 1)

2.8 mg = (3 cos θ - 1)

cos θ = $\frac{2.8}{3}$

θ = cos ⁻¹ (0.933)

θ = 21.09°

Therefore, she will fall to an height of 21.09° only, beyond such an height ; the spring will break.

Katya will not be able to swing to the platform because the angle will be between -21.09° and 21.09° from vertical