Katya is training for an upcoming gymnastics competition and needs to improve her upper body strength. At the moment, she can only support a

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?

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Maris 1 week 2021-07-21T05:07:51+00:00 1 Answers 3 views 0

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    2021-07-21T05:09:23+00:00

    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

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