A ball on the end of a string is whirled around in a horizontal circle of radius 0.300 m. The plane of the circle is 1.50 m above the ground

Question

A ball on the end of a string is whirled around in a horizontal circle of radius 0.300 m. The plane of the circle is 1.50 m above the ground. The string breaks and the ball lands 2.10 m (horizontally) away from the point on the ground directly beneath the ball’s location when the string breaks. Find the radial acceleration of the ball during its circular motion. Magnitude

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King 4 years 2021-09-02T00:23:29+00:00 1 Answers 15 views 0

Answers ( )

  1. niczorrrr
    0
    2021-09-02T00:24:57+00:00
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    Answer:

    48.13 m/s²

    Explanation:

    In this question, circular motion of the ball attached to string turns into trajectory motion.

    Radial acceleration = (v²/r)

    But we need to find the velocity, v, from the ball’s trajectory motion after the string breaks.

    For trajectory motion,

    Range = (initial horizontal velocity) × (time of flight) = (v)(T)

    v = (Range)/T

    But time of flight is related to the vertical height, H, at which the trajectory motion starts through the relation,

    T = √(2H/g) = √(2×1.2/9.8) = 0.553 s

    v = (Range)/T

    v = 2.1/0.553

    v = 3.80 m/s

    Radial acceleration = (v²/r) = (3.8²/0.3)

    a = 48.13 m/s²

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