A student attempts to spin a rubber stopper(m=0.5kg) in a horizontal circle with a radius of 0.75m.If the stopper completes 2.5 revolutions

Question

A student attempts to spin a rubber stopper(m=0.5kg) in a horizontal circle with a radius of 0.75m.If the stopper completes 2.5 revolutions every second,determine the following; a)The centripetal acceleration. b)The centripetal force.​

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Ngọc Diệp 3 years 2021-08-11T18:00:27+00:00 1 Answers 5 views 0

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    2021-08-11T18:02:04+00:00

    Explanation:

    a) ac = r(omega)^2

    where omega = 2.5 rev/s

    r = 0.75 m

    We need to convert rev/s first into rad/s:

    2.5 rev/s × (2pi rad/rev) = 5pi rad/s = omega

    Therefore, the centripetal acceleration ac is

    ac = (0.75 m)×(5pi rad/s)^2

    = 185 m/s^2

    b) Fc = mac

    = (0.5 kg)(185 m/s^2)

    = 92.5 N

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