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31 December, 21:26

A cylindrical 4.00-kg reel with a radius of 0.60 m and a frictionless axle, starts from rest and speeds up uniformly as a 3.00-kg bucket falls into a well, making a light rope unwind from the reel (Fig. P8.36). The bucket starts from rest and falls for 6.00s. (a) What is the linear acceleration of the falling bucket?

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  1. 31 December, 21:34
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    Answer: a=2.45m/s/s

    Step-by-step explanation:

    By newton's second law

    The forces on the bucket are the tension in the string and the weight of the bucket; the bucket accelerates down, so we have

    T - mg = - ma (eq. 1) (m is mass of bucket)

    the tension exerts a torque on the reel of magnitude TR where R is the radius of the reel; this torque causes an angular acceleration A, such that

    torque = I A

    the moment of inertia of a cylindrical disk is 1/2 MR^2 where M is the mass of the reel, so we have

    torque = T R = I A = 1/2 MR^2 A

    or T=1/2 MR A

    the angular acceleration is related to linear acceleration via

    a = RA, so we get

    T=1/2 MR (a/R) = 1/2Ma

    use this in equation (1) to get

    T-mg=-ma

    1/2Ma-mg=-ma

    a (1/2M + m) = g

    a=g / (1/2 M + m)

    using relevant values, we get:

    a=9.8m/s/s / (0.5x4 kg + 2.00kg)

    a=2.45m/s/s
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