A hollow cylinder of mass 2.00 kgkg, inner radius 0.100 mm, and outer radius 0.200 mm is free to rotate without friction around a horizontal

Question

A hollow cylinder of mass 2.00 kgkg, inner radius 0.100 mm, and outer radius 0.200 mm is free to rotate without friction around a horizontal shaft of radius 0.100 mm along the axis of the cylinder. You wrap a light, nonstretching cable around the cylinder and tie the free end to a 0.500 kgkg block of cheese. You release the cheese from rest a distance hh above the floor. If the cheese is moving downward at 4.00 m/sm/s just before it hits the ground, what is the value of hh? Express your answer with the appropriate units.

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Thiên Ân 5 years 2021-07-20T14:34:27+00:00 2 Answers 243 views 0

Answers ( )

  1. thanhcong
    1
    2021-07-20T14:35:37+00:00
    Reply

    Answer: 2.86 m

    Explanation:

    To solve this question, we will use the law of conservation of kinetic and potential energy, which is given by the equation,

    ΔPE(i) + ΔKE(i) = ΔPE(f) + ΔKE(f)

    In this question, it is safe to say there is no kinetic energy in the initial state, and neither is there potential energy in the end, so we have

    mgh + 0 = 0 + KE(f)

    To calculate the final kinetic energy, we must consider the energy contributed by the Inertia, so that we then have

    mgh = 1/2mv² + 1/2Iw²

    To get the inertia of the bodies, we use the formula

    I = [m(R1² + R2²) / 2]

    I = [2(0.2² + 0.1²) / 2]

    I = 0.04 + 0.01

    I = 0.05 kgm²

    Also, the angular velocity is given by

    w = v / R2

    w = 4 / (1/5)

    w = 20 rad/s

    If we then substitute these values in the equation we have,

    0.5 * 9.8 * h = (1/2 * 0.5 * 4²) + (1/2 * 0.05 * 20²)

    4.9h = 4 + 10

    4.9h = 14

    h = 14 / 4.9

    h = 2.86 m

  2. Ladonna
    0
    2021-07-20T14:35:58+00:00
    Reply

    Answer:

    The value of h is 2.86 m

    Explanation:

    Applying the energy conservation:

    E_{p,i} +E_{k,i} =E_{p,f} +E_{k,f} \\mgh+0=0+\frac{1}{2} mv^{2} +\frac{1}{2} Iw^{2} \\I=\frac{m(R_{1}^{2}+R_{2}^{2}) }{2}

    Where

    m = 2 kg

    R₁ = 0.2 m

    R₂ = 0.1 m

    I=\frac{2*(0.2^{2}+0.1^{2}) }{2} =0.05kgm^{2}

    The angular speed is:

    w = v/R₂ = 4/0.2 = 20 rad/s

    h=\frac{\frac{1}{2}mv^{2}+\frac{1}{2}Iw^{2} }{mg} =\frac{\frac{1}{2}*2*4^{2}+\frac{1}{2}*0.05*20^{2} }{2*9.8} =2.86m

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