A 1.9 kg , 20-cm-diameter turntable rotates at 150 rpm on frictionless bearings. Two 520 g blocks fall from above, hit the turntable simulta

Question

A 1.9 kg , 20-cm-diameter turntable rotates at 150 rpm on frictionless bearings. Two 520 g blocks fall from above, hit the turntable simultaneously at opposite ends of a diameter, and stick. What is the turntable’s angular velocity, inrpm, just after this event?

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Neala 3 years 2021-07-27T22:04:34+00:00 1 Answers 13 views 0

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  1. Helga
    0
    2021-07-27T22:06:19+00:00
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    Answer: 73.08 rpm

    Explanation:

    Given

    Mass of the table, m(t) = 1.9 kg

    Radius of table, r = 20/2 cm = 10 cm = 0.1 m

    Mass of the block, m(b) = 500 g = 0.5 kg

    Speed of rotation, w(i) = 150 rpm

    Using conservation of angular momentum, where, initial angular momentum is equal to final angular momentum

    L(i) = L(f)

    I(i)w(i) = I(f)w(f)

    w(f) = I(i)w(i) / I(f)

    also, moment of inertia a disk or cylinder is I = 1/2mr²

    Thus, I(i) = 1/2 mr²

    I(i) = 1/2 * 1.9 * 0.1²

    I(i) = 0.0095 kgm²

    I(f) = Σm(i)r(i)² = I(i) + I(block1) + I(block2)

    I(f) = I(i) + I(a) + I(b)

    I(f) = 0.0095 + m(a)r(a)² + m(b)r(b)²

    Remember, r(a) = r(b) = R

    m(a) = m(b) = M, so that

    I(f) = 0.0095 + MR² + MR²

    I(f) = 0.0095 + 2MR²

    I(f) = 0.0095 + 2 * 0.5 * 0.1²

    I(f) = 0.0095 + 0.01

    I(f) = 0.0195 kgm²

    Substituting for values in the first equation

    w(f) = I(i)w(i) / I(f)

    w(f) = (0.0095 * 150) / 0.0195

    w(f) = 1.425 / 0.0195

    w(f) = 73.08 rpm

    Thus, the turntable’s angular velocity is 73.08 rpm

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