A solid disk and a thin-walled hoop each have a diameter of 8 cm. Both are released from rest at the same time at the top of a ramp that is

Question

A solid disk and a thin-walled hoop each have a diameter of 8 cm. Both are released from rest at the same time at the top of a ramp that is 175 cm high and start rolling down the incline. What is the velocity of each at the bottom of the ramp? What percentage of the gravitational potential energy has been transformed into angular kinetic energy for each? Into what form of energy is the remaining percentage transformed?

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Sigridomena 19 hours 2021-07-21T13:09:14+00:00 1 Answers 0 views 0

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    2021-07-21T13:11:05+00:00

    Answer:

    Explanation:using

    Mgh= 1/2mv²+1/2Iw²

    So for disk

    mgh= 1/2mv²+1/2(1/2mR²)v²/R²

    gh = 0.75v²

    V= 4.8m/s

    Then the translational KE

    = (1/2mv²)/mgh x 100

    = 66.7%

    So rotational energy= 100- 67.7%

    =33.3%

    For hoop

    mgh= 1/2mv²+1/2(1/2mR²)v²/R²

    V= 4.14m/s

    Then the translational KE

    = (1/2mv²)/mgh x 100

    = 50%

    Then the remaining 50% is translational energy

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