Yesterday, I went bowling and while I was waiting for my turn, I decided to compute the total kinetic energy of a rolling ball. Treat the ba

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Yesterday, I went bowling and while I was waiting for my turn, I decided to compute the total kinetic energy of a rolling ball. Treat the ball as a solid sphere with a diameter of 22.0 cm and a mass of 5.00 kg, the ball rolls down the lane without slipping at a speed of 15.0 km/h. What did I find for the total kinetic energy

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Trung Dũng 4 years 2021-08-12T13:19:27+00:00 1 Answers 7 views 0

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  1. thaiduong
    0
    2021-08-12T13:20:58+00:00
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    Answer:

    Ktot = 60.8 J

    Explanation:

    • The total kinetic energy of a rolling ball, is composed by a rotational kinetic energy (due to the rotation around a diameter), plus a traslational kinetic energy, due to the translation of the center of  mass.
    • The rotational kinetic energy can be written as follows:

            K_{rot}= \frac{1}{2}* I * \omega^{2} (1)

    • where I = moment of  Inertia of a solid sphere, and  ω², the square of  the angular speed.
    • If the ball rolls without slipping, there exists a fixed  relationship between the angular speed and the speed of the center of mass, as follows:

           \omega = \frac{v_{CM} }{R} (2)

    • The traslational part, can be expressed as follows:

            K_{trasl} = \frac{1}{2} * m * v_{CM} ^{2} (3)

    • For a solid sphere, like a bowling ball, the moment of inertia about an axis passing through a diameter can be written as follows:

           I =\frac{2}{5} * m * R^{2} (4)

    • Replacing (2) and  (4) in (1), we have:

           K_{rot}= \frac{1}{2}* \frac{2}{5} * m * R^{2} * (\frac{v_{CM} }{R}) ^{2} = \frac{1}{5} * m* v_{CM} ^{2} (5)

    • The total kinetic energy is just the sum of (3) and (5):

           K_{tot} = \frac{1}{5} * m* v_{CM} ^{2} + \frac{1}{2} * m * v_{CM} ^{2} = \frac{7}{10} * m* v_{CM} ^{2} (6)

    • The speed of the center of  mass  is given in km/h, so it is needed to convert it to m/s, as follows:

            v_{CM} = 15.0 \frac{km}{h} * \frac{1000m}{1km} * \frac{1h}{3600s} = 4.17m/s (7)

    • Replacing (7) and m=5.00 kg in (6), we get the total kinetic energy of the rolling ball, as follows:

           K_{tot} = \frac{7}{10} * 5.00 kg * (4.17 m/s) ^{2} = 60.8 J

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