A thin rod of length 0.83 m and mass 110 g is suspended freely from one end. It is pulled to one side and then allowed to swing like a pendu

Question

A thin rod of length 0.83 m and mass 110 g is suspended freely from one end. It is pulled to one side and then allowed to swing like a pendulum, passing through its lowest position with angular speed 5.71 rad/s. Neglecting friction and air resistance, find (a) the rod’s kinetic energy at its lowest position and (b) how far above that position the center of mass rises.

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Ngọc Hoa 6 months 2021-08-18T18:04:02+00:00 2 Answers 7 views 0

Answers ( )

    0
    2021-08-18T18:05:10+00:00

    Answer:

    Explanation:

    length of rod, L = 0.83 m

    mass of rod, M = 110 g = 0.11 kg

    angular speed, ω = 5.71 rad/s

    Moment of inertia of rod about the fixed point

    I = Icm + Mh²

    I = 1/12 M L² + M L²/4

    I = 1/12 x 0.11 x 0.83 x 0.83 + 0.11 x 0.83 x 0.83 / 4

    I = 0.0063 + 0.01895 = 0.0252 kgm²

    a) kinetic energy of rod is given by

    K = 1/2 Iω²

    K = 0.5 x 0.0252 x 5.71 x 5.71 = 0.41 Joule

    (b) The kinetic energy at the bottom  is equal to the potential energy at a height of the center of mass. Let the centre of mass rises upto height h.

    mgh = K

    0.11 x 9.8 x h = 0.41

    h = 0.381 m

    h = 38.1 cm

    0
    2021-08-18T18:05:10+00:00

    Explanation:

    (a)  The given data is as follows.

        Length of the rod, L = 0.83 m

        Mass of the rod, m = 110 g = 0.11   (as 1 kg = 1000 g)

     At the lowest point, angular speed of the rod (\omega) = 5.71 rad/s

    First, we will calculate the rotational inertia of the rod about an axis passing through its fixed end is as follows.

          I = I_{CM} + mh^{2}

            = \frac{1}{12}mL^{2} + m(\frac{L}{2})

            = \frac{1}{12} \times 0.11 \times (0.83)^{2} + 0.11 \times \frac{0.83}{2}

            = 0.00631 + 0.415

            = 0.42131 kg m^{2}

    Therefore, kinetic energy of the rod at its lowest point will be calculated as follows.

                 K = \frac{1}{2}I \omega^{2}

                    = \frac{1}{2} \times 0.42131 kg m^{2} \times (5.71)^{2}

                    = 6.86 J

    Hence, kinetic energy of the rod at its lowest point is 6.86 J.

    (b)   According to the conservation of total mechanical energy of the rod, we have

             K_{i} + U_{i} = K_{f} + U_{f}

               K_{i} = U_{f} - U_{i}

    or,      mgh = K = 6.86 J

    Therefore,      h = \frac{6.86}{mg}

                              = \frac{0.63}{0.11 \times 9.8}

                              = 0.584 m

    Hence, the center of mass rises 0.584 m far above that position.

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