A thin rod of length 1.3 m and mass 88 g is suspended freely from one end. It is pulled to one side and then allowed to swing like a pendulu

Question

A thin rod of length 1.3 m and mass 88 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 2.39 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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Cherry 3 years 2021-08-12T11:25:46+00:00 2 Answers 7 views 0

Answers ( )

    0
    2021-08-12T11:26:54+00:00

    Answer:

    Explanation:

    Given:

    Length, L = 1.3 m

    Mass, M = 88 g

    = 0.088 kg

    Angular velocity, w = 2.39 rad/s

    Inertia of the rod = inertia of a cylinder = 1/12 M × L^2

    Rotational inertia, I = Icm + I

    = 1/12 M × L^2 + M × L/2

    = 0.088 × (1/12 × 1.3^2 + 1.3/2)

    = 0.088 × 0.79

    = 0.05 kgm^2

    Kinetic energy, E = 1/2 × l × w^2

    = 1/2 × 0.05 × 2.39^2

    = 0.142 J

    B.

    Using conservation of total mechanical energy,

    Potential energy = kinetic energy

    M × g × h = 0.142

    h = 0.142/(9.8 × 0.088)

    = 0.164 m

    0
    2021-08-12T11:26:59+00:00

    Answer:

    a) the rod’s kinetic energy at its lowest position is 0.39 J

    b) 0.459 m far above that position the center of mass rises.

    Explanation:

    Given

    L = 1.3m

    m = 88g

    w = 2.39 rad/s

    The rotational inertia of the road about an axis passing through its fixed end is given as follow

    I = Icm + mh^{2}

    \frac{1}{12}mL^{2} + m (\frac{L}{2})^{2}    \\I = \frac{1}{12}0.088*1.3^{2} + 0.088 (\frac{1.3}{2})^{2} = 0.04957 kg*m^{2}

    a)

    The kinetic energy of the rod at the lowest point is given by the following

    K = \frac{1}{2} I w^{2} = \frac{1}{2} * 0.04957 * 4^{2} = 0.39 J

    K is the kinetic energy of the rod at the lowest point

    b)

    From the conservation of the total mechanical energy of the rod

    Ki + Ui = Kf + Uf

    Ki = Uf – Ui + 0

    mgh = K = 0.39

    h = 0.39/mg = 0.39/0.088*9.8 = 0.459 m

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