Question

A hollow spherical shell has mass 8.35 kg and radius 0.225 m. It is initially at rest and then rotates about a stationary axis that lies along a diameter with a constant acceleration of 0.885 rad/s^2.

Required:
What is the kinetic energy of the shell after it has turned through 6.00 rev?

Answers

  1. Answer:

    K = 9.41 J

    Explanation:

    The kinetic energy of a spherical shell is given as:

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

    where I = moment of inertia and

    ω = angular velocity

    Let us find I:

    For a hollow sphere:

    I = \frac{2}{3} MR^2

    where M = mass = 8.35 kg

    R = radius = 0.225 m

    => I = \frac{2}{3} * 8.35 * 0.225^2 = 0.282 kgm^2

    Let us find ω:

    Since angular acceleration is constant:

    \omega^2 = \omega_0^2 + 2\alpha(\theta - \theta_0)\\\\\omega_0 = 0\\\\=>\omega^2 = 2\alpha(\theta - \theta_0)\\\\\theta - \theta_0 = 6 * (2 * \pi) = 37.70 rad\\\\\omega^2 = 2 * 0.885 * 37.70 = 66.729

    Therefore, its kinetic energy is:

    K = \frac{1}{2} * 0.282 * 66.729\\\\K = 9.41 J

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