A solid sphere of mass 2 kg is making pure rolling on a horizontal surface with kinetic energy 2240 J. The velocity of centre of mass of the sphere will be ____ ms–1.
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For rolling motion: • Include both translational and rotational kinetic energies. • Use the pure rolling condition ω = v/r to simplify calculations.
To determine the velocity of the center of mass of the sphere, we start by noting the following: For a solid sphere rolling without slipping, the total kinetic energy is the sum of translational and rotational kinetic energy. Given the kinetic energy \(K = 2240 \, \text{J}\), we can express it as:
\(K = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2\)
For a solid sphere, the moment of inertia \(I\) about its center is \(\frac{2}{5}mr^2\). Also, for pure rolling, the relation between linear velocity \(v\) and angular velocity \(\omega\) is \(v = r\omega\).
Thus, the rotational kinetic energy can be rewritten as: \(\frac{1}{2}I\omega^2 = \frac{1}{2} \times \frac{2}{5}mr^2 \times \left(\frac{v}{r}\right)^2 = \frac{1}{5}mv^2\)
Substituting these into the kinetic energy equation, we get: \(K = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2\)
Thus, the velocity of the center of mass of the sphere is \(40 \, \text{ms}^{-1}\). This solution is consistent with the expected range of 40 to 40, confirming its accuracy.
