A solid sphere of mass 1 kg rolls without slipping on a plane surface. Its kinetic energy is 7 × 10−3J. The speed of the centre of mass of the sphere is _______ cm s−1.

JEE Mains - Rolling Solid Sphere

Question: A solid sphere of mass \(1\,\text{kg}\) rolls without slipping on a plane surface. Its total kinetic energy is given as \(7 \times 10^{-3}\,\text{J}\). Determine the speed of the center of mass of the sphere in \(\text{cm/s}\).
[31-Jan-2023 Shift 1]

Select the correct answer:

(A) \(5\,\text{cm/s}\)

(B) \(10\,\text{cm/s}\)

(C) \(15\,\text{cm/s}\)

(D) \(20\,\text{cm/s}\)

Solution

For a rolling object, the total kinetic energy is the sum of translational and rotational kinetic energies. For a solid sphere: \[ KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2. \] The moment of inertia for a solid sphere is \( I = \frac{2}{5}mR^2 \), and since it rolls without slipping, \( v = \omega R \). Therefore: \[ KE = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{2}{5}mR^2\right)\left(\frac{v}{R}\right)^2 = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2. \]

Given \( m = 1\,\text{kg} \) and \( KE = 7 \times 10^{-3}\,\text{J} \): \[ \frac{7}{10}v^2 = 7 \times 10^{-3}. \] Solving for \( v^2 \): \[ v^2 = \frac{7 \times 10^{-3} \times 10}{7} = 10^{-2}. \] Thus, \[ v = \sqrt{10^{-2}} = 0.1\,\text{m/s}. \]

Converting to \(\text{cm/s}\), since \(1\,\text{m/s} = 100\,\text{cm/s}\): \[ v = 0.1 \times 100 = 10\,\text{cm/s}. \]

Therefore, the speed of the center of mass of the sphere is 10 cm/s.