Question

The total energy of a solid sphere of mass 300 g which rolls without slipping with a constant velocity of 5 ms\(^{-1}\) along a straight line is

• A. 5.25 J
• B. 3.25 J
• C. 0.25 J
• D. 1.25 J
• E. 0.625 J

Hint:

For any rolling object, total energy can be written as \(\frac{1}{2}mv^2 (1 + \frac{k^2}{r^2})\), where \(k^2/r^2\) is 2/5 for a solid sphere, 2/3 for a hollow sphere, and 1/2 for a solid cylinder.

Answer 1

The Correct Option is A

Concept: An object rolling without slipping possesses both translational kinetic energy and rotational kinetic energy.
• Translational KE: \(K_T = \frac{1}{2}mv^2\).
• Rotational KE: \(K_R = \frac{1}{2}I\omega^2\).
• Rolling Condition: \(v = r\omega\).
• Moment of Inertia: For a solid sphere, \(I = \frac{2}{5}mr^2\).

Step 1: Derive the total kinetic energy formula.
\[ K_{total} = \frac{1}{2}mv^2 + \frac{1}{2} \left( \frac{2}{5}mr^2 \right) \left( \frac{v}{r} \right)^2 \] \[ K_{total} = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2 \]

Step 2: Perform the numerical calculation.
Given \(m = 300 \text{ g} = 0.3 \text{ kg}\) and \(v = 5 \text{ ms}^{-1}\): \[ K_{total} = \frac{7}{10} \times 0.3 \times (5)^2 \] \[ K_{total} = \frac{7}{10} \times 0.3 \times 25 \] \[ K_{total} = 0.7 \times 7.5 = 5.25 \text{ J} \]