Question

A ball rolls without slipping. The radius of gyration of the ball about an axis passing through its centre of mass is K. If radius of the ball be R, then the fraction of total energy associated with its rotational energy will be:

• A. \(\frac {K^2+R^2}{R^2}\)
• B. \(\frac {K^2}{R^2}\)
• C. \(\frac {K^2}{K^2+R^2}\)
• D. \(\frac {R^2}{K^2+R^2}\)

Answer 1

The Correct Option is C

To solve this problem, we need to find the fraction of the total energy of a rolling ball that is associated with its rotational energy. Let's go through the steps:

1. When a ball rolls without slipping, it possesses both translational and rotational kinetic energies. The total kinetic energy (\(E_{\text{total}}\)) of the ball is the sum of its translational kinetic energy (\(E_{\text{trans}}\)) and rotational kinetic energy (\(E_{\text{rot}}\)).
2. The translational kinetic energy is given by: E_{\text{trans}} = \frac{1}{2} m v^2, where \(m\) is the mass of the ball and \(v\) is the velocity of the center of mass of the ball.
3. The rotational kinetic energy is given by: E_{\text{rot}} = \frac{1}{2} I \omega^2, where \(I\) is the moment of inertia about the axis through the center of mass and \(\omega\) is the angular velocity.
4. For a ball rolling without slipping, we have the condition: \omega = \frac{v}{R}. Thus, the rotational kinetic energy can be expressed as: E_{\text{rot}} = \frac{1}{2} I \left(\frac{v}{R}\right)^2.
5. The moment of inertia \(I\) of the ball about its center of mass is related to the radius of gyration \(K\) by the equation: I = m K^2.
6. Substituting the expression for \(I\) into the equation for \(E_{\text{rot}}\), we get: E_{\text{rot}} = \frac{1}{2} m K^2 \left(\frac{v}{R}\right)^2
   = \frac{1}{2} \frac{m K^2 v^2}{R^2}.
7. Thus, the total energy is: E_{\text{total}} = E_{\text{trans}} + E_{\text{rot}}
   = \frac{1}{2} m v^2 + \frac{1}{2} \frac{m K^2 v^2}{R^2}.
   = \frac{1}{2} m v^2 (1 + \frac{K^2}{R^2}).
8. The fraction of the total energy that is rotational is given by: \[ \text{Fraction} = \frac{E_{\text{rot}}}{E_{\text{total}}} = \frac{\frac{1}{2} \cdot \frac{m K^2 v^2}{R^2}}{\frac{1}{2} m v^2 (1 + \frac{K^2}{R^2})} = \frac{K^2}{K^2 + R^2} \]

Thus, the correct option is \(\frac {K^2}{K^2+R^2}\).