A hollow spherical ball of uniform density rolls up a curved surface with an initial velocity 3 m/s (as shown in figure). Maximum height with respect to the initial position covered by it will be_____ cm (take, g = 10 m/s²)
For pure rolling motion, mechanical energy is conserved. At the initial position (\( A \)), the ball has kinetic energy due to both translational and rotational motion, and at the maximum height (\( B \)), all the kinetic energy is converted into potential energy. Step 1: Write the conservation of energy equation. The total mechanical energy at \( A \) is: \[ \text{(M.E.)}_A = \frac{1}{2} m v_0^2 + \frac{1}{2} I \omega^2, \] where \( m \) is the mass of the ball, \( v_0 = 3 \, \text{m/s} \) is the initial velocity, \( I \) is the moment of inertia of the hollow sphere (\( I = \frac{2}{3} m R^2 \)), and \( \omega = \frac{v_0}{R} \) is the angular velocity. At the maximum height (\( B \)), all kinetic energy is converted into potential energy: \[ \text{(M.E.)}_B = m g h_{\text{max}}, \] where \( h_{\text{max}} \) is the maximum height. Using energy conservation: \[ \text{(M.E.)}_A = \text{(M.E.)}_B. \] Step 2: Substitute the expressions. \[ \frac{1}{2} m v_0^2 + \frac{1}{2} \left(\frac{2}{3} m R^2\right) \left(\frac{v_0}{R}\right)^2 = m g h_{\text{max}}. \] Simplify: \[ \frac{1}{2} m v_0^2 + \frac{1}{3} m v_0^2 = m g h_{\text{max}}. \] Combine terms: \[ \frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6}. \] \[ \frac{5}{6} m v_0^2 = m g h_{\text{max}}. \] Cancel \( m \) on both sides: \[ h_{\text{max}} = \frac{\frac{5}{6} v_0^2}{g}. \] Step 3: Substitute values. Substitute \( v_0 = 3 \, \text{m/s} \) and \( g = 10 \, \text{m/s}^2 \): \[ h_{\text{max}} = \frac{\frac{5}{6} \cdot (3)^2}{10}. \] Simplify: \[ h_{\text{max}} = \frac{\frac{5}{6} \cdot 9}{10} = \frac{45}{60} = 0.75 \, \text{m}. \] Convert to centimeters: \[ h_{\text{max}} = 0.75 \times 100 = 75 \, \text{cm}. \] Final Answer: The maximum height covered is: \[ \boxed{75 \, \text{cm}}. \]
