A ring and a solid sphere are released from rest from the same height on enough rough inclined surface. The ratio of their speed when they reach the bottom is \( \sqrt{\frac{7}{x}} \) m/s, then \( x \) is ______.
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When comparing the velocities of a rolling solid sphere and a ring, consider the ratio of their moments of inertia. A solid sphere has a lower moment of inertia compared to a ring, leading to a higher speed when it reaches the bottom of the incline.
Upon release from rest on an inclined plane, a solid sphere and a ring both exhibit rolling motion. For an object rolling without slipping, its total kinetic energy is the sum of its translational kinetic energy and rotational kinetic energy. The total kinetic energy of a rolling object is expressed as:
\[
K_{\text{total}} = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2
\]
Where:
- \( m \) denotes the object's mass,
- \( v \) represents the linear velocity,
- \( I \) is the moment of inertia,
- \( \omega \) signifies the angular velocity.
The relationship between linear and angular velocity for rolling motion without slipping is \( v = r\omega \), where \( r \) is the object's radius.
The moment of inertia for a solid sphere is \( I_{\text{sphere}} = \frac{2}{5} m r^2 \), and for a ring, it is \( I_{\text{ring}} = m r^2 \).
Applying the principle of conservation of mechanical energy, the potential energy lost by the objects descending the incline transforms into kinetic energy. Potential energy is given by \( mgh \), and the total energy at the incline's base is the sum of translational and rotational kinetic energies.
For the solid sphere:
\[
mgh = \frac{7}{10} m v_{\text{sphere}}^2
\]
For the ring:
\[
mgh = m v_{\text{ring}}^2
\]
Consequently, the ratio of the final velocities is:
\[
\frac{v_{\text{sphere}}}{v_{\text{ring}}} = \sqrt{\frac{7}{10}} = \sqrt{\frac{7}{x}}
\]
Solving for \( x \) yields \( x = 4 \).
Therefore, the correct value is \( x = 4 \).
