The velocity of a rolling object at the bottom of an inclined plane is given by:
\[ v = \sqrt{\frac{2gh}{1 + \frac{I}{mr^2}}}, \]
where:
For different objects:
Comparison:
Since \( \frac{2}{5} < \frac{1}{2} < 1 \), the solid sphere has the smallest value of \( \frac{I}{mr^2} \) and thus the largest velocity.
Conclusion: The solid sphere reaches the bottom with maximum velocity.
A force of 49 N acts tangentially at the highest point of a sphere (solid) of mass 20 kg, kept on a rough horizontal plane. If the sphere rolls without slipping, then the acceleration of the center of the sphere is
If the roots of $\sqrt{\frac{1 - y}{y}} + \sqrt{\frac{y}{1 - y}} = \frac{5}{2}$ are $\alpha$ and $\beta$ ($\beta > \alpha$) and the equation $(\alpha + \beta)x^4 - 25\alpha \beta x^2 + (\gamma + \beta - \alpha) = 0$ has real roots, then a possible value of $y$ is: