A solid cylinder and a solid sphere, having same mass $$M$$ and radius $$R$$, roll down the same inclined plane from top without slipping. They start from rest. The ratio of velocity of the solid cylinder to that of the solid sphere, with which they reach the ground, will be

We need to find the ratio of the final linear velocity of a rolling solid cylinder to that of a rolling solid sphere when they start from rest and roll down the same inclined plane without slipping.

1. Derive the Velocity Formula for Pure Rolling

When an object of mass $$M$$ and radius $$R$$ rolls down an inclined plane of height $$h$$ without slipping, its initial potential energy is converted into both translational and rotational kinetic energy:

$$Mgh = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2$$

Since the object rolls without slipping, its angular velocity is $$\omega = \frac{v}{R}$$. Substituting this and writing the moment of inertia in terms of its radius of gyration ($$I = MK^2$$):

$$Mgh = \frac{1}{2}Mv^2 + \frac{1}{2}(MK^2)\left(\frac{v}{R}\right)^2$$

$$Mgh = \frac{1}{2}Mv^2 \left(1 + \frac{K^2}{R^2}\right)$$

Canceling out the mass ($$M$$) and isolating the velocity ($$v$$) gives:

$$v = \sqrt{\frac{2gh}{1 + \frac{K^2}{R^2}}}$$

Since both objects fall from the same height ($$h$$), their velocities are inversely proportional to their rolling factor:

$$v \propto \frac{1}{\sqrt{1 + \frac{K^2}{R^2}}}$$

2. Calculate the Rolling Factors

• For a Solid Cylinder:

  The moment of inertia is $$I_{\text{cylinder}} = \frac{1}{2}MR^2 \implies \frac{K^2}{R^2} = \frac{1}{2}$$

  $$1 + \frac{K^2}{R^2} = 1 + \frac{1}{2} = \frac{3}{2}$$

• For a Solid Sphere:

  The moment of inertia is $$I_{\text{sphere}} = \frac{2}{5}MR^2 \implies \frac{K^2}{R^2} = \frac{2}{5}$$

  $$1 + \frac{K^2}{R^2} = 1 + \frac{2}{5} = \frac{7}{5}$$

3. Determine the Ratio ($$\frac{v_{\text{cylinder}}}{v_{\text{sphere}}}$$)

Taking the ratio of their final linear velocities:

$$\frac{v_{\text{cylinder}}}{v_{\text{sphere}}} = \sqrt{\frac{1 + \left(\frac{K^2}{R^2}\right)_{\text{sphere}}}{1 + \left(\frac{K^2}{R^2}\right)_{\text{cylinder}}}}$$

Substitute the calculated values into the ratio expression:

$$\frac{v_{\text{cylinder}}}{v_{\text{sphere}}} = \sqrt{\frac{\frac{7}{5}}{\frac{3}{2}}} = \sqrt{\frac{7}{5} \times \frac{2}{3}} = \sqrt{\frac{14}{15}}$$

Conclusion

The ratio of the velocity of the solid cylinder to that of the solid sphere is $$\sqrt{\frac{14}{15}}$$