A solid cylinder first rolls without sliding & then slides without rolling down the same incined plane. The ratio of velocities in the 2 cases, at the bottom of the plane, is

To understand the ratio of velocities for a solid cylinder rolling without sliding and then sliding without rolling down an inclined plane, we need to analyze both scenarios using the principles of physics, particularly energy conservation and dynamics.

Rolling Without Sliding

When a solid cylinder rolls down an incline without sliding, it experiences both translational and rotational motion. The gravitational potential energy at the top of the incline is converted into kinetic energy at the bottom. The total kinetic energy in this case is the sum of translational kinetic energy and rotational kinetic energy.

• Translational Kinetic Energy (TKE): This is given by the formula \( \frac{1}{2} mv^2 \), where \( m \) is the mass of the cylinder and \( v \) is its linear velocity.
• Rotational Kinetic Energy (RKE): This is given by \( \frac{1}{2} I \omega^2 \), where \( I \) is the moment of inertia and \( \omega \) is the angular velocity. For a solid cylinder, \( I = \frac{1}{2} m r^2 \), and since \( \omega = \frac{v}{r} \), we can substitute to get \( RKE = \frac{1}{2} \left(\frac{1}{2} m r^2\right) \left(\frac{v}{r}\right)^2 = \frac{1}{4} mv^2 \).

The total kinetic energy at the bottom of the incline is then:

\( KE_{total} = TKE + RKE = \frac{1}{2} mv^2 + \frac{1}{4} mv^2 = \frac{3}{4} mv^2 \).

Setting this equal to the potential energy at the top, \( mgh \) (where \( h \) is the height of the incline), we have:

\( mgh = \frac{3}{4} mv^2 \).

By simplifying, we find:

\( v^2 = \frac{4}{3} gh \)

Thus, the velocity \( v \) when rolling without sliding is:

\( v = \sqrt{\frac{4}{3} gh} \).

Sliding Without Rolling

In the case where the cylinder slides down the incline without rolling, all of the gravitational potential energy converts into translational kinetic energy. The rotational kinetic energy is zero since there is no rotation.

Thus, we have:

\( mgh = \frac{1}{2} mv^2 \).

By simplifying, we find:

\( v^2 = 2gh \).

So, the velocity \( v \) when sliding without rolling is:

\( v = \sqrt{2gh} \).

Finding the Ratio of Velocities

Now, we can find the ratio of the velocities at the bottom of the incline for both cases:

\( \text{Ratio} = \frac{v_{rolling}}{v_{sliding}} = \frac{\sqrt{\frac{4}{3} gh}}{\sqrt{2gh}}.

By simplifying this expression, we get:

\( \text{Ratio} = \frac{\sqrt{\frac{4}{3}}}{\sqrt{2}} = \frac{2/\sqrt{3}}{\sqrt{2}} = \frac{2}{\sqrt{6}} = \frac{\sqrt{6}}{3} \).

Final Thoughts

Therefore, the ratio of the velocities of the solid cylinder rolling without sliding to that of sliding without rolling down the same inclined plane is \( \frac{\sqrt{6}}{3} \). This analysis highlights how the type of motion affects the distribution of energy and ultimately the speed at which the object reaches the bottom of the incline.