The topic of this post is Rolling along an inclined plane or ramp and here we will do a will study of the combined rotation and translation of a cylinder along a ramp.

## a study of the combined rotation and translation of a cylinder | Rolling along an inclined plane or ramp

When a cylinder starts rolling from the top of a ramp (inclined plane) then its motion is a combination of translational motion and rotational motion. The potential energy of the cylinder at the top of the ramp is converted to linear kinetic energy and rotational kinetic energy, as it starts rolling to the bottom of the incline.

Let’s express this using the following equation.

**m g h = (1/2) m v ^{2} + (1/2) I ω^{2} **…………. (1)

Here *h* is the height of the ramp, **v = linear velocity of the cylinder when it reaches the bottom of the ramp, **& ω is the angular velocity of the cylinder when it reaches the bottom of the ramp.

Expanding equation number 1,

**m g h = (1/2) m v ^{2} + (1/2) I (v/r)^{2} **

=> **m g h = (1/2) m v ^{2} + (1/2) I (v^{2}/r^{2} )** ………… (1)

Now, this equation leads to the expression of v at the bottom of the ramp as follows:

v = [2mgh / ( m + I/r^{2} )]^{(1/2)} ………………….. (2)

## Hollow cylinder

For a hollow cylinder, Moment of Inertia = I = mr^{2}

Hence, the velocity of a hollow cylinder that reaches the bottom of the ramp becomes **v = (gh) ^{1/2} **

## Solid cylinder

For a hollow cylinder, Moment of Inertia = I = (1/2) mr^{2}

Hence, the velocity of a hollow cylinder that reaches the bottom of the ramp becomes **v = (4gh/3) ^{1/2} **

## Comparison between the velocities of the hollow cylinder and the solid cylinder as the roll down a ramp

**The solid cylinder will be going (4/3) ^{1/2} =1.15 times as fast as the hollow cylinder when they reach the bottom of the incline.**