Understanding Translation and Rotational Energy in Pure Rolling

When an object such as a cylinder, sphere, or wheel rolls without slipping, it exhibits both translational and rotational motion. This condition is known as pure rolling. To understand the energy associated with pure rolling, we need to look at both the translational kinetic energy and the rotational kinetic energy of the object.

Translational Kinetic Energy

Translational kinetic energy ($K_{trans}$) is the energy due to the motion of the center of mass of an object moving with a velocity $v$. It is given by the formula:

$$ K_{trans} = \frac{1}{2}mv^2 $$

where:

• $m$ is the mass of the object
• $v$ is the velocity of the center of mass

Rotational Kinetic Energy

Rotational kinetic energy ($K_{rot}$) is the energy due to the rotation of an object about an axis. It is given by the formula:

$$ K_{rot} = \frac{1}{2}I\omega^2 $$

where:

• $I$ is the moment of inertia of the object about the axis of rotation
• $\omega$ is the angular velocity of the object

Pure Rolling Condition

For pure rolling to occur, the following condition must be met:

$$ v = R\omega $$

where:

• $R$ is the radius of the object

This condition ensures that the point of the object in contact with the surface is momentarily at rest with respect to the surface.

Total Kinetic Energy in Pure Rolling

The total kinetic energy ($K_{total}$) of an object in pure rolling is the sum of its translational and rotational kinetic energy:

$$ K_{total} = K_{trans} + K_{rot} $$

Using the pure rolling condition, we can express the total kinetic energy in terms of either $v$ or $\omega$:

$$ K_{total} = \frac{1}{2}mv^2 + \frac{1}{2}I\left(\frac{v}{R}\right)^2 $$

or

$$ K_{total} = \frac{1}{2}m(R\omega)^2 + \frac{1}{2}I\omega^2 $$

Differences and Important Points

Here is a table summarizing the differences and important points of translational and rotational kinetic energy in the context of pure rolling:

Aspect	Translational Kinetic Energy	Rotational Kinetic Energy
Definition	Energy due to linear motion of the center of mass	Energy due to rotation about an axis
Formula	$K_{trans} = \frac{1}{2}mv^2$	$K_{rot} = \frac{1}{2}I\omega^2$
Dependence on Velocity	Directly proportional to the square of the linear velocity $v$	Proportional to the square of the angular velocity $\omega$
Moment of Inertia	Not applicable	Depends on the shape and mass distribution of the object
Pure Rolling Condition	$v = R\omega$ ensures no slipping	$v = R\omega$ ensures no slipping

Examples

Example 1: Solid Sphere

A solid sphere of mass $m$ and radius $R$ is rolling without slipping. The moment of inertia for a solid sphere about its center is $I = \frac{2}{5}mR^2$. The total kinetic energy is:

$$ K_{total} = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{2}{5}mR^2\right)\left(\frac{v}{R}\right)^2 $$ $$ K_{total} = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 $$ $$ K_{total} = \frac{7}{10}mv^2 $$

Example 2: Hollow Cylinder

A hollow cylinder (like a pipe) of mass $m$ and radius $R$ is rolling without slipping. The moment of inertia for a hollow cylinder about its center is $I = mR^2$. The total kinetic energy is:

$$ K_{total} = \frac{1}{2}mv^2 + \frac{1}{2}mR^2\left(\frac{v}{R}\right)^2 $$ $$ K_{total} = \frac{1}{2}mv^2 + \frac{1}{2}mv^2 $$ $$ K_{total} = mv^2 $$

In these examples, we see how the distribution of mass affects the total kinetic energy of an object in pure rolling motion. The moment of inertia plays a crucial role in determining the rotational kinetic energy, and thus the total kinetic energy of the object.