The energy of an object's translational and/or rotational motion.
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Motivation for Concept
One way to think of energy is the ability to do harm. Things that are very dangerous possess considerable energy. With that definition in mind, the concept of kinetic energy can be motivated by considering the danger posed by moving objects. When a baseball is thrown by a child, it is not very frightening, though it can break a lamp. When a baseball is thrown by a major league pitcher, they can cause considerable injury. This contrast indicates that the energy of motion will depend on speed. Similarly, although a baseball thrown at 60 mph is dangerous, a car driving 60 mph is deadly. This contrast indicates that the energy of motion will depend upon mass.
Mathematical Definition
Translational Kinetic Energy
The kinetic energy of a point particle is given by:
\begin
[ K = \frac
mv^
]\end
Kinetic Energy of a System
Since energy is a scalar, the kinetic energy of a system of point particles is the sum of the kinetic energies of the constituents:
\begin
[ K^
= \sum_
^
\frac
m_
v_
^
]\end
where N is the number of system constituents.
Kinetic Energy of a Rigid Body
Consider a rigid body that can rotate and translate. We begin by treating the rigid body as a collection of point masses that are translating with the center of mass of the body and also rotating about it with angular velocity ω. We therefore write the velocity of each point as a sum of rotational and translational parts:
\begin
[ \vec
_
= \vec
_
+ \vec
\times \vec
_
]\end
where rj is the position of the jth particle measured from the body's axis of rotation passing through the center of mass.
With this split, the kinetic energy of the body becomes:
\begin
[ K = \sum_
\frac
m_
(v_
^
+ \vec
_
\cdot(\vec
\times \vec
_
) + \omega^
r_
^
)]\end
The center term will equal zero, because ω and vcm are constants, so:
\begin
[ \sum_
\frac
m_
\vec
_
\cdot(\vec
\times \vec
_
) =
\frac
\vec
_
\cdot\left(\vec
\times \sum_
m_
\vec
_
\right)]\end
and the sum over mjrj is constrained to equal zero because we have assumed the center of mass is at the position r = 0 in our coordinates. With this realization, and using the definition of the moment of inertia, we have:
\begin
[ K = \frac
m_
v_
^
+ \frac
I_
\omega^
]\end
This result shows that the kinetic energy of a rigid body can be broken into two parts, generally known as the translational part and the rotational part.
Rotational Kinetic Energy
The above formula suggests a definition for the kinetic energy of a rotating body:
\begin
[ K^
= \frac
I\omega^
]\end
Work-Kinetic Energy Theorem
Statement of the Theorem
If all the influences on a point particle are represented as works, the net work done by the forces produces a change in the kinetic energy of the particle according to:
\begin
[ \Delta K = W_
]\end
Derivation of the Theorem
From Newton's 2nd Law for a point particle, we know
\begin
[ \vec
_
= m\frac{d\vec{v}}
]\end
Now suppose that the particle undergoes an infinitesimal displacement dr. Since we want to bring the left side of the equation into line with the form of the expression for work, we take the dot product of each side with the displacement:
\begin
[ \vec
_
\cdot d\vec
= m\frac{d\vec{v}}
\cdot d\vec
]\end
Before we can integrate, we make a substitution. Since v is the velocity of the particle, we can re-express the infinitesimal displacement as:
\begin
[ d\vec
= \vec
dt]\end
Making this substitution on the right hand side of the equation, we have:
\begin
[ \vec
_
\cdot d\vec
= m\frac{d\vec{v}}
\cdot \vec
\:dt = m\vec
\cdot d\vec
]\end
We can now integrate over the path:
\begin
[ \int_
\vec
_
\cdot d\vec
= \frac
m(v_
^
-v_
^
)]\end
which is equivalent to the Work-Kinetic Energy Theorem.