The term kinetic energy indicates the energy possessed by the object due to its motion. Based on the type of motion, it is classified as translational kinetic energy, rotational kinetic energy, and vibrational kinetic energy. Here’s now we are discussing translational kinetic energy in detail.
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What is Translational kinetic energy?
Translational kinetic energy is the energy possessed by the object by virtue of its translational motion.
OR
It is the kinetic energy of the object undergoing translational motion. It depends on the mass of the object and the linear velocity of the center of mass of the object.
The classical expression of the kinetic energy indicates that the translational kinetic energy is directly proportional to the mass of the object and the square of the velocity of the center of mass of the object. It is given by,
`K.E._{\text{translational}}` = `\frac{1}{2}`.m.`v^{2}`
From the above equation of translational kinetic energy, we can say that when the object is at the rest (`v` = 0), it has zero translational kinetic energy.
While accelerating the object from rest, the work done by the force on the object is converted into kinetic energy.
Thus the above expression of translational kinetic energy gives the work done on the object of mass ‘m’ to speed up in the linear path from rest (`v` = 0) to the required velocity.
As the mass is a positive quantity and the square of the velocity remains always positive, thus the value of translational kinetic energy always remains positive.
The above equation is only valid when the object’s velocity is much lower than the speed of light. (`v`<<C), whereas when the speed of the object is close to the speed of light then in such a case the kinetic energy of the object is found by the relativistic expression for the kinetic energy.
Translational kinetic energy equation:
The kinetic energy of the object of mass ‘m’ moving at velocity ‘`v`’ is given by,
K.E. = `\frac{1}{2}`m`v^{2}`
The change in kinetic energy of the object due to the velocity change is given by,
K.E. = `\frac{1}{2}m(v_{f}^{2}v_{i}^{2})`
Where,
`v_{f}` = Final velocity
`v_{i}` = Initial velocity
Derivation:
As per the workenergy theorem, the change in kinetic energy is given by,
K.E. = Work done = F x S
Where
F = Force acting on the object
S = Displacement
One of the newtons equations of translational motion says that,
`v^{2}` = `u^{2}` + 2aS
∴ S = `\frac{v^{2}u^{2}}{2a}`
Where,
u = Initial velocity of the object
`v` = Final velocity of the object
`a` = Acceleration of the object
As per newtons second law of motion, the force acting on the object is given by,
F = m x `a`
Where m = mass of the object
Now put the value of force (F) and displacement (S) in the equation of kinetic energy.
K.E. = F x S
K.E. = (m x `a`) x `\frac{v^{2}u^{2}}{2a}`
K.E. = `\mathbf{\frac{1}{2}}`m(`v`² – u²)
The above equation indicates the change in kinetic energy of the object because of the change in velocity from u to `v`.
For the object starting from rest (u = 0), the translational kinetic energy is given by,
K.E. = `\frac{1}{2}`m(`v^{2}` – 0²)
∴ K.E. = `\mathbf{\frac{1}{2}}`m`v`²
Translational kinetic energy solved examples:
1] A ball of mass 180 gram is released from height h. Just before hitting to the ground, the ball has a velocity of 14 m/s. Find the height h.
Given:
`v_{i}` = 0 m/s
`v_{f}` = 14 m/s
g = 9.81 m/s²
m = 180 gram = 0.180 Kg
Solution:
The change in kinetic energy of the ball is given by,
K.E. = `\frac{1}{2}m(v_{f}^{2}v_{i}^{2})`
K.E. = `\frac{1}{2}`0.180(14² – 0²)
K.E. = 17.64 J
The potential energy of the ball at the initial condition is given by,
P.E. = mgh
P.E. = 0.180 x 9.81 x h
P.E. = 1.7658 h
Just before hitting the ball to the ground, the initial potential energy of the ball gets completely converted into kinetic energy.
∴ P.E. = K.E.
1.7658 h = 17.64
∴ h = 9.989 m
2] The 20kg object is moving in a straight line at 11 m/s. Find the translational kinetic energy of the object and the amount of work required to stop the object.
Given:
m = 20 Kg
`v` = 11 m/s
Solution:
The translational kinetic energy of the object is given by,
K.E. = `\frac{1}{2}mv^{2}`
K.E. = `\frac{1}{2}` x 20 x 11²
K.E. = 1210 J
The work required to stop the object is equal to the initial kinetic energy of the object.
∴ `W_{\text{stop}}` = K.E.
∴ `\mathbf{W_{\text{stop}}}` = 1210 J
Translational kinetic energy vs Rotational kinetic energy:
Sr. No.  Translational kinetic energy  Rotational kinetic energy 

1  It is the energy possessed by the object by virtue of its linear or translational motion.  It is the energy possessed by the object by virtue of its rotational motion. 
2  This term is used in translational or linear motion.  It is used in a rotational motion. 
3  It is directly proportional to the mass of the object and the square of the velocity of the object.  It is directly proportional to the moment of inertia of an object about the center of mass and the square of the angular velocity of an object. 
4  It is given by, K.E. = `\frac{1}{2}`mv²  K.E. = `\frac{1}{2}`Iω² 
5  Example: A car moving in a straight line possesses translational kinetic energy.  Example: The rotating flywheel possesses rotational kinetic energy. 
FAQs:

What does translational kinetic energy depend on?
It depends on the mass and velocity of the object.

How to find Translational kinetic energy?
The translational kinetic energy of an object of mass ‘m’ moving at velocity ‘v’ can be find by,
K.E. = (1/2).mv² 
Can an object have rotational and translational kinetic energy?
Yes, the object can have both translational and rotational kinetic energy. E.g.: A moving wheel of the vehicle possesses both translational and rotational kinetic energy.