Suppose i have a point mass which is moving in a circular path.Since the point mass can only have transnational motion, kinetic energy due to it will be $\frac{ mv^²}{2}$. But what if i observe the motion from the centre of circle, I can write it's kinetic energy as $\frac{I\omega^²}{2}$ which is also equal to $\frac{mv^²}{2}$.Why don't we add these two kinetic energies?


Suppose i have a point mass which is moving in a circular path

Let the mass of the particle be $m$

If we see this particle from a frame of reference located not at the centre of the circle but say somewhere else from where the motion appears to be only translational.

And the velocity of the body with respect to this inertial frame(another assumption) is $\vec{v}$

Then its kinetic energy is $\frac{1}{2}mv^2$

$\because$ The velocity of a body is a frame dependent quantity. We can say that if we change our frame of reference then the velocity of the mass $m$ will also change.

Lets look at the particle from a stationary frame at the centre of the circle.

From this point of view the particle appears to be going around a circle with say some angular velocity $\vec{\omega}$

So from this frame the particle's new velocity($\vec{v_1}$) should be $\vec{r} * \vec{\omega}$ where $\vec{r}$ is the position of the particle with respect to the centre.

So from this frame the particle's kinetic energy will be

$\frac{1}{2}I\omega^2 = \frac{1}{2}mr^2(\frac{v_1}{r})^2 = \frac{1}{2}mv_1^2$

So as you can see the Kinetic energy is different for both the frames

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Point mass in circular path is translational and kinetic energy is


Though, you can write it in terms of its moment of inertia and angular speed as $\frac{I{\omega}^2}{2}$. But the two expressions do not represent kinetic energies of two different motions. They are one and same so don't add them, either you look from center or from somewhere else.

(Note:There is no difference in the frames unless the center and the somewhere else frames are moving relative to each other.)

You can add rotational and translational kinetic energies in case of rigid bodies. A rigid body, say, a stone has both rotational and translational when it is
(1)rotating about a fixed axis passing through it: rotational, while at the same time
(2)the whole body is moving from one place to another: translational.
Note the two different types of motions. Then, the total kinetic energy will be rotational + translational. Rotational kinetic energy will be in terms of the moment of inertia and angular speed because of its rotation and translational kinetic energy will be in terms of its mass and velocity of its center of mass.

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