I always thought that whenever there's rotational kinetic energy, there's also translational (linear) kinetic energy. The opposite is not true. There can be translational kinetic energy without having any rotation. Is this true? I thought it was until I stumbled upon this question today:
A pulsar is a form of a neutron star, the core of collapsed star of between $1.4$ and $3$ solar masses, that rotates rapidly and gives off radio waves. Suppose a star, with radius $7 × 10^8$ m, before supernova rotates at an angular rate of $3 × 10^{-6}$ rad/s, and upon supernova, shrinks in radius to $15$ km. At what rate does it rotate?
They solve it as follows:
Conservation of kinetic energy:
$K_{rot_0} = K_{rot_1}$
$ \frac{1}{2} \times I_0 \times \omega_0^2 = \frac{1}{2} \times I_1 \times \omega_1^2$
with $ I_0 = \frac{2}{5} \times m \times R_0^2 , I_1 = \frac{2}{5} \times m \times R_n^2 $
$ \frac{1}{2} \times \frac{2}{5} \times m \times R_0^2 \times \omega_0^2 = \frac{1}{2} \times \frac{2}{5} \times m \times R_n^2 \times \omega_1^2$
$ \frac{1}{5} \times (7 \times 10^8)^2 \times (3 \times 10^{-6})^2 = \frac{1}{5} \times (15 \times 10^3)^2 \times \omega_1^2$
$\omega_1 = 0.14$ rad/s
Before looking the solution I also accounted for the transitional rotation energy which is why I got stuck and couldn't solve it. How can an object have only rotational kinetic energy? If it's rotating, then it's moving. Or am I missing something?
Best Answer
How can an object have only rotational kinetic energy? If it's rotating, then it's moving. Or am I missing something?
You can have pure rotation of a body without translation and therefore have rotational kinetic energy without translational kinetic energy.
This is because the translational velocity of a body, which results in translational kinetic energy, is the velocity of its center of mass along a straight line relative to some external (to the body) frame of reference. The center of mass of a body undergoing pure rotation does not move at all so there is no translational kinetic energy.
ADDENDUM:
The above being said, the total kinetic energy of an extended object can be expressed as the sum of the translational kinetic energy of the center of mass ($\frac {mv^2}{2}$) and the rotational kinetic energy about the center of mass ($\frac {Iω^2}{2}$). The rotational inertia $I$ (moment of inertia) is analogous to translational inertia $m$ (mass) and the angular velocity $ω$ is analogous to the linear velocity $v$.
For an extended object undergoing pure rotation, the center of mass has no translational velocity, thus the translational kinetic energy component of the total kinetic energy is zero.
One very important final point. The terms “pure rotation” and “pure translation” only have meaning with respect to a defined frame of reference. I’m on a train going at constant velocity in a straight line. I have a globe of the earth resting on a table. I spin it. From my frame of reference on the train, the motion of the globe is pure rotation and it has rotational kinetic energy only. From the frame of reference of someone standing on the track watching the globe go by through the train window the globe has both rotational and translational kinetic energy.
Kinetic energy depends on the frame of reference.
Hope this helps.