Rotational energy conservation

[dusurme]

Hi,
I am a little confused about the energy conservation of a pin ended free falling rod.
When i try to derive energy conservation equation i am not sure including angular and linear velocity at the same time. I try to visualize the problem in the attached picture and put my derivation also.
Any explanation will be appretiated.
Thanks.

 

[MisterX]

This is the kinetic energy in general:

$KE = \frac{1}{2}m v_{cm}^2 + \frac{1}{2}I_{cm}\omega^2$


for this example the velocity of the center of mass
$v_{cm} = \frac{L}{2}\omega$

$KE = \frac{1}{2}m \frac{L^2}{4}\omega^2 + \frac{1}{2}I_{cm}\omega^2$

Notice if the center of rotation is at the end of the rod, the angular velocity $\omega$ is the same as for the center of rotation being the center of mass.

The moment of inertia changes by the parallel axis theorem
$I_0 = I_{cm} + m\left(\frac{L}{2}\right)^2 = I_{cm} + m\frac{L^2}{4}$

So we find that the rotational kinetic energy about the end of the rod is equal to the total kinetic energy from before.
$\frac{1}{2}I_0\omega^2 = \frac{1}{2}\left(I_{cm} + m\frac{L^2}{4}\right)\omega^2 = KE$

 

[dusurme]

MisterX thanks for the answer.
Could you also suggest me an alternative way of determining angular velocity?

 

[lightarrow]

MisterX thanks for the answer.
Could you also suggest me an alternative way of determining angular velocity?

No, he's saying that your solution is incorrect: either you use the kinetic energy of rotation with respect to the centre of mass (and then the moment of inertia is not the one you wrote) *and* the centre of mass' kinetic energy, or the kinetic energy of rotation with respect to the pin, *only* (with the moment of inertia you wrote).

 

[PJC01]

Hello there, thank you for reaching out with your question about rotational energy conservation. I can understand your confusion, as it can be tricky to consider both angular and linear velocity at the same time. However, it is important to include both in the energy conservation equation for a free falling rod with a pinned end.

Let's start by defining some terms. The total energy of a system is the sum of its kinetic and potential energies, which can be represented as E = KE + PE. In the case of a free falling rod, we can break down the kinetic energy into two components: translational kinetic energy (KEt) and rotational kinetic energy (KEr). This is because the rod is both moving linearly and rotating about its pinned end.

To derive the energy conservation equation, we can use the principle of conservation of energy, which states that energy cannot be created or destroyed, only transferred from one form to another. In this case, we can say that the initial total energy (Ei) of the system is equal to the final total energy (Ef). Mathematically, this can be represented as Ei = Ef.

Now, let's consider the initial and final energies separately. At the initial moment, the rod has potential energy (PEi) due to its height above the ground, and both translational and rotational kinetic energies (KEti and KEri) due to its linear and angular velocities. As the rod falls, its potential energy decreases while its kinetic energies increase. At the final moment, just before the rod hits the ground, the potential energy is zero (PEf = 0) and the kinetic energies are at their maximum (KEtf and KErf).

Therefore, we can write the energy conservation equation as:
Ei = Ef
PEi + KEti + KEri = PEf + KEtf + KErf
mgh + 1/2mv^2 + 1/2Iw^2 = 0 + 1/2mv^2 + 1/2Iw^2

Notice that the translational kinetic energies cancel out, as they are equal at both initial and final moments. Also, since the rod is rotating about its pinned end, the moment of inertia (I) is constant and can be factored out. Therefore, the final energy conservation equation is:
mgh = 1/2Iw^2

I hope this explanation helps clarify the energy conservation equation for a pin-ended