Rotational Inertia of a Slender Rod of NON-UNIFORM Mass Density (See Note in Description.)

Derives the rotational inertia of a slender rod of non-uniform mass density. The axis is about the end of rod and perpendicular to it, and the mass density varies as Cx^2 where C is positive constant and x is the distance from the axis. IMPORTANT NOTE: At 3:20 the distance x on the diagram is incorrectly marked as x^2. It should be marked as x.
For a complete index of these videos visit www.apphysicsle...

[@seandimmock5813]

I love these videos. So simple!

[@tarunrao3662]

could you please take few more examples for Moment of Inertia of non uniform mass distribution? ......

[@fadedrai9796]

This video is very helpful, thank you!

[@surendersoundiramourty5045]

If the axis is taken at L/2 what would be the answer?

[@noahrubin3327]

Surender Soundiramourty I believe u could just use the parallel axis theorem

[@s.k.2456]

thanks!

[@AP-sz4hj]

i think in the solution provided by you, distance of dM from axis should be x instead of x^2 please correct me if i am wrong

[@lasseviren1]

You are correct. That is a mistake in this video. Thank you for pointing that out.

[@engineeringlife6478]

I did not get that why you assume lyamda = Cx2 .

[@lasseviren1]

That equation will change with the problem. This problem it happened to be lambda=Cx^2 but in another problem it might be Cx^3 or Dx. Where C and D are constants. Each time, it will be given in the problem.

[@anubhavthakur2985]

You could have substituted Cx^3/3 with M to make the answer more Aesthetically pleasing. Just Saying.

[@korayaydogan7365]

What if the rod is rotating about its center of mass (without the parallel axis theorem) and what would be the bounds in the integral ?

[@lasseviren1]

So this is a much more difficult problem. The center mass is not at the center of the rod. You would first have to use a different integral to find the center of mass of the rod, and then you would have to adjust your bounds so that the distance in the rotational inertia integral was from the axis to the dm. In other words, it's not just a matter of changing your bounds. If you give me the specific problem I might be able to get you started.

[@korayaydogan7365]

Let’s say lambda is Ax and it is rotating about its center of mass. So its mass would be A.l^2/2 and its center of mass is 2l/3 . But I want to solve the problem without the parallel axis thm. I got stuck in the bounds of the integral. I tried -2l/3 to l/3 but the result gets negative. So can you help me about the correct bounds.

[@lasseviren1]

@@korayaydogan7365 I have I = integral of (-2/3L+x)^2*dm where the bounds are from x=0 to x = L. Draw a diagram of the stick along the x axis with the one end at x=0 and the other end at x = L and your axis at x=2/3L. Draw in a little dm on the left of the axis. If x is the location of the dm, then -2/3L+x is its displacement from the axis. By the way, since lambda is dm/dx, dm = (Ax)(dx).

[@korayaydogan7365]

@@lasseviren1 Thank you so much you helped me a lot :)

[@lasseviren1]

@@korayaydogan7365 Glad that helped. All the best in Physics. It looks like you're doing very well.

[@usaball9193]

I have a doubt. When you're taking a small dx out of the rod and calculating it's inertia. Shouldn't the inertia be something like dI=(dm)(x/12)^2 ?, because we're taking a small element of the rod and the rod's moment of inertia is ml^2/12. Please correct me if i'm wrong.

[@senan8941]

We are considering it as a point mass. Thus no point considering it as a rod. I hope this is clear 👍🏻

[@usaball9193]

@@senan8941 Yes I understood, Thanks.

[@pratiknath2532]

What is C here

[@mattya25]

C is an arbitrary constant value. I feel this video could be more complete because the C constant can be eliminated by "solving" for C. If dm = C(x^2)dx, you can integrate from 0->L to find total M. M = C(1/3)(x^3). Thus, C = (3M)/(x^3) = 3M/L^3. Plug that back into the "final" answer and get I = 3/5ML^2. (Turns out the C constant doesn't matter!)

[@catrina6541]

im in love with you