Yes I remember. We are glad the videos are helping.

there is a coefficient of inertia momentum that you can look up on the website: hyperphysics.phy-astr.gsu.edu/hbase/imgmec/mic.png If the rod about the center, I=1/12ML^2

Both methods should work if done correctly.

+Liam Roberts The parallel axis theorem is typically used if the point of rotation if not through the center of mass, and the shape of the objects is not one of the standard ones for which we know the standard equation of the moment of inertia. Here we simply have to add m1R^2 and (1/3) m2L^2 to find the total moment of inertia.

+Michel van Biezen thanks for the reply sir, I appreciate your work. You are helping millions worldwide, cost free, not many people can say they do that.

It is a mathematical equivalence. Both have units of radians/sec.

The mass is considered a "point" mass and therefore its dimensions are negligible compared to the length of the rod. If it wasn't you would have to add half the vertical dimension of the mass.

Gotcha. Thank you for the quick reply! Love your videos.

For the bar, the mass is distributed along its length. For the mass you can consider that a "point mass" and all of the mass is located at the distance L from the point of rotation. I = mR^2 is the standard formula for moment of inertia of a point mass.

+Pritham D'souza The total moment of inertia is the algebraic sum of the moment of inertia of each individual part.

Thanks for your prompt reply. So in a case like this there's no need to use the parallel axis theorem? I would've thought that it would be necessary as the rotation is happening through a pivot point at one end and not the Centre of gravity

This is actually a pendulum and the frequency is that of the swinging back and forth of the pendulum.

Use 1/3 when the bar rotates about its end. Use 1/12 when the bar rotates about its middle (center of mass). See this playlist: PHYSICS 12 MOMENT OF INERTIA

Yes, professor. I understand this part, but in some cases, it is not rotated about its middle, nor the end, according to "the parallel axis theorem", the body's moment of inertia about a parallel axis through the object's center of gravity and the perpendicular distance between the axes. so we should use I=I(center of the mass)+md^2, it that correct? and if the bar rotates about its end and no mass connect to the end of thin rod, is the I = 1/3 ml^2 Thank you very much.

this might be a bit too late, but we're looking at the separate moment of inertia of the block and the rod. We're isolating both components in the system, and adding them back together

CheeseTomatoe The mass at the end of the rod is considered to be a "point" mass, so it does not have significant dimensions.

Michel van Biezen Thank you for the quick reply!!

Michel van Biezen I still don't see why it would be 0.8m

67hoursAndCounting The definition of a point mass is that it has no volume dimensions. Therefore the center of mass of the object at the end of the rod is at 0.8 m from the point of rotation. That is how the problem was defined. We could create a different problem where the mass does have dimension, say a cube 10 cm on its side. Then the center of mass would be 0.8 ,m + 0.05 m from the point of rotation.

Michel van Biezen Ahhh thanks, I didn't realize there was a relationship between the center of mass and the point of rotation quite a speedy reply might I add, I appreciate it

+Dima JA That equation can be derived with the use of differential equations or by using Lagrangian Mechanics (which will give you the differential equations). Once you have the formula, all you have to do is find the total moment of inertia of the pendulum.

uxbeuys

Figured it out (it's not in the video): apparently its a formula that applies to the inertia of a rod, when the axis of rotation is at the end (I = 1/3 mL^2). Hope this helps someone else.

This playlist can help. PHYSICS 12 MOMENT OF INERTIA

For that you need to solve the 2nd order differential equation. 0 = d^2(theta)/dt^2 + b d(theta)/dt + a (theta) and determine the constants a and b.

@@MichelvanBiezen Thank you!