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# More on moment of inertia

In this video David explains more about what moment of inertia means, as well as giving the moments of inertia for commonly shaped objects. Created by David SantoPietro.

## Want to join the conversation?

• why is calculus required for everything •   Calculus is the mathematical technique to deal with values that change and since most real world situations are not static you need to use calculus to describe what happens.
• How does all this play with torque, i mean like we know that its easier to rotate something with a large amount of torque and that would be the further you are from the axis of rotation hence increasing the moment arm, but the further you are from the axis the larger the moment of inertia will become. So its kind of like both of these contradict one another , the torque is saying its easiest to push here, while the inertia is telling us its resisting angular acceleration the most here? • Thats a cool question.

Its probably better for you if you can figure it out your self.... let me give you my answer far below, and take a look after you have given it some more thought...

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OK; you got here... :)

So, you are right, the moment or toques WILL increase with distance from the turning point.

You are also right that the moment of inertia DOES depend on distance of the mass from a turning point. BUT... the moment of inertia is a property of the body. It will not change as you change the position of your force.

In much the same way that the mass of a body will not change as you change the direction or magnitude of a force acting on it.

Does that make sense??

How did you do?
• At the formula of the moment of inertia. He said he used calculus
to derive the formula I=1/3ml^2. However, when i tried deriving it using the indefinite integral. I got I= 1/3ml^3. Could you please explain how you got
I=1/3ml^2? • I think you probably assumed that I=∫mr^2 dr =∑m_ir_i^2. But look, the masses in the sum are getting smaller and smaller as you increase the number of parts you divide the rod into (eventually reaching infinitely many, infinitely small point masses - our goal). On the other hand, in the integral, each point mass has the mass m. It would imply that the whole rod has infinitely high mass, which makes no sense.

My solution is as following - introduce a linear density σ such that σ=m/L, so m = σL. This way, every infinitely short part of the rod will have an infinitely small mass, but together, their masses will sum up to m. Consequently, I = ∫(from 0 to L) σr^2 dr = 1/3 σL^3 = 1/3 (σL)L^2 = 1/3 mL^2.
• What will be moment of inertia of rod whose axis of rotation is somewhere between the centre and the edge?
In this case, do we have to consider the larger length or the smaller length? • If we consider a door rotating about it's hinge, isn't it more difficult to rotate it near its axis than from its end point? Will that not mean that moment of inertia is greater near its axis ? • The reason that it takes more force near the hinge is that the torque that you exert is the product of the force (perpendicular to the door) exerted times the torque arm (the distance from the hinge). The same torque requires more force if the distance is less. So the angular acceleration is lower (for the same moment of inertia) when the torque is lower.
• Hi! So at when you wrote the formula for the rod of constant density rotating around the axis at one end, I had expected it to be (mL^2)/4.
I thought that since the center of mass was at the exact center of the body, we could consider the system to be a rope with a body of mass 'm' situated at distance 'L/2'. That way it would be (mL^2)/4.
Could anyone tell me where I went wrong? Ik he said that we were supposed to use calculus, but I can't understand why my explanation isn't correct. • It's because of L square. If there was just L (not-squared), you would be right. I am sure you will understand this when you learn calculus, but let me give you an example.

Consider just 3 point masses for simplicity. One is at the axis of rotation (like the left end of the rod), the third is at the distance L from the axis, and the second point lies exactly between them.
1*----2*----3*
Each has a mass m, so the total mass is obviously M=3m and m = 1/3 M.

Following your way of thinking, the mean distance from the axis of rotation is L/2 (equal to (0 + L/2 + L)/3), so the moment of inertia would be
I = ML^2/4.

However, let's now consider these points separately. The moment of intertia of the first point is i1 = 0 (as the distance from the axis is 0). Of the second point: i2 = m (L/2)^2 = mL^2/4. Of the third point: i3 = mL^2. The total moment of inertia is just their sum (as we could see in the video): I = i1 + i2 + i3 = 0 + mL^2/4 + mL^2 = 5mL^2/4 = 5ML^2/12.

The result is clearly different, and shows you cannot just consider the mass of an object to be concentrated in one point (like you did when you averaged the distance). You can also check that it would not matter if the distance wasn't squared - then your method would produce correct results.

I hope this clears things up. Cheers
• question-we have a beam , o is axis of rotation f1 is, first force's position direction is upward while f2 is second force's position . whole of beam from f1 to f2 is massless so can we say f1=f2?

----o----------f1---------------------------------f2- • are we saying the beam is in equilibrium??

If so, then no. We can not say f1 - f2.

We can say that the turning effect due to f1 = - (minus) turning effect due to f2

and the turning effect (or moment) of each force is given by = Force x distance from turning point

when we are told a beam is massless, it means the weight of the beam is tiny compared with the other forces so you can neglect it.

ok??
(1 vote)
• I need to learn the proofs of the differential and integral calculus for these results?
Can anyone provide me with them ?
(1 vote) • If radius is bigger, than according to moment of inertia it is more difficult to rotate but we also know that the bigger the radius the more the torque ........... than the simple question is how radius effects the rotation........and why?
(1 vote) • Moment of inertia and torque are completely independent of each other.
If you spread out the mass distribution further away from the axis of rotation, the moment of inertia will increase. If you apply a force further away from the axis of rotation, the torque will increase. Even though both depend on radius, the torque and moment are completely independent (ex. you can apply the force anywhere or the mass can be distributed anywhere). 