# Why is rotational momentum I=mr²?

1. Jan 1, 2010

### tkdiscoverer

why is rotational momentum I=mr²???

I was studying for the AP physics and got stuck with this definition...

...So why is I = mr²???

I looked at almost everywhere but couldn't find the answer....Please help this poor guy T_T...

p.s: I looked at the internet and people just 'sways' away from answering...just answers like "just get comfortable with it" ...----> I don't want this kind of answer.... so please don't answer like that, please

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Sorry... I meant rotational inertia;;;got messed up in my head -_-;;;

Last edited: Jan 1, 2010
2. Jan 1, 2010

### Astronuc

Staff Emeritus
Re: why is rotational momentum I=mr²???

Think about centripetal acceleration, or the forces on a mass moving in circular motion, and the meaning of torque and moment of inertia.

http://hyperphysics.phy-astr.gsu.edu/hbase/mi2.html#rlin

3. Jan 1, 2010

### Staff: Mentor

Re: why is rotational momentum I=mr²???

That's the rotational inertia for a point mass at a distance r from an axis. To see that this definition makes some sense, read this: http://hyperphysics.phy-astr.gsu.edu/HBASE/mi2.html#rlin".

Edit: Astronuc beat me to it!

Last edited by a moderator: Apr 24, 2017
4. Jan 1, 2010

### Valle

Re: why is rotational momentum I=mr²???

Hey tkdiscoverer

I am not sure what AP physics is, assuming it's a course of sorts, I don't know how comfortable you are with calculus, tensors and the like. I'm assuming you aren't. Instead I will try to resolve, what seems to me as mixed used of concepts.

First, I don't know what texts you've been reading and what terms they are using, but I have never heard of "rotational momentum". The term I use for the rotational analog of linear momentum, is angular momentum.

But this dosn't seem to be what you're asking for. The formula, or rather, definition, you present is that for the moment of inertia for a point-mass about an axis. Here $$m$$ is the mass, $$r$$ is the perpendicular distance from the point to the axis and $$I$$ is the moment of inertia.

This physical quantity, the moment of inertia, has to do with the mass distribution about an axis, and is a quantity that arise when studying dynamics of rotational motion. Somewhat like inertial mass in linear dynamics.

Consider for example a point-mass rotating in the $$xy$$-plane a distance $$r$$ about a point (i.e. an axis $$z$$ perpendicular to the plane), the point-mass is subject to a resulting force $$F$$, which has a component tangent to the circular path $$F_{tan}$$. Lets evaluate the torgue about that axis.

$$\tau_z = F_{tan} r = m a_{tan} r = m \alpha_z r^2 = m r^2 \alpha_z = I \alpha_z$$

The torgue about $$z$$ and the angular acceleration about $$z$$ is proportional. And by comperision with Newtons second law $$F=ma$$ you see that the moment of inertia is the rotational analog of the inertial mass.

My suggestion would be to review the chapter in your book. Keep in mind that the moment of inertia is rotational inertia, i.e. resistance to change in rotational motion when subject to torgue.

Edit: Damn, guess I was too slow, just had to try writing different formulae with LaTeX :P

Last edited: Jan 1, 2010
5. Jan 1, 2010

### D H

Staff Emeritus
Re: why is rotational momentum I=mr²???

So far the discussions have centered on torque. I'll take a different approach, one centered on angular momentum.

There is a fundamental problem with a torque-centric point of view: It doesn't always work. The hyperphysics page on Newton's Second Law for Rotation, http://hyperphysics.phy-astr.gsu.edu/HBASE/n2r.html#n2r, caveats this analog to Newton's Second Law with (emphasis mine)
It is not as general a relationship as the linear one because the moment of inertia is not strictly a scalar quantity. The rotational equation is limited to rotation about a single principal axis, which in simple cases is an axis of symmetry.

In short, the rotational analog of Newton's second law is not universally true. (In a similar vein, F=ma is not universally true. Newton's second law is F=dp/dt. The rotational analog of this is τ=dL/dt, not τ=Iα.)

This alone suggests taking a step back and looking at the bigger picture. In the case of translational motion, the bigger picture is to look at linear momentum. Linear momentum is a conserved quantity in an isolated system. This makes linear momentum an extremely useful quantity in physics, more basic than forces. Physics education pretty much throws the concept of forces out the window after freshman physics. The same cannot be said regarding momentum.

Angular momentum is another one of those quantities that is conserved in an isolated system. The angular momentum of a point mass with respect to some central point is defined as

$$\vec L\equiv \vec r \times \vec p$$

where $\vec r$ is the displacement vector from the central point to the point mass and $\vec p$ is the point mass' linear momentum with respect to the central point: $\vec p = m\vec v$. If the point mass is viewed as being at the end of a massless rod of length r that is rotating about the central point with some angular velocity ω, the velocity of the point mass is

$$\vec v = \vec{\omega} \times \vec r$$

The angular momentum of the point mass can thus be re-expressed as

$$\vec L = \vec r \times (m\vec {\omega}\times \vec r)$$

The magnitude of this vector is

$$L = mr_{\perp}^2 \omega$$

where $r_{\perp}$ is the distance between the axis of rotation and the point mass.

6. Jan 2, 2010

### Lsos

Re: why is rotational momentum I=mr²???

"I" is how hard it is to get something to spin.

Obviously, the more massive the thing is, the harder it is to get spinning, hence the "m". Also, the further the mass is away from the center axis, the harder it is to get spinning, hence the "r".

But think about what happens when we increase the "r". Imagine two wheels. Imagine one has an "r" twice as big as the other. If you put them side by side, you'll quickly realize that the one that has a 2x bigger "r" is actually not twice as big...it's MUCH bigger. 4 times bigger, in fact, hence, 4x harder to get spinning, and hence why the "r" is squared.

Edit: I just realized you were talking about rotational momentum, or inertia. Well, it's the same thing, just backwards. The harder it is to get the thing to spin, the harder it is to also stop it's spinning (momentum)...