Moment of inertia maths

[Tony]

I was doing some swatting up about inertia matching and i'm really struggling over this equation>

 

if you double the mass for the same size and shape, you double the moment of inertia, but if you keep the mass constant while doubling the size, the moment of inertia goes up by a factor of four.

 

Where does the factor of "four" come from?

[Sagitar]

I was doing some swatting up about inertia matching and i'm really struggling over this equation>

 

if you double the mass for the same size and shape, you double the moment of inertia, but if you keep the mass constant while doubling the size, the moment of inertia goes up by a factor of four.

 

Where does the factor of "four" come from?

 

It's because the formula for moment of inertia includes the square of the radius. If you double the radius, then square it the increase in moment is four.

 

The question is, where does the square of the radius come from.

 

Imagine a particle of mass m rotating in a circular path of radius r around a pivot point. The mass is subject to continual acceleration because it is changing direction continually. Call the angular acceleration A. At any instant, the particle is subject to linear acceleration which is equal to the angular acceleration times the radius of rotation. i.e. if the angular acceleration is constant and we double the radius, the linear acceleration doubles.

 

So linear acceleration a = Ar

 

The force needed to maintain this acceleration (from Newton) equals the mass m times the acceleration a

 

So force F = ma

 

But force doesn't help us, in a rotating system what we need to know is torque. Torque is force times the radius at which the force is acting so we add the radius r to both sides of the force equation and get

 

Torque = Fr = mar

 

Now we need to convert for angular acceleration by substituting Ar for a and this gives us

 

Torque = Fr = mr x Ar or mAr²

 

The torque so described, is the moment of inertia - changing the mass gives a linear effect, but changing the radius gives a change defined by a square law.

 

I hope that helps . . . . .

[Tony]

Blimey that sure beats what i found....

 

I = Kmr2 where I is the moment of inertia, m is the mass, and r is the object's radius. The additional parameter K is a numeric value that depends on how its mass is distributed. When most of the mass is concentrated near the axis, such as for a solid ball, K is low. If it's mostly far from the center, such as for a mass on the end of an arm, K is much larger.

[Sagitar]

Blimey that sure beats what i found I = Kmr2

 

It's the same formula. All they have done is substitute a constant K for my angular acceleration symbol.

 

It's all a huge over-simplification anyway. If you double the diameter of a wheel while keeping its mass constant, you cannot assume that it's MofI will increase by four. The "r" in the equation in this case is not the radius of the wheel, but the radius of the centre of rotational mass.

 

The position of the centre of rotational mass (the centre of percussion) will depend upon the profile of the wheel and the distribution of material within it. Effectively, you would have to do an integration for the infinite number of individual particles that make up the wheel. The MofI of a rotating body can be calculated practically by measuring the torque required to accelerate that body at a given rate. e.g. a bit like the Fletchers trolley experiments in schoolboy physics.

 

I am probably teaching granny to suck eggs, but you also have to remember that the MoI calculated relates only to rotation about a particular axis. For example an object such as a scaffold pole will have a high moment of inertia if you rotate it end to end, but a much lower MofI when rotated about its longitudinal axis. Similarly, increasing the width of a wheel may have a bigger effect on its MofI when rotating it about a transverse axis than when rotating it around the normal rolling axis.