Scaling Rotational Inertia - from Model to Full Size

[hitspace]

Homework Statement

If we multiply all the design dimensions of an object by a scaling factor (f), its volume and mass will be multiplied by f^3. a) By what factor will its moment of inertia be multiplied? b) IF a 1/48 scale model has rotational kinetic energy of 2.5 Joules, what will be the kinetic energy for the full-scale object of the same material rotating at the same angular velocity?

Homework Equations

KE = 1/2 (I) w^2
I_Disk = (1/2) M (r^2)

The Attempt at a Solution

Wasn't really sure how to approach this problem as my understanding of moment of inertia, is that it changes according to the dimensions of an object. I decided to try this problem anyway assuming a solid disk spun about the central axis. I attempted part b) wherein I set up the equations as follows

if 2.5 = .5 [ .5 (M)r^2 ] w^2
then .5 [ .5 (48M)(48r)^2 ] w^2 = 48^3 {.5 [ .5 (M)r^2 ] w^2 }
I got the answer 2.5 * 48^3 = 2.76 E6 Joules.

My solutions manual says I should have multiplied by 48^5 instead of to the third power.

Any ideas?

 

[gneill]

Think about how the mass scales. Does the density of the material change?

 

[hitspace]

Do you mean in the sense that if I played around with some numbers for example..

D= M/V

M = D(l^3) then an object 48x the dimensions would be D(48L)^3 = 48^3 M?

If I plug it back into the equation I see how I would then get (48^5)(2.5J).

Am I on the right track with this line of thinking?

 

[gneill]

Do you mean in the sense that if I played around with some numbers for example..

D= M/V

M = D(l^3) then an object 48x the dimensions would be D(48L)^3 = 48^3 M?

If I plug it back into the equation I see how I would then get (48^5)(2.5J).

Am I on the right track with this line of thinking?

Right. If the scaled objects are made from the same material then the density of the material remains the same. So if the overall volume changes by some factor (the cube of the scaling factor) then the mass changes by the same factor.

 

[hitspace]

Thank you so much. I really appreciate the help. The solutions manual says this is the right answer, but I am curious. How would this work on an object with different dimensions for moment of inertia. This whole process we went through to solve this problem seems unique to this particular moment of inertia equation I_Disk = .5(M)(r)^2

 

[gneill]

All moments of inertia boil down to a mass multiplied by a distance squared. For the same material, mass scales as the cube of the linear dimensions. So for a real object where all the linear dimensions scale at the same time and by the same amount, the moment of inertia (and related properties) will scale as the fifth power of the scale factor.

The only time you need to worry is when only certain dimensions are scaled. For example, if some object is made out of a thin sheet of steel and the scaled version is just a larger version cut from the same sheet material, then only two of the three dimensions are scaling and you need to go back to the derivations to see what dimensions actually scale in the moment of inertia.

 

[hitspace]

Thank you! That makes excellent sense.