How do you find the moment of inertia for wierdly shaped objects?

[jjellybean320]

i need to find the moment of inertia for a cardboard boomerang. what equation could i use?

 

[ideasrule]

If it's so weirdly shaped that you can't model it with a mathematical function, you're out of luck. You'll have to use numerical integration to find the moment of inertia.

 

[gneill]

Moment of inertia about what axis? Have you considered an experimental approach? You might find its center of gravity by suspending it from two separate points and see where the vertical plumb lines intersect. You could suspend it from a point as a physical pendulum and time its period,...

 

[jjellybean320]

its just 3 rectangular pieces of cardboard pieced together at one end of each piece to make a flat boomerang. each blade of the boomerang is 120 degrees apart.

the axis of rotation is through the center of the boomerang

 

[jjellybean320]

Moment of inertia about what axis? Have you considered an experimental approach? You might find its center of gravity by suspending it from two separate points and see where the vertical plumb lines intersect. You could suspend it from a point as a physical pendulum and time its period,...

can you explain how to find moment of inertia from knowing the center of gravity? I know how to find center of gravity but i don't know how that relates to finding a numerical value for the moment of inertia.

and can you explain how i can get a value of moment of inertia through knowing the period?
thanks

 

[gneill]

can you explain how to find moment of inertia from knowing the center of gravity? I know how to find center of gravity but i don't know how that relates to finding a numerical value for the moment of inertia.

and can you explain how i can get a value of moment of inertia through knowing the period?
thanks

See, for example,

http://hyperphysics.phy-astr.gsu.edu/hbase/pendp.html

 

[jjellybean320]

See, for example,

http://hyperphysics.phy-astr.gsu.edu/hbase/pendp.html

do i hang the boomerang at its center with the 3 blades spanning outward?

also do you know of a mathematical way i could use to check my moment of inertia value?

 

[gneill]

do i hang the boomerang at its center with the 3 blades spanning outward?

also do you know of a mathematical way i could use to check my moment of inertia value?

Hang it from anywhere BUT the center of mass (maybe the end of one of the spokes). You want there to be a distance L between the center of mass and the point of support.

To calculate the moment of inertia value you'll need either a set of mathematical functions that describe the geometry, or some other form of mathematical "image" of the object, such as laying a rectangular grid over the shape and assigning a mass value to each square that is "occupied" by boomerang and zero to the rest. Then apply the defining formulas for calculating moment of inertia to the resulting matrix of values.

 

[ideasrule]

The easiest non-experimental way to do this is probably to take a photo of the boomerang from high up (to minimize distortion), blanking out the non-boomerang regions of the photo, and writing a program to compute the moment of inertia of the remaining area. Note that if you use the pendulum approach, air resistance will be a big issue for a cardboard boomerang, so you might not get accurate results.

 

[jjellybean320]

if the moment of inertia = m r^2
could i just find the moment of inertia of each of the 3 blades separately and then multiplying by 3?

so for example if each blade weighed 2 grams and was 15 cm long (i'm rotating around the end of the blade) then the moment of innertia would be (15^2)*2

would that work if i wanted simply an approximate value?

 

[gneill]

Note that if you use the pendulum approach, air resistance will be a big issue for a cardboard boomerang, so you might not get accurate results.

It might not be too bad if the oscillation amplitude is kept small. One could also "scale" the problem by producing a more massive version in another material (do they still have metal working shop classes?).

I fear that if air resistance is an issue for small amplitude oscillations in a quiet environment, they will prove insurmountable for a (relatively) high velocity launch -- accuracy of the angular momentum value will be rendered moot by air resistance effects that scale non-linearly with velocity.

 

[gneill]

if the moment of inertia = m r^2
could i just find the moment of inertia of each of the 3 blades separately and then multiplying by 3?

so for example if each blade weighed 2 grams and was 15 cm long (i'm rotating around the end of the blade) then the moment of innertia would be (15^2)*2

would that work if i wanted simply an approximate value?

Look up some moment of inertia expressions for different geometrical shapes. mr² pertains to particular configurations only. You might also want to familiarize yourself with what is known as the "Parallel Axis Theorem", which tells you how to calculate the moment of inertia about an axis that is different from (but parallel to) one that you have the formula for.

See, for example,

http://en.wikipedia.org/wiki/List_of_moments_of_inertia

http://en.wikipedia.org/wiki/Parallel_axis_theorem

 

[jjellybean320]

thank you. i will talk to my physics teacher tomorrow.

 

[gneill]

Can you provide an image of the boomerang you have in mind so that we might be better able to advise on methodology? Approximate physical dimensions would also be a help.

 

[jjellybean320]

Can you provide an image of the boomerang you have in mind so that we might be better able to advise on methodology? Approximate physical dimensions would also be a help.

each blade is approximately 5 inches by 1.25 inches. and it is made from a christmas card.

 

[gneill]

Judging by the symmetry of the figure, if the blades really are rectangular then you're in luck. You can work out the moment of inertia of a single blade about the center of rotation (which should coincide with the center of mass of the whole assembly) and multiply by three. Look up the expression for the moment of inertia of a thin rectangular plate about its center. Determine how far your center of mass of the "plate" is from the center of rotation and apply the parallel axis theorem.

 

[jjellybean320]

thank you gneill so much!

 

[jjellybean320]

http://www.livephysics.com/tables-of-physical-data/mechanical/moment-of-inertia.html

the rectangular plate equation on that page is
(1/12)*Mass*(a^2+b^2)
where a and b are the length and width

but this equation is if the axis is through the center.

if the axis is at one end like my boomerang, would i just substitute (1/12) with (1/3)?

 

[cosmik debris]

each blade is approximately 5 inches by 1.25 inches. and it is made from a christmas card.

It's funny what passes for a boomerang these days :-)

 

[gneill]

http://www.livephysics.com/tables-of-physical-data/mechanical/moment-of-inertia.html

the rectangular plate equation on that page is
(1/12)*Mass*(a^2+b^2)
where a and b are the length and width

but this equation is if the axis is through the center.

if the axis is at one end like my boomerang, would i just substitute (1/12) with (1/3)?

No. Look up the Parallel Axis Theorem as I suggested. You need to be precise about where the center of rotation is with respect to the center of the plate -- in your case it won't be precisely at the end of the plate, but a bit in from the end. Measure and be sure.

 

[jjellybean320]

No. Look up the Parallel Axis Theorem as I suggested. You need to be precise about where the center of rotation is with respect to the center of the plate -- in your case it won't be precisely at the end of the plate, but a bit in from the end. Measure and be sure.

the center of rotation is about 1 cm from the center of the plate. i looked at the wikipedia page for the parallel axis theorem and it was really confusing. I've only taken ap calculus ab.

 

[gneill]

the center of rotation is about 1 cm from the center of the plate. i looked at the wikipedia page for the parallel axis theorem and it was really confusing. I've only taken ap calculus ab.

That seems rather close to the center of mass of the plate. From your image it looked like it was more like 1cm from one end (the end where the blades all come together in the center), so about 5.4cm from the center of mass of the plate.

The parallel axis theorem states, in essence, that given a known moment of inertia I about some axis, then if you move the axis of rotation some distance r from that axis, being sure that the new axis is parallel to the old axis, then the moment of inertia about that new axis is given by I_new = I + M*r². M is the mass of the object.

The formula that you found for the moment of inertia for the plate about its center should be 'shifted' by adding M*r² to it, where r is the distance from the center of the plate to the actual axis of rotation.

 

[jjellybean320]

thank you. I understand it so much better now. And you were right, it is about 5.4 cm between the center of rotation and the center of the plate. I simply misread what you wrote.