Moment Of Inertia Problem using Algebra

[jaychouf4n]

Homework Statement

1a . By dividing a uniform narrow bar of length l into a number of small masses, find the moment of inertia of the bar rotating about an axis perpendicular to the bar and passing through its end

1b. use the parallel axis theorem and the result above to find the moment of inertia about an axis parallel to that above passing through the centre of mass of the bar

1c. calculate the moment of inertia of the turbine blade. Each blade has mass m and length l

2. a car with its door open moves off from rest with an acceleration of 10ms-2. how long will it take the door to close

Homework Equations

The Attempt at a Solution

I think i got 1 a) right

I=m(l/10)^2+m(2l/10)^2...+ml^2
=m/100(l^2+(2l)^2...(10l)^2)
=385l^2/100

For the rest I have not really any idea, so any tips would be great

thx thanks =D

 

[Dick]

That's a start on a). But shouldn't m for each part be m/10? And what happened to m in your final formula? And why just divide it into 10 parts? Why not into 100 or 1000 etc? I think you are actually supposed to set up an integral.

 

[jaychouf4n]

I don't think I can use calculus on this problem because we haven't learned it yet.

I only know difrrentiation, not much integration =/.

On the m, forgot that XD

Is it possible to use mathematical induction to solve the infinite series if i use n lengths?

 

[Dick]

You can write out the formula for a general value of n. Then find or derive a formula for the sum of the first n squares and you can write out an explicit formula. Then take the limit as n goes to infinity.

 

[rohanprabhu]

You can write out the formula for a general value of n. Then find or derive a formula for the sum of the first n squares and you can write out an explicit formula. Then take the limit as n goes to infinity.

yup.. if you divide it into 'n' small parts.. you get the lengths as: $\frac{l}{n}$, $\frac{2l}{n}$ etc. Since you need to square that.. take $n^2$ and $l^2$ common. Then you'll have to solve for the series:

$$\frac{l^2}{n^2}\left(1 + 2^2 + 3^2 + 4^2 + ...\right)$$

use the formula for sum of squares and use $n \rightarrow \infty$ and you shall get your answer. This however, will be much easier if worked with an integral..

 

[jaychouf4n]

So I get

lim n-->infinity = 3l^2

 

[rohanprabhu]

So I get

lim n-->infinity = 3l^2

well.. this is your radius of gyration.. Multiply with the mass and get moment of inertia:

$$I = \frac{3Ml^2}{2}$$

For the second question.. you know the moment of inertia for an axis [from the first question] and there is another axis at a distance l/2 from the first one and parallel to the first one. Can you use the parallel axis theorem now?

 

[rohanprabhu]

ok.. we seem to have made a huge mistake here...

the sum:

$$\frac{l^2}{n^2}\left(1 + 2^2 + 3^2 + 4^2 + ...\right)$$

tends to infinity as $n \rightarrow \infty$. As Dick said, the mass of each such small element will be M/n and not 'M', hence the sum we need to evaluate is:

$$\frac{M}{n}\frac{l^2}{n^2}\left(1 + 2^2 + 3^2 + 4^2 + ...\right)$$

which would give you:

$$\frac{Ml^2}{3}$$

this is the required moment of inertia...

 

[jaychouf4n]

Oh thank you

i think its cos i forgot the 6 as the denominator =/

parallel axis theorem is I=I(centre of mass)+Mr^2 I think

so taking that I you get

Ml^2/3=I(cm)+Mr^2

I(cm)=Ml^2/3-Mr^2

 

[rohanprabhu]

Oh thank you

i think its cos i forgot the 6 as the denominator =/

parallel axis theorem is I=I(centre of mass)+Mr^2 I think

so taking that I you get

Ml^2/3=I(cm)+Mr^2

I(cm)=Ml^2/3-Mr^2

For most of the part, you have done it right.. but what is 'r' here?

 

[Dick]

Ok so far. What the relation between l and r?

 

[jaychouf4n]

i think r is the distance from the original point to the centre so l/2

 

[rohanprabhu]

i think r is the distance from the original point to the centre so l/2

well.. you got it.. just put it in the equation and solve.. you should have ur solution.. :D

 

[jaychouf4n]

cheers everyone =]

especially rohanprbhu :D