Center of mass of a non-uniform rod

[Linus Pauling]

1. A straight rod has one end at the origin and the other end at the point (L,0) and a linear density given by \lambda=ax^2, where a is a known constant and x is the x coordinate. Since this wire is not uniform, you will have to use integrtation to solve this part. Use M=\int_0^L dm to find the total mass M. Find x_cm for this rod.



2. X_cm = (1/M)Integral(x dm)



3. To obtain M, I did a*Integral(x^2 dx) from 0 to L, obtaining M = (1/3)aL^3

I then did x_cm = (1/M)*a*Integral(x^3) from 0 to L, obtaining:

(3/4)(a^2/L)

Apparently the answer does not depend on a

 

[Linus Pauling]

Ok, I am confident that my calculation of M is correct

M = (1/3)aL^3

But what is the next integral I need to do? I know it's (1/M)*Integral(x dm)

where dm = ax^2 dx. What is x? I'm going psycho.

 

[Doc Al]

I know it's (1/M)*Integral(x dm)

where dm = ax^2 dx. What is x?

What do you mean, what is x? It's the x-coordinate along the wire, just like before. Just write dm in terms of dx (like you just did) and you'll have what you need to integrate.

 

[Linus Pauling]

I do not understand. The length of the rod is L. So if I integrate L*ax^2 dx from 0 to L I obtain (1/3)aL^4.

Dividing by M=(1/3)aL^3 I obtain L, which is incorrect.

?

 

[Doc Al]

The length of the rod is L. So if I integrate L*ax^2 dx from 0 to L I obtain (1/3)aL^4.

Why are you integrating that?

x_cm = (1/M) ∫ x dm, just like you stated in your last post. Just write dm in terms of x, which you also stated in your last post.