# Center of Mass by integration

1. Dec 17, 2003

### smeagol

Show that the center of mass of a uniform semicircular disk of radius R is at a point (4/(3*Pi))R from the center of the circle.

well I know I am suppose to find this by integration. By this equation

$$M\vec{r_{cm}}=\int\vec{rdm}$$

But, I am not sure how to find dm in this case...

do I divide mass by area? or by circumference?

and since it's a disk, and it was 2 variables, would I have to integrate for 2 variables? (x and y)?

2. Dec 17, 2003

### NateTG

In general, CM calculations end up being double or triple integrals, but this one is pretty straightforward since the disk has constant density.

I'd be inclined to cut the half-disk into parralel strips and integrate along the CM of each strip.

Since the density is uniform, you might as well assume that it's one.

3. Dec 17, 2003

### smeagol

so I divide M by $$2R\sin{\theta}$$?

4. Dec 18, 2003

### HallsofIvy

Staff Emeritus
Since you are talking about a semicircular disk, I would be inclined to use polar (actually cylindrical coordinates).

The differential of area in polar coordinates is r dr d&theta; so the differential of volume in cylindrical coordinates is r dr d&theta; dz.

Taking the density to be &rho;(r,&theta;,z), dm= &rho;(r,&theta;z) dr d&theta; dz.

The mass, for semicircular disk of radius R and thickness h would be M= &int;(z=0 to h)&int;(&theta;=0 to &pi;)&int;(r= 0 to R) &rho;(r,&theta;,z)r dr d&theta; dz.

If &rho; is a constant, this is just &rho;&pi;r2h.

Since x= r cos&theta;, the formula for the xcm would be Mxcm= &int;(z=0 to h)&int;(&theta= 0 to &pi;)&int;(r= 0 to R)(r cos &theta;)(&rho;(r,&theta;,z)r dr d&theta; dz)=
&int;(z=0 to h)&int;(&theta= 0 to &pi;)&int;(r= 0 to R)(&rho;(r,&theta;,z)r2cos&theta; dz d&theta; dz.

Since y= r sin&theta;, the formula for ycm would be Mycm= &int;(z=0 to h)&int;(&theta;= 0 to &pi;)&int;(r= 0 to R)(&rho(r,&theta;,z)r2sin&theta; dz d&theta; dz.

The formula for zcm would be Mzcm= &int;(z=0 to h)&int;(&theta;= 0 to &pi;)&int;(r= 0 to R)(&rho(r,&theta;,z)zr dr d&theta; dz.

Of course, if &rho; is a constant, from symmetry (cosine is an even function) xcm= 0 and zcm= h/2.

5. Dec 20, 2003

### smeagol

Ok, I got that one.

Moving on, the next problem is confusing me as well.

A baseball bat of length L has a peculiar linear density (mass per unit length) given by $$\lambda=\lambda_0(1+x^2/L^2)$$

so what I've done is
$$\int_{0}^{L}x\lambda_0(1+x^2/L^2)dx$$

which gives
$$\frac{3\lambda_0L^2}{4}$$

so I know that M * x_cm = that but how do I find M so I can find x_cm?

6. Dec 20, 2003

### krab

C'mon. You know the mass per unit length. How can you not know the mass?

$$M=\int_0^L\lambda dx$$

7. Dec 21, 2003

### HallsofIvy

Staff Emeritus
krab: Watch it now! You're starting to sound like me!

8. Dec 21, 2003

### HallsofIvy

Staff Emeritus
By the way, why were these posted under the "classical physics" forum? The look like fairly standard calculus homework problems.

9. Dec 21, 2003

### smeagol

well I got it from a "physics for scientists and engineers" from tipler. So I figured it would fit better under physics.