- #1

- 45

- 0

The uniform semicircle has radius R and mass M. I am getting the wrong answer but I can't see where I am going wrong. Any help would be appreciated.

My solution:

The centre of mass satisfies ##M\mathbf{r_{cm}}=\int \mathbf{r} dM##. In order to use polar coordinates, we will consider an element of area ##dA## and sum over all elements in the semicircle.

We have ##dA=\frac{1}{2}R^2 d\theta## and since the semicircle is uniform we have ##\frac{M}{A}=\frac{dM}{dA}## where ##A## is the area of the semicircle, ie ##dM=\frac{2M}{\pi R^2} dA##. Also, the position of this area along the semicircular arc we are sweeping out is ##R\cos \theta \mathbf{i}+R\sin \theta \mathbf{j}##. Substituting this into the integral we get

##M\mathbf{r_{cm}}=\int (R\cos \theta \mathbf{i}+R\sin \theta \mathbf{j}) (\frac{2M}{\pi R^2}(\frac{1}{2}R^2 d\theta))##

##=\frac{MR}{\pi} \int \cos \theta \mathbf{i}+\sin \theta \mathbf{j} d\theta##

and I simplified this to get ##M\mathbf{r_{cm}} = \frac{2RM}{\pi} \mathbf{j}##.

Where have I gone wrong?