Problems

1. Let us evaluate the center of mass of a continuous rod of length $L$ and total mass $M$, to make sure it is in the middle:
   

   $$M \vec{X}_{\rm cm} = \int \vec{x} dm = \int_0^L \lambda x dx$$ (4.10)

   
   
   where
   

   $$M = \int dm = \int_0^L \lambda dx = \lambda L$$ (4.11)

   
   
   (which defines $\lambda$, if you like) so that
   

   $$M \vec{X}_{\rm cm} = \lambda \frac{L^2}{2} = M \frac{L}{2}$$ (4.12)

   
   
   and
   

   $$\vec{X}_{\rm cm} = \frac{L}{2}.$$ (4.13)

   
   

   Gee, that was easy. Let's try a hard one.

   

2. Let's find the center of mass of a circular wedge. It is two dimensional, so we have to do it one coordinate at a time. We start from the same place:
   

   $$M X_{\rm cm} = \int x dm = \int_0^R \int_0^{\theta_0} \sigm... ... \int_0^R \int_0^{\theta_0} \sigma r^2 \cos \theta dr d\theta$$ (4.14)

   
   
   where
   

   $$M = \int dm = \int_0^R \int_0^{\theta_0} \sigma dA = \int_0^... ...\theta_0} \sigma r dr d\theta = \sigma \frac{R^2 \theta_0}{2}$$ (4.15)

   
   
   (which defines $\sigma$, if you like) so that
   

   $$M \vec{X}_{\rm cm} = \sigma \frac{R^3\sin\theta_0}{3}$$ (4.16)

   
   
   from which we find (with a bit more work than last time but not much) that:
   

   $$\vec{X}_{\rm cm} = \frac{2R^3 \sin\theta_0}{3R^2\theta}.$$ (4.17)

   
   

   Amazingly enough, this has units of $R$ (length), so it might just be right. To check it, do $\vec{Y}_{\rm cm}$ on your own!