Torque due to Gravity

Question

I have learned that a torque on an object due to the gravity can be calculated as if the gravity acts on the center of mass of the object when the object is near the surface of the earth.

Now I want to prove it...

So, I have $$\tau=\int d\tau=\int r(dmg)sin\theta=g\int rsin\theta dm.$$

Then, how do you obtain $$\int rsin\theta dm= mr_{cm}sin(\theta_{cm}),$$ where $r_{cm}$ is the distance between an axis of rotation and the center of mass, and $\theta_{cm}$ is the angle between the two vectors $\vec{r_{cm}}$ and $m\vec{g}$ at the center of mass??? Would $$\int r dm=mr_{cm}$$ be helpful to prove this?

Would the integral be dependent on the object I use, or would there be a general way to prove this?

Answer 1

The proof is most easiest if we use the vector notation. We have

$$\vec \tau = \int {d\vec \tau } = \int {(\vec r \times dm\vec g)} = \left( {\int {\vec r dm} } \right) \times \vec g$$

where I have used the assumption that near the earth $\vec g$ is constant. Now according to the definition of center of mass we have,

$${{\vec r}_{cm}} = \frac{1}{M}\int {\vec rdm}$$

Therefore, we find

$$\vec \tau = M{{\vec r}_{cm}} \times \vec g = {{\vec r}_{cm}} \times M\vec g$$

which is the desired result.