Small confusion understanding center of gravity?

Question

I'm reading Sears/Zemansky book on physics, here:

The referenced figure is:

$\quad\quad\quad\quad\quad\quad\quad\quad$

I'm a bit confused, the center of gravity is given by $\tau = r_{cm} \times Mg$ and the torque is given by $\tau =r\times F$ where $r$ is a vector from the origin to where the force acts while $F$ is the force. In this case, where is the origin? In the point of suspension?

Trying to understand the "If it were anywhere else..." part: As the mug is in equilibrium, I think we must have $\sum \tau =0$, in this case, I think the only torque acting is $\tau = r_{cm} \times Mg$, I guess this means that $\tau = (x,y,z)\times (0,0,Mg)= (-\text{Mg} y,\text{Mg} x,0)=0$ and hence the center of gravity must be in the origin? This doesn't seem to make sense: Either this is wrong or the previous assumption is wrong?

Answer 1

Take a look at this pic :

Keep cup firmly at point $A$ as it is drawn, then your fingers will feel torque : $$ \tau = \vec{r} \times m\vec{g} $$

Answer 2

The text says "with respect to the point of suspension".

We have two forces involved:

• The holding-up force acting at the suspension point.
• The weight acting from the centre-of-gravity.

By imagining the centre-of-rotation at the suspension point, we only have the weight to worry about (because a force can't cause a torque about the point where it acts; the perpendicular distance to this point is zero in the torque formula). Since the weight pulls vertically downwards and since we don't see any rotation, it must act in a point somewhere directly, vertically below the suspension point. Somewhere on the vertical line through the suspension point.

Were it acting in any other point, then it would cause a torque about the suspension point (because there would be a non-zero perpendicular distance for the torque formula), which would make the cup rotate. This doesn't happen, so the weight must be acting on this vertical line through the suspension point.

By moving the suspension point (by holding the cup somewhere else) we get a new vertical line on which the weight must act somewhere. Since we expect the centre-of-gravity to be a fixed point regardless of orientation (it is a constant of the object), we can then compare those two lines. The centre-of-gravity must be at their intersection.

Answer 3

First of all the torque caused by gravity on an object is not $\vec{r}_{cm}\times M\vec{g}$ it is only in uniform gravitational filed see the torque on each infinitesimal element of the object , $dm$ , is $d\vec{\tau} = \vec{r}\times\vec{F}_{g}$ where $\vec{F}_{g} $ is the gravitational force on $dm$ which in general depends on the inverse square of the element's distance to the object excerting the gravitational force on it , e.g. the earth but considering a uniform gravitational filed so $\vec{F}_{g} $ would be $dm$ times $\vec{g}$ ,the torque would be : $$d\vec{\tau} = \vec{r}\times{dm\vec{g}}$$ now summing (integrating) over all these elements of the object we would have (note that $\vec{g}$ is constant):

$$\int{d\vec{\tau}} = \left ( \int {\vec{r}dm} \right )\times\vec{g}$$ the first term in the right side is just $M\vec{r}_{cm}$ therefore the total gravitational torque on the body , $\vec{\tau}_{g}$ , would be : $$\vec{\tau}_{g} = \vec{r}_{cm}\times{M\vec{g}}$$

That's why they call it "center of gravity" since in general it does not coincide with the center of mass of the body , but since $\vec{F}_{g}$ does not change much in a (sufficiently small) object it's a pretty good approximation to write $ \vec{\tau}_{g} =\vec{r}_{cm}\times{M\vec{g}}$ when calculating the torque caused by gravity.

Now back to your question, you can choose the origin for calculating torque and angular momentum to be any ("inertial") point in space that you want , but since the pivot force is unknown it is better to eliminate its effect in the net torque by choosing our origin right there ($\vec{r}_{pivot} = 0$ so ... ) now the only force that has a torque is gravity , which is $\vec{\tau}_{g} =\vec{r}_{cm}\times{M\vec{g}}$ now $\vec{r}_{cm}$ is nonzero so the only possible way for zero torque is for $\vec{r}_{cm}$ and $M\vec{g}$ (force of gravity ) to be on the same line (forming a zero angle) by hanging the objects from two different points you will find two intersecting lines that will intersect at the center of gravity.