# Chapter 11: Torque

## 11.3 Center of mass

Try to hold a meterstick from any random spot, and it will probably fall due to the force of gravity pulling on it—gravity applies a torque. However, if you hold it in one particular location, it will not rotate. The force of gravity has not ceased to exist, but by changing the axis of rotation, you can make it so gravity no longer exerts any torque on the meterstick.

The center of mass is the point where an object will balance. In terms of torque, placing the fulcrum (axis of rotation) at the center of mass results in no net torque on the object due to it’s own weight. In other words, we can treat torque due to an object’s own weight as if gravity is acting on the object at the center of mass.

This actually holds for more than just torque! In general, objects behave as if all their mass were concentrated at the center of mass. In this video, you’ll see several that rotate as they are thrown. However, when the center of mass is highlighted, you see that it takes a single smooth path. (This is how we got away with representing our whole system as a single point when we made free body diagrams in chapter 9. That point that we drew represented the center of mass of our system.)

If an object’s mass is uniformly distributed—the assumption we will often (thought not always) be making in this class—the center of mass is at the geometric center of the object. Note that this does not need to be physically located on the object. For example, the center of mass of a doughnut is located in the center of the hole.

#### Example 11.4

You are walking along in the woods, and find a stick that is 50 cm long. When you support the stick by placing your finger 20 cm from the left end, the stick balances.

1. Is the mass of the stick uniformly distributed?
2. How far from the left end is the center of mass?
3. How far from the right end is the center of mass?

The only forces acting on the stick are its weight (the force of gravity of the Earth pulling down on the stick) and the normal force of your finger pushing up. Your finger acts as the axis of rotation. There is no torque exerted by your finger, since the normal force is acting at the axis of rotation. The stick remains balanced because the force of gravity exerts zero torque around your finger. So, the distance from the center of mass to the axis of rotation must be zero. In other words, the location of the center of mass of the stick is where you placed your finger.

1. Since the center of mass of the stick is not an equal distance from either end, we know the stick has a nonuniform mass distribution.
2. We were given the distance from the center of mass to the left end of the stick directly: it is 20 cm.
3. To find the distance from the center of mass to the right end of the stick, we need to do a little bit of geometry:

$(50\ \textrm{cm}) – (20\ \textrm{cm}) = 30\ \textrm{cm}$

The center of mass of the stick is 30 cm from the right end.