## SAT Physics Subject Test

**Chapter 6 ****Curved and Rotational Motion**

**CENTER OF MASS**

The center of mass is the point where all of the mass of an object can be considered to be concentrated; it’s the dot that represents the object of interest in a free-body diagram.

For a homogeneous body (that is, one for which the density is uniform throughout), the center of mass is where you intuitively expect it to be: at the geometric center. Thus, the center of mass of a uniform sphere or cube or box is at its geometric center.

If we have a collection of discrete particles, the center of mass of the system can be determined mathematically as follows. First, consider the case where the particles all lie on a straight line. Call this the *x-*axis. Select some point to be the origin (*x* = 0) and determine the positions of each particle on the axis. Multiply each position value by the mass of the particle at that location, and get the sum for all the particles. Divide this sum by the total mass, and the resulting *x* value is the center of mass:

The system of particles behaves as if all its mass, *M* = *m*_{1} + *m*_{2} + … + *m** _{n}*, were concentrated at a single location,

*x*

_{cm}. The subscript cm in

*x*

_{cm}stands for center of mass.

In the example above, =−8. The system behaves as if an object with mass 8 kg were 8 units to the left of the origin.

If the system consists of objects that are not confined to the same straight line, use the equation above to find the *x*-coordinate of their center of mass, and the corresponding equation

to find the *y*-coordinate of their center of mass.

**Another Instance of Newton’s Second**

From the equations for the center of mass, we see that

**F**_{net} = *M***a**_{cm}

In other words, the net (external) force acting on the system causes the center of mass to accelerate according to Newton’s second law.

If the net external force on the system is zero, then the center of mass will not accelerate.

6. Object *A*, of mass 5 kg, and object B, of mass 10 kg, hang from light threads from the ends of a uniform bar of length 18 and mass 15 kg. The masses *A* and *B* are at distances 6 and 12, respectively, below the bar. Find the center of mass of this system.

Here’s How to Crack It

The center of mass of the bar alone is at its midpoint (because it is uniform), so we can treat the total mass of the bar as being concentrated at its midpoint. Constructing a coordinate system with this point as the origin, we now have three objects: one with mass 5 at (–9, –6), one with mass 10 at (9, –12), and one with mass 15 at (0, 0).

**Choice of Origin**

Remember, you can

choose your origin

anywhere, but a good

strategy is to use the

location mentioned in the

problem. If it asks, “How

far from the left is the

center of mass?” choose

the left as your origin.

We figure out the *x*- and *y*-coordinates of the center of mass separately.

Therefore, the center of mass is at

(*x*_{cm}, *y*_{cm}) = (1.5, –5)

relative to the midpoint of the bar.

**Questions 7-8**

A man of mass 70 kg is standing at one end of a stationary, floating barge of mass 210 kg. He then walks to the other end of the barge, a distance of 90 meters. Ignore any frictional effects between the barge and the water.

7. How far will the barge move?

8. If the man walks at an average velocity of 8 m/s, what is the average velocity of the barge?

Here’s How to Crack It

7. Since there are no external forces acting on the man + barge system, the center of mass of the system cannot accelerate. In particular, since the system is originally at rest, the center of mass cannot move. Letting *x* = 0 denote the midpoint of the barge (which is its own center of mass, assuming it is uniform), we can figure out the center of mass of the man + barge system:

So the center of mass is a distance of 11.25 meters from the midpoint of the barge, and since the man’s mass is originally at the left end, the center of mass is a distance of 11.25 meters to the left of the barge’s midpoint.

When the man reaches the other end of the barge, the center of mass will, by symmetry, be 11.25 meters to the *right* of the midpoint of the barge. But, since the position of the center of mass cannot move, this means the barge itself must have moved a distance of 11.25 + 11.25 = 22.5 meters to the left.

8. Let the time it takes the man to walk across the barge be denoted by *t*; then *t* = . In this amount of time, the barge moves a distance of 22.5 meters in the *opposite* direction, so the velocity of the barge is