G9-4a. The Concept of Center of Mass and its Use
In all the problems we've done up to this point, we've treated the objects in the problem as if they were point masses. In drawing force diagrams, we've represented the objects as point masses on which the forces act. While this makes for some convenience in solving problems, physics demands more than convenience in order to justify using any particular method. The method must provide results that describe how objects actually move and interact. As you might have suspected, there is justification as long as one picks the right point to use to represent the object or a collection (system) of objects. That point is called the center of mass of the object.
There are several ways to determine where the center of mass of an object is. Two of them are operational; that is, one must perform an experiment. One method is intuitive, and the other requires calculation using a formula. Here are the methods:
The intuitive method: In cases where the object has geometric symmetry and its mass is distributed uniformly, then the center of mass is the geometric center. For example, the center of mass of a sphere or cube of uniform density is the center of the sphere or cube.
Operational methods: If you can balance the object on a point support, then the center of mass is either at that point or on a vertical line that runs through the point. An alternative method is to hang the object from a string. The center of mass will be on the extension of the vertical line formed by the string. Suspend the same object from a different point to form a second vertical line that intersects the first. The intersection is the location of the center of mass. These methods are demonstrated in one of the center-of-mass videos.
Formulaic method: The most general form of this method uses calculus and requires integrating the position vector over the mass distribution. In cases where the object can be represented as a system of point objects,the following formulas are used:
where represents the sum of the products of the masses and corresponding x-coordinates of each of the point objects, and represents the sum of the masses of the point objects.
In all the problems that we will do, we'll treat the objects as 2-dimensional and just use the x- and y-equations above. Use of the formulas requires that one has set up the origin and directions of the coordinate axes. For convenience in dealing with a system of two or more point objects, the origin is typically selected as the position of one of the objects so that the mixi and miyi products for that object will be 0.
Calculation of the Center of Mass
Examples of the use of the formulas follow.
A barbell has a 10.0 kg mass on one end and a 4.0 kg mass on the other. The masses are separated by a very strong bar a distance of 1.5 m apart. A weight lifter wants to hold the barbell above her head with a single hand placed in such a location as to balance the weight and make it as easy as possible to lift the barbell.
For a 2-dimensional example, consider problem 9-43 on page 277 of your text. Read the problem statement now. The selection of the coordinate system is key to making that problem simple. With the origin at the center of the sulphur molecule, that molecule will not contribute to the numerator of the center of mass formula. With the two oxygen molecules oriented symetrically about the +y axis, we can see that xcm must be 0. That only leaves one to calculate ycm. We leave it to you to complete the problem for practice.
Motion of the Center of Mass
In the barbell example above, if the bar were supported to the left of the center of mass, the barbell would rotate clockwise about the point of support. If the bar were supported to the right of the center of mass, the rotation would be counterclockwise. Knowing the location of the center of mass helps us to determine how an object will behave when not supported at that point. Later we will see how to deal with that situation. (It involves the definition of the concept of torque, which we take up in Chapter 11.)
A situation that we're in a position to consider now is that of 1-dimensional collisions of two objects. By treating the objects as a system, we can determine the center of mass of the system and how the center of mass moves. We find that the center of mass moves with uniform, constant velocity before, during, and after the collision as long as momentum is conserved.
Think of the center of mass of the system as representing the location and mass of the entire system. If the momentum of the system is conserved and the masses don't change, then the velocity of the center of mass can't change. The following argument will show how the former statement follows from the physics that you already know. We'll limit the argument to a collision of 2 objects in 1-dimension; however, the argument can easily be extended to any number of objects in 2 or 3 dimensions.
Let's apply the formula for the center of mass of a system of two objects to two instants of time that are an interval Δt apart.
Subtract the first formula from the second.
Divide both sides of the equation by Δt.
The expressions are average velocities. However, we can take the limit as Δt approaches 0 to obtain instantaneous velocities.
With the substitutions , we have the following.
where the upper-case symbol Vcm represents the velocity of the center of mass of the system.
Bring the sum of the masses to the left-hand side and to obtain
, where Msys has been substituted for the mass of the system.
The quantity on the right is the total momentum of the system. If the momentum of the system is conserved, MsysVcm must remain constant. If Msys is constant, then the center of mass velocity is also constant.
Let's take the argument a step further. We've said previously that the total momentum of a system is conserved if the net, external force Fnet,ext on the system is 0. Let's see how this condition arises by determining an expression for the acceleration of the center of mass of the system. We can provide a simplification in the notation by introducing the definition of the derivative from calculus. Applied to this situation, we have for the velocities,
We take the derivative of both sides of the equation to obtain
or in terms of acceleration
, where Acm represents the acceleration of the center of mass of the system.
Multiply both sides of the equation by the sum of the masses to obtain
The product on the left is the net, external force on the system. Thus, we have
The only accelerations are those that occur during the collision. At all times during the collision, Newton's 3rd Law tells us that the colliding objects exert equal and opposite forces on each other. Assuming that all other forces acting on the objects add to 0, we must have . Substituting into the equation above gives
The above being the case, the acceleration of the center of mass Acm of the system must be 0. Hence, the velocity of the center of mass is constant. This completes the argument.
For an illustration of the motion of the center of mass in a collision, click on this link. The center of mass is represented as a black dot. Also click on Show Graph to display the velocity vs. time graphs for each object as well as for the center of mass. Note that the latter is a horizontal line. (Ignore the bumps and wiggles, which are artifacts of the resolution of the graphic.)
Rotation about the Center of Mass
The net, external force on a system is equal to the product of the total mass of the system and the acceleration of the center of mass. In the argument above for a 1-dimensional collision of 2 objects, we took Fnet,ext to be 0 so that we could say the momentum of the system was conserved. However, we can also apply the general relationship Fnet,ext = MsysAcm to systems on which the net, external force is not zero. We can describe the motion of such systems, at least in part, by the motion of the center of mass. For, example, the photo to the right shows a hammer tossed into the air. The position of the center of mass of the hammer is indicated by the black dot. (This was determined by suspending the hammer by a string.) If you traced the motion of the center of mass, you would see that the path was parabolic, as expected for the motion of a point mass in a uniform gravitational field. Describing the motions of other points of the hammer is determined by employing the physics of rotational dynamics, which is the subject of Chapter 11.
Solving Elastic Collision Problems Using the Center-of-Mass Method
In the guide, Solving One-Dimensional Elastic Collisions, you were shown a method for determining the final velocities of two colliding objects for which momentum and kinetic energy were conserved. There is another method that involves much less algebra and employs the concept of the center-of-mass velocity. Basically, the method amounts to using the fact that in an elastic collision of two objects, the velocity at which an object in the system approaches the center of mass is equal to the opposite of the velocity at which that same object separates from the center of mass. Click on the link below to see how the method works.
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