I now discuss a little about calculation of the centre of mass of a mass distribution. Calculation of the centre of mass is similar to calculating the centroid of an area (lecture 7), except that the area is now replaced by mass. For finite masses at given positions, the definition of centre of mass given above is used directly. For a mass distribution in three-dimensions, we calculate all three components of the poison of the centre of mass. These are given as

where dm is a small mass element at the position (x,y,z) in the mass distribution (see figure 4 below).

We are now going to change the topic a bit and ask how we describe a system where a large force acts for very short durations. A cricket bat striking a ball, a hammer hitting a nail, a person jumping on a floor and coming to sudden stop and a carom striker hitting a coin, or collisions in general, are examples of such forces in operation. In these cases it is not meaningful to talk about the force as a function of time because the time span over which the force acts is very-very short. Further, the force varies a great deal over this short time-interval, as I show in an example below. It is therefore better to describe the overall impact of the force in terms of the momentum change it causes to the system. This is given by the integral of the force over the time that it operates. Thus describes the effect of the force on the system. The integral is known as the impulse and denoted by the symbol J. Obviously the momentum change of a system equals the impulse given to it. We now discuss these ideas with the help of an example, that of a ball hitting a wall or any other hard surface.

Let us ask what happens when a ball hits a wall or we jump on the floor. If the ball hitting the wall reflects back, that means that the wall has applied a force on the ball so that

If the time of contact between the ball and the wall is seconds then the average force is

But the real force varies greatly from the average force. We show that now. Take the model of the ball as following Hooke's law so that if it is compressed by x by the wall, it applies a force kx on the wall and consequently experiences an equal force in the opposite direction (see figure 5 below).

Since the force on the ball follows Hooke's law, the ball performs a simple harmonic motion, its compression is given by , where A is the maximum compression and . From time t = 0 , when the ball comes in and touches the wall, it takes   time (half a cycle) before leaving the wall. The force during this time is given as

Since for a hard ball k is very large, . So by the time the ball comes back, the force varies with time as shown in the figure 6 below. Here the maximum force F_max is given by kA and . In the figure we show both F_max and F_average . The latter is calculated as

or

So you see that over this short period force varies a great deal and is hardly ever near the average force that we calculated. The discussion above has been in terms of a model of the force; the exact force will be different this model and so the variation could be even larger than that shown. It is in such situations, when a strong force is applied over a very short time period, that it is much more meaningful to talk of the total momentum change of a particle than the force . Further, in such cases, we generally observe only the initial & final momentum and are hardly concerned about the finer details. It is this change

In the momentum that is known as the impulse. So in the ball rebounding from a hard surface with the same speed as it comes in with, the impulse is , where is the initial momentum of the ball. So instead of talking of the force applied by the ball on the surface, we say that the ball has imparted momentum to the surface it hit. The amount of momentum transferred is equal to the impulse. This has interesting application in calculating the force on a surface when there are many-many particles continuously hitting a surface, for example molecules in a vessel hitting its walls from inside.

We show two situations in figure 7 below. The upper figure shows the variation of force on a wall when particles hit a surface at some time interval. The lower one, on the other hand, shows the situation when particles hit continuously. In the first case the force on the surface due to the particles hitting it varies pretty much like the force due to each particle itself. In the second case, however, the force at any instant is given as the sum of the forces applied by each particle at that time. This gives an almost constant force F_many  as shown in the figure. The value of this force is calculated as follows. Let each particle hitting the surface impart an impulse J to it. If on an average there are n particles per second hitting the surface, then in time Δt the momentum transferred to the surface will be (nΔt)J. The force F_many  will then be given as

Since , the force above can also be written as

Thus when a stream of particles hits a surface, the force applied by them to the surface equals the number of particles striking in time Δt times the average force applied by each one of them, a result that you could have anticipated. This is precisely what happens when a jet of water or flowing mass hits another object.