Delightful Dynamics
In dynamics, we're concerned with bodies in motion. Specifically, we're interested in bodies that have a net force acting on them, since statics can also be applied to bodies at constant velocity (no net force). This can include falling objects, objects colliding, a ball being thrown, etc. Let's get into some examples to show the basic principles at work.
Let's start with the same ball from the first example of my last post, and instead of having it nicely resting on a table we drop it onto the table. There are three events here:
- The ball falling under the influence of gravity
- The ball colliding with the table
- The stuff that happens after the ball hits the table
Let's take it one event at a time. Our first event is the ball falling under the influence of gravity. We'll ignore drag (the force exerted by the atmosphere on the ball), since really it would only be an additional force acting against the acceleration of the ball. The steps to analyzing a dynamics problem are very similar to a statics problem, with one big difference: we allow there to be a net force on the system.
For our falling ball, the free-body diagram (FBD) looks like:
 |
| FBD for the falling ball |
The ball has weight W, a corresponding mass m, and we take the acceleration due to gravity to be g. So the sum of forces is:
\[\sum{F_y} = W = mg \]
This seems pretty boring, but really is the definition of the weight of an object: the weight is the mass of the object times the acceleration due to gravity. In other words: weight is the force exerted on a body due to gravity. What is more interesting is: how does the position of the ball change with time while falling?
First let's go back to the basic concepts of position, velocity and acceleration. Velocity is the change in position with time and acceleration is the change in velocity with time. To pull out some calculus-speak: velocity is the time derivative of position, and acceleration is the time derivative of velocity (and the second time derivative of position). A derivative really is just a generalization of the idea of a slope of a curve, so we can also talk about this graphically:
 |
| Really, derivatives aren't scary. Just think: slope or rate of change. |
There's another important calculus concept that goes along with derivatives: the concept of the integral. An integral can be thought of as the area under a given curve, but also it is the inverse of a derivative. In other words, if you integrate the derivative of a function, you get back the function (plus some important generic constants called constants of integration that will become important). So to put the relationships between position, velocity and acceleration into the language of mathematics:
\[\mathrm{position:}\:x(t)\]
\[\mathrm{velocity:}\:v(t) = \frac{dx(t)}{dt} \sim \frac{x(t_2) - x(t_1)}{t_2 - t_1}\]
\[\mathrm{acceleration:}\:a(t) = \frac{dv(t)}{dt} = \frac{d^2 x(t)}{dt^2} \sim \frac{v(t_2) - v(t_1)}{t_2 - t_1}\]
\[v(t) = \int{a(t)\:dt}\]
\[x(t) = \int{v(t)\:dt} = \int{\int{a(t)\:dt}dt}\]
If we take the acceleration, a(t), to be constant (a), we can then derive the following relationships:
\[a(t) = a\]
\[v(t) = \int{a\:dt} = at + C_1\]
\[x(t) = \int{v(t)\:dt} = \int{(at + C_1)\:dt} = \frac{1}{2}at^2 + C_1t + C_2\]
If we define the initial velocity \(v(0) = v_0\) and initial position \(x(0) = x_0\), then we have the following:
\[a(t) = a\]
\[v(t) =at + v_0\]
\[x(t) = \frac{1}{2}at^2 + v_{0}t + x_0\]
We can apply these equations to our falling ball if we know its acceleration, initial position and initial velocity. If the ball was being held a distance h above the table, at the moment of release its velocity is zero (\(v_0 = 0\)) and its initial position is \(x_0=h\) if we take the x-axis in the vertical direction, with the origin at the top of the table. This selection of axis means that we have an acceleration of -g (here the negative denotes the acceleration acts downward), and since we have a zero initial velocity this results in a downward (negative) velocity.
Substituting these values into the above equations, we arrive at the equations which govern the motion of the ball as it falls, called its equations of motion:
\[a(t) = -g\]
\[v(t) =-gt\]
\[x(t) = -\frac{1}{2}gt^2 + h\]
These equations are valid until the moment of contact with the table (x(t) = 0). But what happens if we decided to toss the ball upward a bit when we let go of it? In effect, we would have given it a non-zero initial velocity, let's call it u. So our FBD looks the same, but now our equations of motion look like:
\[a(t) = -g\]
\[v(t) =-gt + u\]
\[x(t) = -\frac{1}{2}gt^2 + ut + h\]
In words, this means that now the acceleration due to gravity also has to fight the initial positive velocity. It also means that the time it takes to fall is greater than the case with no initial velocity. We can see this in the displacement equation above. Because the acceleration term in the displacement (\(\frac{1}{2}gt^2\)) must be larger since there's an additional positive term (\(ut\)) it has to counteract. Conversely, if the initial velocity were negative (downward), the fall time would be shorter.
What happens when the ball hits the table? Let's look at a diagram for the moment just before the ball contacts the table, and just after:
At the moment just before contact, the ball has a kinetic energy of \(E_i = \frac{1}{2}mv_i^2\). The initial velocity, \(v_i\), can be determined from the height the ball was dropped, based on the above equations. We can use the conservation of energy principle to determine how the velocity changes. Since we've made the table rigid, impenetrable and immobile, it can't absorb any energy from the ball and we can reasonably assume the mass of the ball doesn't change, so we can set the kinetic energy of the ball just prior to the collision equal to the kinetic energy just after:
\[E_i = E_f\]
\[ \frac{1}{2}mv_i^2 = \frac{1}{2}mv_f^2\]
\[v_i^2 = v_f^2\]
Which is satisfied by both of the following solutions:
\[v_i = v_f\]
\[v_i = -v_f\]
So which is the correct answer? The first corresponds to the ball continuing to fall downward, through the table, which we can rule out because we assumed the table to be rigid and unable to be penetrated. So the velocity must change directions, causing the ball to bounce off the table: the result we'd expect intuitively. But we've swept something under the rug here: conservation of momentum. The principle of conservation of momentum requires that the initial and final values of momentum for an isolated (or closed) system are the same. However, our ball isn't an isolated system, the wall exerts a force on it (\(F_{r}(t)\)) causing the change in its momentum (due to the change in velocity). The influence of the force on the momentum is called an impulse, and is a consequence of Newton's laws of motion. So our momentum conservation equation for the ball would look like the following:
\[p_f - p_i = J = \int{F_r(t)dt}\]
\[mv_f - mv_i = m(v_f-v_i) = J\]
Because we already determined the change in velocity (\(v_f = -v_i\)), we can determine the impulse the table exerted on the ball:
\[J = m(v_f-v_i) = -m(2v_i) = -2mv_i\]
Let's look at the moment of contact a bit closer. The above analysis doesn't involve any time scale, and in effect what we've done is allow the ball to instantaneously change direction because of its rigidity. But nothing is actually perfectly rigid, and during the contact some level of deformation occurs. This takes time, and that is the time over which the impulse we just discussed acts. A much more physically reasonable description of a perfectly elastic ball hitting the table is the following:
\[v(t) = \int{a\:dt} = at + C_1\]
\[x(t) = \int{v(t)\:dt} = \int{(at + C_1)\:dt} = \frac{1}{2}at^2 + C_1t + C_2\]
If we define the initial velocity \(v(0) = v_0\) and initial position \(x(0) = x_0\), then we have the following:
\[a(t) = a\]
\[v(t) =at + v_0\]
\[x(t) = \frac{1}{2}at^2 + v_{0}t + x_0\]
We can apply these equations to our falling ball if we know its acceleration, initial position and initial velocity. If the ball was being held a distance h above the table, at the moment of release its velocity is zero (\(v_0 = 0\)) and its initial position is \(x_0=h\) if we take the x-axis in the vertical direction, with the origin at the top of the table. This selection of axis means that we have an acceleration of -g (here the negative denotes the acceleration acts downward), and since we have a zero initial velocity this results in a downward (negative) velocity.
Substituting these values into the above equations, we arrive at the equations which govern the motion of the ball as it falls, called its equations of motion:
\[a(t) = -g\]
\[v(t) =-gt\]
\[x(t) = -\frac{1}{2}gt^2 + h\]
These equations are valid until the moment of contact with the table (x(t) = 0). But what happens if we decided to toss the ball upward a bit when we let go of it? In effect, we would have given it a non-zero initial velocity, let's call it u. So our FBD looks the same, but now our equations of motion look like:
\[a(t) = -g\]
\[v(t) =-gt + u\]
\[x(t) = -\frac{1}{2}gt^2 + ut + h\]
In words, this means that now the acceleration due to gravity also has to fight the initial positive velocity. It also means that the time it takes to fall is greater than the case with no initial velocity. We can see this in the displacement equation above. Because the acceleration term in the displacement (\(\frac{1}{2}gt^2\)) must be larger since there's an additional positive term (\(ut\)) it has to counteract. Conversely, if the initial velocity were negative (downward), the fall time would be shorter.
What happens when the ball hits the table? Let's look at a diagram for the moment just before the ball contacts the table, and just after:
 |
| Diagram of ball just before and just after impact with the table |
\[E_i = E_f\]
\[ \frac{1}{2}mv_i^2 = \frac{1}{2}mv_f^2\]
\[v_i^2 = v_f^2\]
Which is satisfied by both of the following solutions:
\[v_i = v_f\]
\[v_i = -v_f\]
So which is the correct answer? The first corresponds to the ball continuing to fall downward, through the table, which we can rule out because we assumed the table to be rigid and unable to be penetrated. So the velocity must change directions, causing the ball to bounce off the table: the result we'd expect intuitively. But we've swept something under the rug here: conservation of momentum. The principle of conservation of momentum requires that the initial and final values of momentum for an isolated (or closed) system are the same. However, our ball isn't an isolated system, the wall exerts a force on it (\(F_{r}(t)\)) causing the change in its momentum (due to the change in velocity). The influence of the force on the momentum is called an impulse, and is a consequence of Newton's laws of motion. So our momentum conservation equation for the ball would look like the following:
\[p_f - p_i = J = \int{F_r(t)dt}\]
\[mv_f - mv_i = m(v_f-v_i) = J\]
Because we already determined the change in velocity (\(v_f = -v_i\)), we can determine the impulse the table exerted on the ball:
\[J = m(v_f-v_i) = -m(2v_i) = -2mv_i\]
Let's look at the moment of contact a bit closer. The above analysis doesn't involve any time scale, and in effect what we've done is allow the ball to instantaneously change direction because of its rigidity. But nothing is actually perfectly rigid, and during the contact some level of deformation occurs. This takes time, and that is the time over which the impulse we just discussed acts. A much more physically reasonable description of a perfectly elastic ball hitting the table is the following:
- Ball makes first contact with the table at a single point at velocity \(v_i\), at time \(t = 0\).
- Ball begins to deform, and decelerates. Contact area increases in size.
- Ball's speed hits zero, deformation stops at time \(t = t_{mid}\).
- Ball begins to rebound from the deformation, and begins to accelerate away from table. Contact area decreases in size.
- Ball has returned to original shape and is traveling away from the table at velocity \(v_f\), at time \(t = t_f\).
\[\bar{a} = \frac{(v_f-v_i)}{t_f - 0} = \frac{-2v_i}{t_f}\]
Here it becomes clear how important the contact time is. Longer contact time is lower average acceleration, shorter corresponds to a higher average acceleration. By multiplying by the ball's mass, we arrive at the average force on the ball:
\[\bar{F} = m\frac{-2v_i}{t_f}\]
And since we defined impulse earlier as the time integral of the contact force (area under the force-time curve) we can calculate an average impulse by multiplying the average contact force by the total contact time, \(\Delta t\):
\[\bar{J} = \bar{F}\Delta t = \bar{F} t_f = \left(m\frac{-2v_i}{t_f}\right) t_f = -2mv_i\]
So we find that the impulse we calculated earlier is simply the average impulse during the ball-table contact, taking place over some arbitrary time.
In my next post in this series, we'll get into some mechanics of materials and look more at stuff that deforms!
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