Sine and Cosine Functions applied to motion
Very Short Preliminary Activity with TIMER (should take 5 minutes or less once you get the TIMER loaded)
This exercise can be put off until you are near a computer. However it is best done before some of the problems that follow. If you can't do it before starting the problems, at least imagine doing it, actually doing the 8-counts and clicking an imaginary mouse, and making your best estimate of the time intervals.
Click the mouse as you start an 8-count, doing your best to count at the same rate you used in class. Complete four 8-counts and click the mouse again. Note the time interval required to complete your set of four 8-counts.
Repeat four more times.
Report your five time intervals in the first line below, separated by commas:
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Based on your results, how long does your typical 8-count last?
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Based on your result, what is the time interval of each of your counts?
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If you counted the motion of a ball down the ramp, completing two 8-counts and 1-2-3-4-5 of a third, how long would you conclude the ball spend moving down the ramp? Based on the TIMER data you reported above, what do you think is the percent uncertainty in your result?
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From analyzing the v0, vf, `dt trapezoid we find that whenever the v vs. t trapezoid is an accurate representation of the behavior of the object, the acceleration is uniform and we have
and
These equations apply as long as the straight line segment accurately represents the motion of the object during the interval.
These equations involve five quantities: v0, vf, a, `ds and `dt.
For example if we know a, v0 and `dt:
The first equation includes the quantities `ds, vf, v0 and `dt. Knowing a, v0 and `dt, we would have the values of three of the four quantities in that equation, and would therefore be able to easily substitute the known values to find `ds.
The second equation includes the quantities vf, v0, a and `dt. Knowing a, v0 and `dt we could easily substitute to find the value of vf.
Of course we don't really need the equations to solve this situation. Knowing a, v0 and `dt, we can use a and `dt to find the change in velocity, which we can then add to v0 to obtain the final velocity. Having found the final velocity we can average it with the initial velocity to obtain the average velocity (this works because the v vs. t graph is a straight line, i.e., acceleration is uniform), which we then multiply by the known `dt to find `ds.
As another example, if we know a, v0 and vf:
The first equation includes our known quantities v0 and vf, but also includes the unknown quantities `dt and `ds. We can't solve a single equation for two different unknowns, so the first equation isn't helpful.
The second equation does contain all three of our known quantities, plus the quantity `dt. So we will be able to use this equation to find `dt:
- vf = v0 + a `dt. Subtracting v0 from both sides we get
- a `dt = vf - v0. Dividing both sides by a we get
- `dt = (vf - v0) / a.
If we substitute our values for v0, vf and a into this last form of the equation, we easily find `dt.
As another example, suppose we know v0, a and `ds.
The first equation contains v0 and `ds, but it also contains the unknown quantities vf and `dt.
The second equation contains v0 and a, but it also contains the unknown quantities vf and `dt.
Neither equation can be solved by itself; you can't solve a single equation with two unknowns.
However we do have two equations in the two unknowns vf and `dt. Given two equations in the same two unknowns, we have a system of simultaneous equations, and we can solve the system.
To save time, we go ahead and solve the system symbolically for the variables vf and `dt:
... identifying quantities, units
... problems involving various combinations, using eqns and/or using definitions
These equations should not prevent us from directly reasoning out as much of the solution as possible, using the seven quantities vf, v0, vAve, `dv, a, `dt and `ds and the definitions of average velocity and acceleration.
Most combinations of three of the five variables can be reasoned out without resorting to either of these equations. Only two combinations (namely, (v0, a `ds) and (vf, a, `ds)) cannot be directly reasoned out using the definitions. In those cases, we would require several steps to solve the two simultaneous equations. We can take care of that complication now so we don't have to worry about it later. We will use the process of eliminating one variable from the system, in the first case eliminating vf, in the second case eliminating `dt:
We can pretty easily eliminate vf from the two equations:
The equation vf = v0 + a `dt gives us an expression to substitute for vf in the equation `ds = (vf + v0) / 2 * `dt. We get
`ds = ( (v0 + a `dt) + v0 ) / 2 * `dt. Simplifying the numerator (v0 + a `dt) + v0 we get
`ds = ( 2 v0 + a `dt) / 2 * `dt. Dividing the numerator by 2, being sure to use the distributive law, we obtain
`ds = ( (2 v0) / 2 + a `dt / 2) * `dt = (v0 + 1/2 a `dt) * `dt. Multiplying through by `dt we get
`ds = v0 `dt + 1/2 a `dt^2. We will refer to this as the third equation of uniformly accelerated motion.
We can eliminate `dt from the two equations.
We solve the first for `dt:
a = (vf - v0) / `dt so
`dt = (vf - v0) / a.
We now substitute this expression for `dt in the second equation `ds = (v0 + vf) / 2 * `dt. We obtain
`ds = (v0 + vf) / 2 * (vf - v0) / a.
The right-hand side can be written (vf + v0) ( vf - v0) / (2 a). Since (vf - v0) ( vf + v0) = vf^2 - v0^2, our equation becomes
`ds = (vf^2 - v0^2) / (2 a).
We could leave the equation as it is, but you will see later that there is good reason to rearrange it as follows:
Multiply both sides by 2 a to get
2 a `ds = vf^2 - v0^2.
Then solve for vf^2, obtaining
- vf^2 = v0^2 - 2 a `ds.
We will refer to this as the fourth equation of uniformly accelerated motion.
Our v vs. t trapezoid therefore gives us four equations that apply to uniformly accelerated motion on an interval:
University Physics students:
We can do something similar with the equations we derived previously using integration. Integrating a(t) = a = constant, assuming velocity v0 and position x0 at clock time t = 0, we get the two equations
- v(t) = v0 + a t
- x(t) = x0 + v0 t + 1/2 a t^2.
We can easily eliminate a between these equations to obtain
- x(t) - x0 = (v(t) - v0) / 2 * t
We can also eliminate t between the equations, obtaining
- v(t) = v0^2 + 2 a ( x(t) - x0).
Our four equations are therefore
- x(t) - x0 = (v(t) - v0) / 2 * t
- v(t) = v0 + a t
- x(t) = x0 + v0 t + 1/2 a t^2.
- v^2(t) = v0^2 + 2 a * (x(t) - x0).
Our second equation can be rearranged to the form x(t) - x0 = v0 t + 1/2 a t^2, so using s(t) = x(t) - x0 this equation becomes s(t) = v0 t + 1/2 a t^2. If we use s(t) for the expression x(t) - x0, our four equations become our four equations can therefore be written
- s(t) = (v(t) - v0) / 2 * t
- v(t) = v0 + a t
- s(t) = v0 t + 1/2 a t^2
- v^2(t) = v0^2 + 2 a s(t).
The expression s(t) is understood to represent the displacement x(t) - x0 since initial clock time t = 0. s(t) can always be replaced by x(t) - x0, should it be convenient to do so. If x0 = 0, then we can replace s(t) with x(t).
If we are talking about motion on an interval, then s(t) is just the displacement `ds from the beginning of the interval to the end, and v(t) is just the final velocity on the interval. The clock time t becomes the time duration of the interval, which we call `dt. Thus, using the symbols `ds, vf and `dt, our equations become
- `ds = (v0 + vf) / 2 * `dt
- vf = v0 + a `dt
- `ds = v0 `dt + 1/2 a `dt^2
- vf^2 = v0^2 + 2 a `ds.
These are the same as the equations obtained previously from the v vs. t trapezoid.
The three forms of the four equations as given above are all equivalent. One form can be obtained from and of the other two.
You should be able to quickly derive the third set of equations from the v0, vf, `dt trapezoid.
You should be able to derive the first set using integration.
You should understand how the two are connected, so that given one form you can always write down the other.
Sine and cosine functions in analysis of motion
If x(t) = A sin(omega * t), then x(t) describes a motion of amplitude A and period 2 pi / omega. An ideal pendulum or a mass on an ideal spring will move in such a manner. Much more about that later in the course.
If x(t) = A sin(omega * t) then
v(t) = x ' (t) = omega A cos(omega * t)
describes the corresponding velocity as a function of clock time, and
a(t) = v ' (t) = -omega^2 A sin(omega * t)
describes the corresponding acceleration.
For future reference: This motion would be called simple harmonic motion, and it results whenever an object experiences a net force toward some equilibrium position, with the net force being directly proportional to the distance from that position.
Related note: One student asked whether optics is a second-semester topic because it is difficult. Any topic can be difficult to just about any degree, depending on how deeply we go into it. However second-semester topics aren't delayed until second semester because they are difficult. They are delayed because they rely on very basic first-semester topics like motion, force, energy and momentum, as well as on other first-semester topics such as simple harmonic motion, and/or gravitation, and/or angular dynamics and/or others. The course gets progressively more difficult if early topics aren't mastered; with good mastery the topics get progressively easier.
Note on questions (accessible to University Physics students)
Two vectors are parallel if their cross product is zero, or if their dot product is equal to plus or minus the product of their magnitudes, or if one is a scalar multiple of the other (which leads to three simultaneous equations in a single unknown; the value of the unknown must be the same for all three of the equations). The simplest test for orthogonality is whether the dot product is zero.
An object thrown up in the air will return to its original vertical position with a speed equal to its original speed, provided only gravity acts on it. If other forces such as air resistance act on it, this is no longer the case.
To find the final velocity without actually measuring it, you could integrate the acceleration function to get the change in velocity. If you know the original velocity you can then add the change to the original velocity to get the final velocity. If it's a uniform acceleration situation, then if you know three of the quantities v0, a, `dt and `ds, you can either reason out the final velocity or use the equations to find it.
The x and y components of the length of the rubber band are both displacement; neither is a force. The x and y components of the force vector are both forces; neither is a displacement. The tension of a rubber band has a magnitude (e.g., how many Newtons of force) and a direction. The direction of the tension force will be the same as that of the displacement vector from one end of the rubber band (the one at which you wish to find the tension vector) to the other. To find a unit vector in the direction of the displacement vector, divide the components of the displacement vector by its magnitude (equal to the length of the rubber band). To get the force vector, multiply this unit vector by the magnitude of the force.