PHY 121/123 LECTURE NOTES - Week 2

Last week we considered problems of motion with constant speed along a path, or constant velocity in up to two dimensions. This week, we go one step further, to consider motion with constant acceleration in one or two dimensions. A little bit of calculus sets the stage for the discussion. The solution to any problem involving constant acceleration may be written:

$${\bf \vec{x}}(t) = {\bf \vec{x}}_0 + {\bf \vec{v}}_0t+(1/2){\bf \vec{a}}_0t^2 .$$

where the subscript "₀" indicates that a quantity is a constant:

${\bf \vec{x}}_0={\bf \vec{x}}(t=0)$, etc. Now let's use differentiation to verify that this indeed corresponds to constant acceleration:

$${\bf \vec{v}}(t)=d{\bf \vec{x}}/dt= {\bf \vec{v}}_0+(1/2)(2t){\bf \vec{a}}_0={\bf \vec{v}}_0+{\bf\vec{a}}_0t.$$

Because the velocity is changing linearly with time, this is by definition an example of constant acceleration, but let's just differentiate once more to see it another way, getting

$${\bf\vec{a}}(t)={\bf\vec{a}}_0.$$

We need to look at the meaning of the vector notation: If we describe vectors by components, then a time derivative just means we should differentiate the components, e.g., (a₁,a₂)=(dv₁/dt,dv₂/dt).

Armed with the general formulas above, we can solve any problem involving constant acceleration. Problems can be simple or compound. In a simple problem, we are given enough information to determine directly all except one quantity in the formulas, and then we find the remaining ``unknown'' quantity from the formula. For example, we could be given a displacement, initial velocity, and time interval, and then have to compute the acceleration: John jumped off a diving board with negligible initial velocity, and hit the water 3 m below in 0.775 s. What was the average acceleration in the vertical direction? We are not concerned at all with horizontal motion in this problem. We aren't even told if John landed directly below the end of the diving board or not. The problem therefore is in just one dimension, the vertical. This gives us

$$3 \ {\rm m} = (1/2) a (.775 \ {\rm s})^2,$$

or

$$a=2\times 3/0.6 = 10 \ {\rm m/s^2} .$$

Here is another example in one space dimension, again with gravity, again using the standard `lecture' value for g, 10 m/s². Suppose I throw a ball straight up with initial speed $v_0=20 \ {\rm m/s}$, how long will it take for the ball to return to my hand at the same height it left? The height as a function of time, starting with my hand as the zero point, is

y=v₀t-(1/2)gt².

We want to find when the ball returns to my hand, and therefore when this expression vanishes. One solution is t=0, but this is trivial - just the moment of release when the ball is thrown up. Leaving this time out, we may divide the expression by t, leaving an equation for the time of catching,

$$t_c = 2v_0/g= 40 \ {\rm (m/s)^2/(10\ m/s^2) = 4 \ s} .$$

Suppose we ask a different question, how long does it take for the ball to stop moving up and start moving down? In this case, we use the equation for velocity in the vertical direction, v_y = v₀-gt =0, or t = v₀/g = 2s. Notice that the time for the ball to go up and come back down is exactly twice the time to go up, or in other words, the time to go up is the same as the time to come back down. This is no accident, and we'll talk about that more later on.

Now we come to a problem in 2 space dimensions, the legend of William Tell. He was standing with his bow and arrow 50 m away from his son, who had an apple on top of his head, at the same height as father William's arrow would start. Assuming the arrow is released at 20 m/s, can William aim it so it will hit the apple, and certainly not hit his son? This is a compound problem. First, consider motion of the arrow in the horizontal direction. If the arrow is gliding through the air, we can neglect frictional force, which means that the horizontal motion goes at constant velocity v_x. The travel time is related to the distance D= 50 m by

$$D=v_x \ t ,$$

with $v_x=v_0{\rm cos}\theta$, and $\theta$ the angle above the horizontal direction at which William aims his arrow.

On the other hand, the same time t is determined by the vertical motion, because the arrow has to fall back down to its initial height if it is going to hit the apple, rather than going above it, or (much worse!) below it into William's son. This is where the compound nature of the problem appears. Fortunately, we already know how long it will take for the arrrow to go up and then down - that's the same as the vertical motion of the ball we just discussed. All we need is to know the initial vertical velocity of the arrow, but that is just $v_{y0} =v_0 {\rm sin} \theta$. Using our established formula for what we called t_c above, we get

$$D=v_0^2 \ 2{\rm sin \theta cos\theta}/g =(v_0^2/g){\rm sin 2\theta}.$$

Plugging in the numbers, and assuming the maximum value 1 for the sine of the angle, we get $D ={\rm (20 \ m/s)^2/10 \ m/s^2 = 40 \ m}$, which means that the arrow cannot get more than 40 m from William, or still 10 m before arriving at the apple, before falling below the correct height. If William wants to hit the apple, he either has to move 10 m closer, or shoot the arrow about 10 % faster. Since the sine is one for an angle of 90^o, the maximum D is achieved for an angle of 45^o, halfway between the horizontal and vertical directions.

Another comment on this problem, and projectile motion more generally: As the horizontal displacement is proportional to the time, while the vertical displacement involves $-(1/2)\ g t^2$, therefore the vertical displacement can be written

$$y-y_0=-(1/2) \ g ((x-x_0)/v_{x0})^2 + v_{y0} ((x-x_0)/v_{x0}),$$

yielding a parabolic curve of vertical versus horizontal positions.

Going back to a notion discussed in the CAPA but not really in the class for this week, we can see an interesting example of how a `non-calculus' approach actually sneaks in some calculus: Average velocity is defined by

$${\bf \vec{v}}_{av} = \Delta {\bf \vec{x}}/\Delta t.$$

Now, because by definition ${\bf \vec{v}} = d{\bf \vec{x}}/dt$, it follows that $\Delta {\bf \vec{x}} = \int_{t_0}^{t_0+\Delta t}dt {\bf \vec{v}}(t)$. This means that the integral is nothing but the time-averaged velocity ${\bf \vec{v}}_{av}$ multiplied by the time interval $\Delta t$, so that the integral in fact is defined in these books in terms of the average velocity, without ever using the words `calculus' or `integral'!

A word about the labs. We encourage you to do your best to finish your lab report before leaving the two hour lab. To make this possible, you need to be `on top' of each lab before you start. That means reading the lab manual description in advance, looking over lecture notes and textbook to be sure you understand what is going on, and asking for help in the help room if something still seems mysterious. That means you'll be ready to start on the lab after maybe one or two questions, and thereby get your measurements finished quickly.

Advantages of finishing the report while still in the lab:

1. You will be graded more generously, because there clearly was not time to do a very polished report. You can concentrate on the essentials.

2. You can be sure that the TA received your report - No question of its getting lost in the return boxes.

3. Perhaps most important - if you find a big discrepancy while writing things up, you still have a chance to look at the apparatus, try again, and get advice from the TA, so that you can fix the problem. Once you are out of the lab, this becomes impossible.

For the third week of classes, everyone will be doing Lab 2. Unlike Lab 1, which involves an indirect measurement of the acceleration of gravity g using the behavior of a simple pendulum and theory we shall only be discussing after quite a few more weeks, this time you will be making a direct measurement of g using the basic equation for motion with constant acceleration. A transparent ruler with equally spaced clear and covered patches is dropped through a gate where an electric eye gives a signal for each start and stop of a transparent part passing the eye. The automatically recorded signals tell you how far the ruler gets after each time interval, and also measure how the velocity changes with time. Plugging into the basic formula gives you a measurement of g.