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Massachusetts Institute of Technology - Physics Department
Physics - 8.01 Assignment #1 Fall 1999
SOLUTIONS
by David Pooley — [email protected]
Problem 1.1 (Estimates and Uncertainties – Ohanian Question 1.1)
For the first two objects, I estimated the error by using half the smallest division on my wooden ruler, 0.5 mm. For the last two objects, I used a metal tape measure which can expand and contract with changing temperature, just as the wooden ruler can with changing humidity. For large distances, such as a desk, it can produce a noticeable error, which I estimated at ±2 mm.
Object Estimate by Eye Measured Value Comments Mug (height) 22 cm 15.75 ± 0.05 cm I didn’t start off too hot CD Case 15 cm 14.20 ± 0.05 cm a bit better Desk 1.6 m 1.524 ± 0.002 m not bad for something so big Ruler 30.5 cm 30.48 ± 0.05 cm I am the greatest!
Problem 1.2 (Fundamental Units – Ohanian Question 1.14)
As Ohanian says, “position, time, and mass give a complete description of the behavior and the attributes of an ideal particle.” Therefore, we must have units of each of these ([L], [T], and [M]) in some combination in any system of units that we use. If we take length, mass, and density as our fundamental units, we have
[length] → [L] , [mass] → [M] , [density] →
[M]
[L]^3
As you can see, we have no units of time [T] so we cannot take these as the three fundamental units. If, however, we take length, mass, and speed, we have
[length] → [L] , [mass] → [M] , [speed] →
[L]
[T]
Here, we have all the necessary units, and this is a valid choice for the three fundamental units. It is important to remember that we must be able to “separate” the fundamental units from each other. For example, mass and speed alone contain all the necessary units but there would be no way to separate [L] from [T] with just these two. By having length, mass, and speed, we can extract each fundamental unit on its own (for example, by dividing length by speed to get [T]).
Problem 1.3 (Thickness of a Sheet of Paper)
a) Thickness = 10.0 mm b) Absolute uncertainty = ± 0 .5 mm c) Relative uncertainty = absolute uncertaintymeasurement = 100 .5 mm.0 mm = 5% d) There’s 85 pages, so the thickness of one page is 10.^0 ± 850 .5 mm = 118 ± 6 μm. e) Absolute uncertainty = 6 μm
f ) The relative uncertainty is unchanged – 5%. g) Different students will apply different amounts of pressure when measuring the thickness. Another factor is the humidity of the room in which the measurement is done, as this will affect the thickness of the paper. Also, there is a possibility that the paper is slightly different from book to book.
Problem 1.4 (Relative Uncertainties)
The relative uncertainty is the absolute uncertainty divided by the value of the measure- ment. Let’s pick the antelope as our bone. It has a measured thickness of 18.3 mm with an uncertainty of 1.0 mm; therefore, its relative uncertainty is 1 .0 mm 18 .3 mm
The student’s length (using Zach from the 10 AM lecture) was 183.2 cm with an uncertainty of 0.1 cm. The absolute uncertainties have roughly the same value (in the case of the antelope, it’s exactly the same) in both cases. However, the relative uncertainty in Zach’s length is 0 .1 cm 183 .2 cm
Why are the two relative uncertainties so very different? Because Zach is much longer than the antelope’s femur is thick. They both have the same absolute uncertainty because they both come from the same source, namely, human error in “eye-balling” a ruler or meter stick. The relative uncertainties, however, are two orders of magnitude different!
Problem 1.5 (Distant Quasar – Ohanian Problem 1.8)
To figure out the distance on a map, all you need to do is multiply the actual distance (12. 4 × 109 ly) by the scale of the map (1 : 1. 5 × 1020 ).
(actual distance) = 12. 4 × 109 ly ·
- 46 × 1015 m 1 ly
= 1. 17 × 1026 m
(map distance) = (actual distance) · (scale) = 1. 17 × 1026 m ·
1. 5 × 1020
= 7. 82 × 105 m = 782 km = 486 miles = roughly the distance between Cambridge, MA and Richmond, VA
Problem 1.6 (Distances on Earth – Ohanian Problem 1.10)
r
d
r Equator
Pole s
North
φ=π/
The distance from the pole to the equator measured along the sur- face of the earth is the length of the arc s in the figure. An easy way to remember how to compute arclength is to realize that the ratio of the arclength s to the whole circumference is the same as the ratio of the subtended angle φ to 2π. s 2 πr
φ 2 π s = rφ = 6. 37 × 106 m ·
π 2 = 1. 00 × 107 m = 6220 miles The distance d along a straight line is given by the Pythagorean theorem. d = r
2 = 9. 01 × 106 m = 5600 miles
Problem 1.10 (Position, Velocity, and Acceleration) a) x(t) = 16 − 12 t + 2t^2 (see plot)
b) By differentiating, v(t) = −12 + 4t (see plot)
c) Another differentiation yields a(t) = 4 (see plot)
d) Using the equation in part b): v(0) = −12 m/s, v(2) = −4 m/s, v(4) = 4 m/s
e) Using the equation in part c): a(0) = 4 m/s^2 , a(2) = 4 m/s^2 , a(4) = 4 m/s^2
f ) Set v(t) = −12 + 4t = 0 to find that t = 3 s. Now, plug this t into the equation for x(t) to find the position of the object when the velocity is 0. To summarize x(3) = −2 m, v(3) = 0 m/s, and of course a(3) = 4 m/s^2.
g) The average velocity between two times is de- fined as v¯t 1 ,t 2 ≡ x(t 2 ) − x(t 1 ) t 2 − t 1
→ ¯v− 1 , 3 = x(3) − x(−1) 3 − (−1)
= −8 m/s
h) We use the same formula
¯v 0 , 6 = x(6) − x(0) 6 − 0
= 0 m/s
i) The average speed s is the total distance trav- eled over the time taken. The total distance is found by adding up each one-way segment of the complete journey. Looking at the plot, we can see that the object traveled from x = 16 m to x = −2 m and back to x = 16 m (this is why plots are useful). The total distance is 36 m and the time taken is 6 s. Therefore, the average speed is s = (36 m)/(6 s) = 6 m/s.
0
2
4
6
8
10
12
14
16
18
0 1 2 3 4 5 6
x (m)
t (s)
x(t) = 16 − 12 t + 2t^2
0
4
8
12
0 1 2 3 4 5 6
v (m/s)
t (s)
v(t) = −12 + 4t
0
1
2
3
4
5
0 1 2 3 4 5 6
a (m/s
2 )
t (s)
a(t) = 4
j) What does it mean to reverse direction? Of course, it means that you’re going one way, and then you turn around and go the other way. If you’re going in the negative direction, you then start going in the positive direction and vice versa. In other words, your velocity changes sign. The point at which this happens is at zero (this is an important concept). You can also think about it in terms of the calculus. If you reverse direction, then your position will be a local minimum or maximum at that point. To find where this happens, you take the derivative of position and set it equal to zero. This is exactly the same as setting velocity equal to zero since velocity is the derivative of position! ¿From part f ), we know that v(3) = 0 so the object reverses direction at t = 3 s.
Problem 1.11 (Car Crash and Seat Belts – Ohanian Problem 2.35)
This is a tough problem because the answer requires many steps, and at the end it is essential to ask the all-important question: “Does my answer make sense?” First of all, what exactly does it mean to crash into something? It means that two objects (e.g., a passenger and a dashboard) come into contact, i.e., that their positions are equal at the same time. This is the first step – describing the position of each object during the period of deceleration. We know that for uniform acceleration, x(t) = x 0 + v 0 t +
at^2 and now we must figure out what x 0 , v 0 , and a are for the passenger and for the dashboard. We need to set up a coordinate system. Imagine freezing time right as the car begins to crash. We’ll put our origin at the position of the passenger. His initial conditions are x 0 = 0 and v 0 = 50 km/h = 14 m/s. The dashboard is 0.6 m in front of the passenger and has initial conditions x 0 = xsep = 0.6 m and v 0 = 14 m/s. What about acceleration^1? Since the dashboard is attached to the car, it will accelerate uniformly at a = −200 m/s^2 , but, since the passenger isn’t wearing a seat belt and therefore isn’t connected to the car, he experiences no acceleration as the car begins to crash; he “flies out of his seat.” He of course accelerates quite quickly when he hits the dashboard. Until then, he continues to travel forward with velocity v 0. The equations of position for the passenger (p) and dashboard (d) are xp(t) = 0 + v 0 · t +
· 0 · t^2 = v 0 · t
xd(t) = xsep + v 0 · t +
· (−a) · t^2
Okay, so we’ve described the positions, but we need to keep in mind that these equations are only valid during the period of uniform acceleration, i.e., from when the car starts accelerating until the time it comes to rest. We have position as a function of time x(t), and we can take a derivative to get velocity as a function of time v(t). So, let’s solve for the specific time, tbang when the position of the passenger is equal to the position of the dashboard. Then we can plug that into our equations for v(t) and find the (almost) final answer. xp(tbang) = xd(tbang) v 0 · tbang = xsep + v 0 · tbang −
· a · (tbang)^2
0 = xsep −
· a · (tbang)^2
tbang =
√ 2 xsep a
√ 2 0 .6 m 200 m/s^2
(^1) It may sound awkward at first to speak of something crashing into a wall as “accelerating,” but this is the language of physics. Remember, to “accelerate” merely means to change velocity, whether it be increasing velocity or decreasing velocity. Non-physicists generally use the term “decelerate” to indicate that an object’s speed is decreasing. For example, suppose the car were traveling backward with velocity v = −14 m/s and then crashed until its velocity were v = 0 m/s. Strictly speaking, it’s velocity increased, but because its speed decreased there might be confusion over whether to call it an “acceleration” or a “deceleration.” We overcome the confusion in 8.01 by calling ALL changes in velocity “accelerations.”
Equator
South Pole
φ
r
R = 6357 km
l start
l circ
s = 10 km
Assuming the Earth is a perfect sphere, we can use the figure on the left to figure out the latitude lcirc 1 of the circle of circumference 10 km. From the figure, we see that this leads to the condition 2 πr = 10 km, where r = 6357 km · sin(90◦^ − lcirc 1 ). Solving for lcirc 1 , we find that circle 1 is at lcirc 1 = 89◦ 59 ′ 08 ′′^ South. (The error introduced by our assumption that the Earth is a perfect sphere is less than 1′′.) Now we need to find the latitude lstart 1 of the circle 10 km north of this. From the figure and what we know of arclength, we find that 10 km 2 π · 6357 km
φ 360 ◦^ → φ = 5′ 24 ′′ The latitude lstart 1 of the starting point is then lcirc 1 − φ. lstart 1 = 89◦ 53 ′ 44 ′′^ South This procedure can be repeated for the second circle of circumference 5 km to find lstart 2 = 89 ◦ 54 ′ 10 ′′^ South, and so on. There are an infinite number of these circles, each with an infinite number of points.
Problem 1.13 (Human Femur)
a) I measured d to be 1. 6 ± 0 .2 cm and l to be 24. 5 ± 0 .5 cm. The errors are large because I was not very confident in my ability to measure the image accurately. The ratio is d/l = 1. 6 / 24 .5 = 0.065 To get the uncertainty when we’re dividing measured quantities, we can use one of two methods which are equivalent. Remember that these are simplified methods that we use in 8.01; later in your MIT career, you’ll probably take more formal courses that will teach how to arrive at a final uncertainty which is derived from several quantities. Method One: Figure out the maximum that the computed value can be, i.e., add the error in the numerator, and subtract the error in the denominator: 1 .6 + 0.2 cm
- 5 − 0 .5 cm
1 .8 cm 24 .0 cm
Then subtract the computed value without uncertainties from this to get the range of uncertainty:
- 075 −
1 .6 cm 24 .5 cm
Method Two: When multiplying or dividing measured quantities, add the relative un- certainties of each quantity to get the total relative uncertainty: 0 .2 cm 1 .6 cm
0 .5 cm 24 .5 cm
Multiply this by the computed value without uncertainties to get the total absolute uncertainty: 14 .5% ·
So the final answer is d/l = 0. 065 ± 0. 010. b) To average we just add up all the quantities and divide by the total number of quantities. When adding or subtracting numbers with uncertainties attached, we add the absolute uncertainties together (at least, we do this in 8.01).
d/lavg =
d/lavg = 0. 087 ± 0. 006
c) The following interesting note is from Professor Bernie Burke.
Among mammals, d/l appears to be approximately constant. This is not what one would expect from Galileo Galilei’s reasoning as discussed in lec- tures. This follows from a correct and complete analysis of the scaling for failure of columns. Columns can fail two ways: by having a load so great that the material yields (this is what we evaluated in lectures), or by having a load so great that the column buckles (not considered by Galileo). The point at which the material yields is described by the yield modulus. The yield modulus for well-aged concrete is about 2000 pounds per square inch, and it may yield (i.e. crumble) sooner if it has only been poured for a few days; mild steel is 25 times stronger, and there are special alloys that have a yield modulus 50 times that of concrete. Another form of columnar failure, on the other hand, is determined by the elastic modulus, sometimes called Young’s modulus. Imagine a thin metal column under a load. Theoretically, if it were perfectly vertical, it could sustain a load up to the yield point. If it is not in stable equilibrium, however, any sideways displacement, no matter how small, will develop into a bend, and the column acts like a spring. Try this by pushing down on a vertical plastic ruler; you will see the bending, and you will feel the spring action. The greater the load, the more the sideways bend. If the column is too slender, the bend will increase uncontrollably, and the column will buckle. The great mathematician Leonard Euler tackled this problem. We suggest you look up on the web “Euler Buckling”.
It is unclear why the value of d/l is significantly lower for humans than for the aver- age four-legged mammal. Apparently, we are still far enough from the danger zone of buckling that it does not pose a problem.
