Physics I 070831
The following was given as an in-class assessment. Another copy was handed out with the assignment to complete it to the best of your ability over the weekend. Have it ready to submit when you come to class on 9/5/07.
Washer hanging from thread:
Sketch a washer hanging from a thread in each of the following three configurations; this means that you will make three different sketches:
- Hanging at its equilibrium point
- While swinging, at a point which displaced to the right of equilibrium at a distance equal to about 10% of its length
- While swinging, at a point which displaced to the right of equilibrium at its maximum distance, which is a distance equal to about 20% of its length
Assume that the only forces acting on the washer are the Earth’s gravitational pull and the tension in the thread.
Sketch the gravitational force vector and the vector representing the tension acting on the washer, for each of the three sketches.
- For each sketch make the gravitational force vector the same length, about a couple of inches long, and give the tension vector a length which is reasonably consistent with that of the force vector (e.g., if you think the tension is more than the gravitational force your tension vector should be accordingly longer, if less than the gravitational force then accordingly shorter; if the same or about the same as the gravitational force then the tension vector should have about the same length as the gravitational force vector).
- For each sketch estimate the angle of each vector, as measured counterclockwise from a direction parallel to the positive x axis. You will estimate six angles.
- For each sketch, sketch vectors to show how much ‘up’ and how much ‘over’ there is in the tension vector. That is, sketch the vertical and the horizontal components of the tension vector.
- For each sketch, sketch and label the vector that you believe represents the net resultant of the gravitational and tension forces—i.e., the single net force which you believe is equivalent to the action of these two forces. If you think the net force for any of your sketches is zero, then simply write ‘net force zero’.
If in the second sketch the pendulum is swinging out away from equilibrium, then is the velocity of the washer in pretty much the same direction as the net force, or pretty much in the direction opposite that of the net force? Would the pendulum be speeding up or slowing down? Explain how the answers to these two questions are related.
Is the net force constant over the swing of the pendulum, or variable? If it is variable, describe in your own words how it varies.
Would the velocity vs. clock time graph for the motion of the pendulum be a straight line or not? Why or why not?
What do you think the velocity vs. clock time graph would look like over an entire cycle of the pendulum’s motion?
Ball Rolling down InclineSketch the gravitational force vector and the normal force vector for a ball on each of two inclines, one inclined about 10 degrees with respect to horizontal, and other about 30 degrees, with the incline being downward to the right.
- For each sketch, construct an x-y coordinate plane through the center of the ball, with the x axis horizontal and the y axis vertical. Estimate the angle of each vector in each of your sketches, as measured counterclockwise from the positive x axis.
- For each sketch, sketch vectors to show how much ‘up’ and how much ‘over’ there is in the normal force vector. That is, sketch the vertical and the horizontal components of the normal force vector.
- For each sketch, add a vector which depicts what you think would be the net resultant of the gravitational and tension forces. Keep everything to a consistent scale, with longer vectors indicating greater forces and shorter vectors depicting lesser forces. Estimate the angle of this vector with the positive x axis.
Repeat this exercise, but this time sketch your x axis parallel to the incline and your y axis perpendicular to the incline.
Answer the following questions for the ball on the incline:
- Does the speed of the ball have any influence on the gravitational force?
- Does the speed of the ball have any influence on the normal force?
- Does the speed of the ball have any influence on the net force?
- Does the position of the ball on the incline have any influence on the net force?
- Will the velocity vs. clock time graph for this motion be a straight line? Why or why not?
Ball falling to floor
Assume the ball rolls off the edge of the incline and falls to the floor.
- Sketch the ball after it has left the edge of the incline, at some instant while it is falling toward the floor. List the forces that are acting on the ball at this instant, and sketch a vector depicting each of these forces.
- Add to your sketch a vector which you think depicts the velocity of the ball at this instant. Estimate the angle of this velocity vector with respect to the direction of the positive x axis. Sketch vectors indicating how much of this velocity is in the horizontal direction, and how much in the vertical direction.
- If the ball travels 25 cm in the horizontal direction as it falls to the floor, and if it takes .4 seconds to fall to the floor after leaving the edge of the ramp, then what is its average horizontal velocity? If the floor is 90 cm below the edge of the ramp, then what is its average velocity in the vertical direction? What would be its vertical velocity when it hits the floor? How fast do you think the ball is moving when it hits the floor (it has both a vertical and a horizontal velocity when it hits the floor)?
- List the forces acting on the ball at the instant just before it touches the floor.
- List the forces acting on the ball while it is in contact with the floor.
- List the forces acting on the ball at the top of its first bounce after hitting the floor.
Velocity vs. clock time for ball on ramp and for pendulum
Sketch a velocity vs. clock time graph for a ball which travels from rest through a displacement of 30 cm in 3 seconds, assuming that the graph is a straight line.
- On your sketch indicate the initial, final and average velocities of the ball.
- What is the average rate of change of the ball’s velocity with respect to clock time during this interval?
- Explain how you applied the definition of the average rate of change of A with respect to B in order to find the average velocity and in finding the average rate of change of the velocity with respect to clock time.
* Sketch a velocity vs. clock time graph for a pendulum as it swings from its leftmost position to its equilibrium position.
- Since the net force on the pendulum decreases from the leftmost to the equilibrium position, the slope of your graph must decrease; the slope will always be positive, the graph will continue to increase, but the rate of increase will decrease.
- This will cause the graph to be increasing, but concave downward.
- If the average velocity of this pendulum is 40 cm / sec, what do you think will be its final velocity (make the most reasonable estimate you can; an estimate is your only option since it would require calculus to arrive at a completely accurate answer)?
Results from Ball and Marble Down Ramp, Timing by Pendulum
There was little difference in the quality and consistency of the data for the steel ball and for the pendulum. In both cases the average of the velocities obtained by the various groups was about 24 cm / sec and the average of the accelerations about 19 cm/sec^2 (and in both cases the standard deviations appeared to be on the order of 3 cm/s in the case of average velocities, or 4 cm/s^2 in the case of accelerations).
An average velocity of 24 cm/s for a ball starting from rest, assuming a straight-line v vs. t graph, implies a final velocity of 48 cm/s.
Most groups reported that as it falls to the floor the ball has a horizontal range of about 20 cm. It is possible to analyze the motion of a ball which falls about 95 cm from the edge of a table to the floor, starting with a speed of 48 cm/s.
[you aren't expected to understand the details at this point but for future reference they are given here in a smaller font:
The slope is .05, so the tangent of the angle with horizontal is .05.
For small slopes the percent difference between tangent and the sine of the angle is very small, and the cosine is very close to 1. So the initial horizontal and vertical velocities of the projectile are about 48 cm/s and 2.5 cm/s, with the initial vertical velocity downward.
With initial velocity 2.5 cm/s and acceleration 980 cm/s^2 the final vertical velocity is found by the fourth equation of uniformly accelerated motion to be 430 cm/s downward, so the average vertical velocity is 220 cm/s.
To fall 95 cm at average velocity 220 cm/s takes .44 seconds.
In .44 seconds, moving at a constant horizontal velocity of 48 cm/s, the ball will travel 21 cm.]
The range of actual results is probably consistent with an uncertainty in average velocity which is no more than +- 2 cm/s, and an average acceleration which is uncertain to within no more than +- 2 cm/s^2. The percent uncertainty is therefore no more than about +- 2 units out of about 20, or around +- 10%.
These results are very good considering that the time was measured using a hand-held pendulum, which without considerable practice and feedback can be difficult to synchronize. However, when synchronized and refined slightly, the pendulum gives better results than a hand-held stopwatch, and the results obtained for this experiment. Results for this experiment are probably as good as those that would be obtained using stopwatches.
Acceleration of gravity
Two dense and palpably massive objects dropped from the same moderate height at the same instant will strike the level floor at the same instant.
A dense object and a loosely wadded-up piece of paper were also released from a height of about 2 meters; the dense object 'beat' the paper wad to the floor by about 10 cm, or .1 meter, which is about 5% of the distance fallen. This difference is less than most people would expect, and reflect a significant (i.e., measurable) difference due to air resistance, but not a large effect.
An object which falls to the floor having been given an initial velocity in the horizontal, but not the vertical, direction will hit the floor at the same instant as an object dropped from the same height at the same instant. That is, horizontal velocity does not affect vertical motion.
A washer dropped from a height of about 2 meters fell to the floor in what we estimated to be about 1/2 second. While falling the only significant forces acting on the object were its weight (i.e., the gravitational force exerted by the Earth) and air resistance; we assume that air resistance is negligible. Since a wadded-up piece of paper experienced only a small change in its motion due to air resistance, we are justified in regarding the influence of air resistance on a dense object (in this case a washer) as being negligible.
From these results we estimate the acceleration due to gravity:
An object which falls 2 meters in 1/2 second is moving with average velocity 2 m / (1/2 s) = 4 m/s.
If the object starts from rest then a linear v vs. t graph implies a final velocity of 8 m/s.
Acceleration is rate of change of velocity with respect to clock time, which by the definition of rate is change in vel / change in clock time. The velocity changes from 0 to 8 m/s, a change of 8 m/s, in 1/2 s. So the acceleration is
accel = change in vel / change in clock time = (8 m/s - 0 m/s) / (.5 s) = 8 m/s / (.5 s) = 16 m/s^2.
We didn't actually measure the time of fall. It might just as well have been .4 sec or .6 sec as .5 sec. The distance of fall was also a bit less than 2 m (probably more like 1.8 m). A fall of 1.8 m from rest in .6 sec gives us an acceleration of 10 m/s^2 (you can repeat the preceding analysis with these numbers to verify this result).
The point is that a little difference in our time and distance estimates give us a pretty big change in our estimate of the acceleration.
Careful measurements with good instruments tell us that the acceleration due to gravity at sea level is about 9.81 m/s^2, or 981 cm/s^2.
The acceleration of gravity differs slightly from one point on the Earth to another, due mostly to changes in altitude and in the radius of the Earth (greater at equator than at poles) but also due to changes in the density of the Earth's crust and to the local countours of the Earth.
The acceleration of gravity decreases by about 3.05 * 10^-6 cm/s^2 for every meter above sea level. This rate of change does not vary by more than 1% from the lowest point in the ocean to the top of the highest mountain on Earth.
This acceleration is the same for any object which falls freely in the vicinity of the surface of the Earth. This result has been verified by a very large number experiments and applications. VHCC is about 600 meters above sea level. The acceleration of gravity at this altitude is about .003 m/s^2 less, or .3 cm/s^2 less, than at sea level. However this small difference is insignificant compared to the differences at various latitudes in the effective radius of the Earth.
You don't need to know the following at this point. It is included for at this point general interest and enrichment for interested students. Some of this analysis will be done later in the course, but much of it applies only to University Physics courses.
Gravity is an inverse-square force exerted between masses. Its strength is inversely proportional to the square of the distance between the masses. So at 3 times the separation the force would be 1 / 3^2 = 1/9 as great.
The acceleration of gravity at the surface of a planet of mass M and radius R is G M / R^2, where the approximate value of G is 6.67 * 10^-11 N m^2 / kg^2. The most accurate value determined for this constant is 6.67428 * 10^-11 N m^2 / kg^2. G is called the Universal Gravitational Constant. The unit N m^2 / kg^2 is equivalent to m^3 / (kg s) = m^3 kg^-1 s^-1 (you will soon be able to verify the equivalence of these units, which is not particularly difficult using only very basic knowledge from Algebra I), and this is the unit as given by many sources.
The radius of the Earth varies between the poles and the equator; it averages about 6370 km but varies by as much as 20 km or so.
The rate at which the quantity G M / r^2 changes with respect to r is given by the function - 2 G M / r^3 (this result is derived by a simple involving first-semester calculus; the details are beyond the scope of the Principles of Physics and General College Physics courses). If we plug in the values for G, M and r we get the average rate -.0035 (m/s^2) / m.
The value of G can be measured with good accuracy in the lab. We know the radius of the Earth. We can determine g as accurately as we wish by determining, to sufficient accuracy, the period of a pendulum of known length. From the value of G, our pendulum data and the radius of the Earth we can determine the mass of the Earth.
At the altitude of the Space Station the acceleration of gravity is nearly 1 m/s^2 less than on the surface of the Earth. Since the Space Station is in free fall as it orbits, occupants do not directly experience the effects of gravity.
At the distance of the Moon the acceleration of gravity is only about .0025 m/s^3, about 1/4000th that at the surface of the Earth.