"Things should be made as simple as possible, but not any simpler."
-- Einstein
We will be discussing the test extensively as it draws nearer. For now, I remind you that it will be held:
Tuesday, November 1
6 - 7:30 PM
If you have a demonstrable conflict with that day and time, you should see Dr. Savaria in MP129E immediately.
We began with a little calculus, the derivatives of the sine and cosine, which we need now, although MAT135 doesn't get to this until next week.
We then discussed Chapter 9 - Impulse and Momentum. This is just "mining" Newton's Laws for new forms which will be useful to us. We applied this to a tennis serve and also discussed the damage caused to people in collisions.
We used a Flash animation to discuss the derivatives of the sine and cosine. As is common, the physicists are ahead of the mathematicians: MAT135 gets to this next week (§3.4 of Stewart). |
The above photographs are of Amin, a PHY138 student
from two years ago, serving a tennis ball. |
In order to maximise the speed of the serve, Amin needs to get the part of the racket that contacts the ball going at the highest possible speed. He does this with four lever arms:
Note that at contact, all four of these levers are at the maximum values as he "uncoils" into the serve. For each, the speed is given by:
r
where the angular speeds and radii are different numbers for each of the four rotations. Of course, except for the fourth rotation, the pivot point is also moving at the moment of impact.
Thus the speed of the racket as it contacts the ball is as large as possible. One of today's suggested problems explores the tennis server further.
When a collision occurs with a duration of less than about 100 ms, then the damage caused to the body is roughly proportional to the total impulse.
For longer duration collisions, it is the magnitude of the force of the impulse that is proportional to the damage that can be caused. The is because it is the acceleration of the body that causes the damage. For up to a few seconds, the body can withstand accelerations of 15 g; longer duration accelerations of this magnitude can cause the blood to stop circulating. Larger accelerations than 15 g can cause damage such as tearing the aorta away from the heart. Yuk!
This is why modern cars have crumple zones. We illustrated with 2 movie clips.
The first clip is of an older car, without crumple zones. You may access it by clicking on the button to the right. The file size is 118k, and will appear in a separate window. To view the clip requires that you are able to view mov format movies, such as with the QuickTime player. QuickTime is available free from http://www.apple.com/quicktime/. |
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The second clip is of a more modern car. You may access the clip by clicking on the button to the right. The file size is 441k. To view the clip requires that you are able to view rm format movies, such as with the Real Media player. That player is available free from: http://www.realnetworks.com/. |
The clips are by the U.S. National Highway Traffic Safety Administration, and are used with permission. The original site of the clips is: http://www.nhtsa.dot.gov/cars/testing/ncap/Videos.html.
For the second part, the correct answer is 3: the ball that sticks and is captured exerts the least force on the block. Here is how you did on this question before and after some class discussion:
Before Discussion | After Discussion | |
---|---|---|
2: bounces off | 50% |
20% |
3: captured (correct) | 50% |
80% |
One of the keys to this question is that momentum is a vector. I urge you to understand why 3 is the correct answer.
Pdf version of the PowerPoint on the side screens. | |
Today's Journal. |
In Class 8 on Wednesday, October 5, I screwed up the massive string case. The correct value of the force that the string S exerts on block B, FS on B = T = (mA + mS) F / mtot. |
I have corrected the PowerPoint pdf linked to from Class 8's summary. I have also corrected the Journal linked to from Class 8's summary; the correction is in purple.
Before I made the corrections, it was: FS on B = T = (mA + mB) F / mtot.
Recall the final step of the official Problem-Solving Strategy is Assess. A moment doing that would show clearly that this is not correct: the answer must depend on the mass of the string and must reduce to the massless string result when mS is zero.
The arrows let you jump to the previous/next class summaries. |