The 2nd midterm will be on Friday October 26th in class at 9:00 AM.

There will be 3 questions.

Emphasis on

- Energy
- Momentum
- Rotational Motion
- Equilibrium

Exam is cumulative. No fluids. No integration (ie. you won't be asked to derive any moments of inertia or centers of mass).

2 Practice exams and review questions are posted and previous years exams are posted below.

There will be a review session on Tuesday at 10:00AM in B131.

You may bring one handwritten sheet of notes which can have anything you want on them, and you can write on both sides of the sheet of paper.

You should bring a calculator. Any kind of calculator is acceptable. It should be able to perform scientific functions (especially trig), but does not need to have graphing capabilities. Your calculation device should not allow you to access the internet, send and receive messages or play Angry Birds.

The 2nd midterms was clearly more challenging than the first, with some elements that were quite difficult. The class average was 76%. Here is the histogram of scores. You can find your score on Blackboard in the Grade Center.

Everyone can have a bad day. It's a shame when it happens to be on a midterm exam. That's why I actually use something I call the Adjusted Total Midterm Score (ATMS) in calculating your grades. You can find your ATMS score on Blackboard in the Grade Center.

How does this work? In the syllabus it is advertised that 10% of the grade comes from Midterm 1 and 15% comes from Midterm 2. Rather than this the ATMS means your final grade will actually use

5% Midterm 1 10% Midterm 2 10% Best of Midterm 1 Score or Scaled Midterm 2 Score

For the scaled Midterm 2 Score, I scaled up the scores of Midterm 2 so they had a similar mean and variance to Midterm 1. Here is the histogram of the ATMS. As you can see this helps to stop people from dropping off the bottom of the curve if they performed badly in one exam, while still rewarding those who did well on both exams. Note that grades in this class are awarded on absolute, not relative performance, there is no fixed number of each grade that will be awarded, or in other words, you only need to worry about high your own score is, not what anybody else's is. I will prepare grade estimates based on work so far in the next couple of days, so you can have a clearer idea of how you are doing.

Here are grade estimates based on the current weighted totals in Blackboard. As you can see the class is doing VERY well. A couple of caveats. The one kind of assessment that is not included is lab reports. These tend to have a slightly negative effect on the average weighted total compared to homework, attendance, labs etc. So your weighted total might fall a little after we have those scores…or if course it might go up if you do a really great job. The most likely cause of changes in scores will be the final exam. I will be aiming for an average score on that exam closer to the midterm 2 average than the midterm 1 average, so I am anticipating that this will reduce weighted totals somewhat, and you still need to work very hard for the rest of the semester if you want an A. Still you should all be happy about your progress, and if you do all manage to do well in the final, and other work we still have left it looks like I will be handing out quite a lot of high grades this semester!

a) $\Delta h= 1-\cos50^{o}=0.357\mathrm{m}$

$KE=-mg\Delta h=0.1\times 9.8\times0.357=0.35 \mathrm{J}$

b) All of the kinetic energy is transferred to the second ball so

$KE=0.35\mathrm{J}$

c) KE at top of ramp is $0.35-0.1\times9.8\times0.1=0.252\mathrm{J}$

For a rolling ball $KE=\frac{1}{2}m(1+\frac{2}{5})v^{2}$ so

$v^{2}=\frac{0.252\times 2\times 5}{0.1\times7}=3.6$

$v=1.9\mathrm{m\,s^{-1}}$

d) $L=I\omega$

$I=\frac{2}{5}mr^{2}=\frac{2}{5}\times0.1\times.02^{2}=0.000016\mathrm{kg\,m^{2}}$

$\omega=\frac{v}{r}=\frac{1.9}{0.02}=95\mathrm{s^{-1}}$

$L=0.00152\mathrm{kg\,m^{2}\,s^{-1}}$ in to the page

e) First we need to find the time it takes until the ball hits the ground

$0=0.1+1.9\sin 15^{o}t-\frac{1}{2}gt^{2}$

$4.9t^{2}-0.49t-0.1=0$

$t=0.2 \mathrm{s}$

$x=1.9\times\cos 15^{o}t$

$x=0.37\mathrm{m}$

a) $I=2\times2\times1.2^{2}+2\times3\times(\frac{1}{12}\times1.1^{2}+0.65^{2})+\frac{1}{2}\times50\times0.1^2$

$I=5.76+3.14+0.25=9.15\,\mathrm{kg\,m^{2}}$

b) $\sum \tau =2 \times 100 \times 1.2 =240\,\mathrm{N\,m}$

$\alpha=\frac{\sum \tau}{I}=\frac{240}{9.67}=26.23\,\mathrm{s^{-2}}$

c)$\omega=24.82\times0.5=13.115\mathrm{s^{-1}}=125.24\mathrm{rpm}$

d)$L=I\omega=9.15\times13.115=120\,\mathrm{kg\,m^{2}\,s^{-1}}$

e)$I=2\times5\times0.1^{2}+\frac{1}{2}\times50\times0.1^{2}=0.35\,\mathrm{kg\,m^{2}}$

f) Angular momentum is conserved so

$\omega=\frac{120}{0.35}=342.9\mathrm{s^{-1}}=3274\mathrm{rpm}$

g) $v=\omega r=342.9\times0.1=34.29\,\mathrm{m\,s^{-1}}$

a) Torques around hinge must balance so

$8\times0.25=2m\times 0.5$

$m=\frac{8\times0.25}{2\times0.5}=2\mathrm{kg}$

b) Balance of vertical forces

$8g-4g-F_{Hy}=0$

$F_{Hy}=4g=39.2\mathrm{N}$ up

c) Balance of horizontal forces

$F_{Hx}=0\mathrm{N}$

d) $I=\frac{1}{3}\times8\times0.5^{2}=0.67\,\mathrm{kg\,m^{2}}$

e) For the still attached mass

$2a=T_{2}-2g$

For the pulley

$\frac{1}{2}\times2\times0.15^2\times \frac{a}{0.15}=(T_{1}-T_{2})0.15$

$T_{1}-T_{2}=a$

For the arm

$0.67\times \frac{a}{0.5}=8\times9.8\times0.25-0.5T_{1}$

$2.68a=39.2-T_{1}$

Combining the first two equations

$3a=T_{1}-2g$

And then using it with the third

$2.68a=39.2-3a-19.6$

$5.68a=19.6$

$a=3.45\,\mathrm{ms^{-2}}$ up

e) $T_{1}=3\times3.45+19.6=29.95\,\mathrm{N}$

$T_{2}=T_{1}-a=29.95-3.45=26.5\,\mathrm{N}$

Fall 2010 Midterm 2. Note this has 4 problems instead of 3. That was a mistake that won't be repeated! Also question 3 was obviously way too hard.