# Midterm 2 information

The 2nd midterm will be on Monday October 31st in class at 8:30 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. The practice exams have 4 problems, but the actual exam will only have 3. You can also take a look at last year's midterm. That also has 4 problems and was WAAAYY to long. So much so that I had to rescale the results on the exam (I graded the exam as if it was out of 80 and converted that to a percentage). I will try to avoid having to do that this year by asking 3 relatively sane questions. Solutions to last years exam and the average score the class achieved on each problem in the exam are below.

In addition to focusing on exam style problems in this weeks recitation, I will do a review lecture on Friday morning and there will be a review session from 3-5pm in P119.

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.

## Midterm 2 Scores

Overall the class did very well and the average score was 74%.

5% Midterm 1/20% Midterm 2 or 15% Midterm 1/10% Midterm 2

which ever is higher. The distribution of these scores is below.

Here are rough estimates of how grades would be assigned if I were to do it today (which I won't). There is still plenty of opportunity to improve your grade!

## Fall 2011 Midterm 1 Q1 Solution (Ave: 25.6/35)

A. $I=\frac{2}{3}\times\frac{2}{5}mR^{2}+\frac{1}{3}\times\frac{2}{3}mR^{2}=\frac{22}{45}mR^{2}$

B. $KE=\frac{1}{2}mv^{2}+\frac{1}{2}I\omega^{2}=\frac{1}{2}mv^{2}+\frac{1}{2}\frac{22}{45}mR^{2}\frac{v^{2}}{R^{2}}=\frac{67}{90}mv^{2}$

$\frac{67}{90}mv^{2}=mgh$

$v^{2}=\frac{90}{67}g10\sin30^{o}$

$v^{2}=65.82$

$v=8.11\mathrm{ms^{-1}}$

C. From right hand grip rule $\vec{L}$ is in to page.

D. Momentum is conserved in the collision. We consider the conservation of momentum equation for the center of masses of the objects.

$m_{A}v_{A}=m_{A}v'_{A}+m_{B}v'_{B}$

As the collision is elastic kinetic energy is conserved. The first object is a rolling object, the second one is not.

$\frac{1}{2}(1+\frac{22}{45})m_{A}(v_{A}^{2}-v'_{A}^{2})=\frac{1}{2}m_{B}v'_{B}^{2}$

by combining these two equations

$(1+\frac{22}{45})(v_{A}+v'_{A})=v'_{B}$

(Note if you neglect rotation then you just get rid of the $\frac{22}{45}$ and I only deducted a couple of points if your did this and then everything else correctly.)

The momentum equation can be written as

$\frac{m_{B}}{m_{A}}=\frac{v_{A}-v'_{A}}{v'_{B}}=\frac{v_{A}-v'_{A}}{(1+\frac{22}{45})(v_{A}+v'_{A})}$

Now as the first object rolls halfway back up the hill is has half the kinetic energy, which means that

$v'_{A}=-\frac{1}{\sqrt{2}}v_{A}$

$\frac{m_{B}}{m_{A}}=\frac{v_{A}-v'_{A}}{(1+\frac{22}{45})(v_{A}+v'_{A})}=\frac{\frac{v_{A}}{v'_{A}}-1}{(1+\frac{22}{45})(\frac{v_{A}}{v'_{A}}+1)}=\frac{\sqrt{2}+1}{(\sqrt{2}-1)(1+\frac{22}{45})}=3.915$

If you left off the rotational kinetic energy then just working from the equations in the lecture

$\frac{m_{B}}{m_{A}}=\frac{v_{A}-v'_{A}}{(v_{A}+v'_{A})}=\frac{\frac{v_{A}}{v'_{A}}-1}{(\frac{v_{A}}{v'_{A}}+1)}=\frac{\sqrt{2}+1}{\sqrt{2}-1}=5.83$

E. With rotational kinetic energy

$v'_{B}=(1+\frac{22}{45})(v_{A}+v'_{A})=(1+\frac{22}{45})(1-\frac{1}{\sqrt{2}})8.11=3.54\mathrm{ms^{-1}}$

Without

$v'_{B}=(v_{A}+v'_{A})=(1-\frac{1}{\sqrt{2}})8.11=2.38\mathrm{ms^{-1}}$

F. $\frac{1}{2}m_{B}v'_{B}^{2}=m_{B}gh$

$h=\frac{v'_{B}^{2}}{2g}=0.64\mathrm{m}$

## Fall 2011 Midterm 2 Q2 Solution (Ave: 22.27/30)

A. Sum of torques on beam around hinge

$0.8\mathrm{m}\times T\sin30^{o}-x\times 4\mathrm{kg}\times g-0.4\mathrm{m}\times 2\mathrm{kg}\times g=0$

$x=\frac{0.8\mathrm{m}\times 50\mathrm{N}\sin30^{0}-0.4\mathrm{m}\times 2\mathrm{kg}\times g}{4\mathrm{kg}\times g}=0.31\mathrm{m}$

B. Sum of horizontal forces on beam

$F_{HH}-T\cos30^{o}=0$

$F_{HH}=43.3\mathrm{N}$ in the opposite direction to the tension.

Sum of vertical forces on beam

$F_{HV}-4g-2g+T\sin30^{o}=0$

$F_{HV}=33.8\mathrm{N}$ up

$F_{NET}=\sqrt{43.3^2+33.8^{2}}=54.93\mathrm{N}$

Direction of force on beam by hinge $\tan^{-1}=\frac{-43.3}{33.8}=-52^{o}$

But the force on the hinge is directed in the opposite direction $180^{o}-52^{o}=128^{o}$

C. Sum of torques on beam around hinge

$0.8\mathrm{m}\times T\sin\phi-0.8\mathrm{m}\times 4\mathrm{kg}\times g-0.4\mathrm{m}\times 2\mathrm{kg}\times g=0$

$\sin\phi=\frac{4\mathrm{kg}\times g+\frac{2\mathrm{kg}}{2}\times g}{50\mathrm{N}}$

$\phi=78.5^{o}$

## Fall 2011 Midterm 2 Q3 Solution (Ave: 26.32/35)

A. $\frac{1}{3}Ml^{2}=\frac{1}{3}2\times0.38^{2}=0.096\mathrm{kgm^{2}}$

B. $mgh_{CM}=2\times9.8\times0.19=3.724\mathrm{J}$

C. Previous potential energy gain plus potential energy needed to raise bullet by $30\mathrm{cm}$ ($0.02\times g\times0.3=0.0588\mathrm{J}$), which is $3.783 \mathrm{J}$.

D. $I_{B}=0.02\times0.3^{2}=0.0018\mathrm{kgm^{2}}$

E. Conservation of angular momentum for collision

$0.0018\omega=(0.096+0.0018)\omega'$

Conservation of mechanical energy for upswing of arm

$\frac{1}{2}(0.096+0.0018)\omega'^{2}=3.783$

$\omega'=8.8 \mathrm{rad/s}$

$\omega=\frac{0.096+0.0018}{0.0018}8.8$

$\omega=478 \mathrm{rad/s}$

$v=\omega r=478\times0.3=143.4 \mathrm{ms^{-1}}$

## Fall 2010 Midterm 2 Q1 Solutions (Ave: 12.5/20)

a) $v=at$

$1=a\times0.5$

$a=2\mathrm{ms^{-2}}$

$Fr-Tr=\frac{1}{2}Mr^2\alpha$ → $F-T=\frac{1}{2}Ma$

$T-mg=ma$ → $T=mg+ma$

$F=\frac{1}{2}Ma+mg+ma$

$F=0.5\times 0.2\times 2+1\times(9.8+2)=12\mathrm{N}$

b)$\alpha=\frac{a}{r}=\frac{2}{0.1}=20\mathrm{rad\,s^{-2}}$

c) At constant speed $a=0$ so $F=mg=1\times9.8=9.8\mathrm{N}$

d) $\omega=\frac{v}{r}=\frac{1}{0.1}=10\mathrm{rad\,s^{-1}}=10\times\frac{60}{2\pi}=95.5\mathrm{rpm}$

## Fall 2010 Midterm 2 Q2 Solutions (Ave 17.8/25)

a) Start from projectile motion

$y=\frac{1}{2}gt^{2}$ → $t^{2}=\frac{2}{g}$ ($y=1$)

Use conservation of energy to get the horizontal velocity

$\frac{1}{2}mv^{2}+\frac{1}{2}I\omega^{2}$ → $\frac{1}{2}mv^{2}+\frac{1}{2}mfv^{2}$

where we have expressed the moment of inertia as $I=fmr^{2}$

$\frac{1}{2}m(1+f)v_{x}^{2}=mgh$ → $v_{x}^2=\frac{2gh}{1+f}$

$x^2=v_{x}^2\frac{2}{g}$ → $x^{2}=\frac{4h}{1+f}$ → $f=\frac{4h}{x^2}-1$

$f=\frac{4\times0.09375}{0.5^2}-1=\frac{1}{2}$ so we should choose a disk

b) g cancels out in the above equation so your choice should not change.

c) Total kinetic energy is the same as the potential energy lost $mgh=0.1\times9.8\times1.09375=1.07\mathrm{J}$

d) Rotational kinetic energy is only gained while the object is rolling. You can calculate it a few ways, for example, $\frac{1}{4}mv_{x}^{2}$ or $\frac{1}{3}0.1\times9.8\times0.09375=0.031\mathrm{J}$

## Fall 2010 Midterm 2 Q3 Solutions (Ave: 7.6/30..Ooops)

a) Conservation of Momentum

$mv=mv'+\Delta p_{rod}=mv'+I\frac{\omega'}{r}$

This comes from the fact that $\vec{L}=\vec{r}\times\vec{p}$ and as $\vec{r}$ and $\vec{p}$ are at right angles to each other $L=rp$ so $\Delta p =\frac{\Delta L}{r}$ aand $\Delta L= I\omega'$

Taking in to account the moment of inertia of the rod is $I=\frac{1}{12}Ml^{2}$

$mv=mv'+\frac{1}{12}Ml^{2}\frac{\omega'}{r}$

This is not enough to solve! You also need to use conservation of energy

$\frac{1}{2}mv^{2}=\frac{1}{2}mv'^{2}+\frac{1}{2}\frac{1}{12}Ml^{2}\omega'^{2}$

Rearrange the two equations

$mv-mv'=\frac{1}{12}Ml^{2}\frac{\omega'}{r}$

$\frac{1}{2}mv^{2}-\frac{1}{2}mv'^{2}=\frac{1}{2}\frac{1}{12}Ml^{2}\omega'^{2}$ → $m(v-v')(v+v')=\frac{1}{12}Ml^{2}\omega'^{2}$

Divide the second equation by the first

$v+v'=\omega'r$ → $\omega=\frac{v+v'}{r}$

b) Now to get v' we can substitute our formula for $\omega'$ in to our first equation

$m(v-v')=\frac{1}{12}M\frac{l^{2}}{r^{2}}(v+v')$

Group terms with $v$ on one side, terms with $v$' on the other and the rearrange to find

$v'=\frac{(m-\frac{1}{12}M\frac{l^{2}}{r^{2}})}{(m+\frac{1}{12}M\frac{l^{2}}{r^{2}})}v$

c) Inspection shows that when $m$ is much less than $\frac{1}{12}M\frac{l^{2}}{r^{2}}$ the ball will have negative $v'$, when it is greater it will have positive $v'$. We are looking for when it is zero. The smallest value $\frac{1}{12}M\frac{l^{2}}{r^{2}}$ can take is when $r=\frac{l}{2}$ so we want to find the mass ratio when

$m-\frac{1}{12}M\frac{l^{2}}{l^{2}/4}=m-\frac{1}{3}M=0$

which is $\frac{m}{M}=\frac{1}{3}$

d) When the ball hits the tip of the rod at the mass ratio above $v'=0$ which means from our first result $\omega=\frac{v+v'}{r}$, $\omega'=\frac{v}{r}$ and as the tangential velocity of the tip of the rod is $\omega'r$ we find that the tangential velocity of the tip of the rod is $v$, or $10\mathrm{ms^{-1}}$

## Fall 2010 Midterm 2 Q4 Solutions (Ave 15.3/25)

a) Balance of vertical forces $T\sin\theta-mg+F_{Fr}=0$

Balance of horizontal forces $T\cos\theta-F_{N}=0$

Balance of torques around end of rod on wall $mg\frac{l}{2}=T\sin\theta l$ → $mg=2T\sin\theta$

Combining the vertical force equation and the torque equation gives

$T\sin\theta=F_{Fr}=\mu F_{N}$

Now use the horizontal force equation

$T\sin\theta=\mu T\cos\theta$

$\tan\theta=\mu=0.2$

$\theta=11.3^{o}$

b) $T=\frac{mg}{2\sin\theta}=\frac{5\times9.8}{2\sin(11.3^{o})}=126\mathrm{N}$

c) $F_{N}=T\cos\theta=126\cos(11.3^{0})=123.6\mathrm{N}$