# Differences

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 phy141:examprep:m2 [2010/11/02 22:11]mdawber phy141:examprep:m2 [2010/11/04 19:21] (current)mdawber Both sides previous revision Previous revision 2010/11/04 19:21 mdawber 2010/11/02 22:11 mdawber 2010/11/02 22:07 mdawber 2010/10/30 22:48 mdawber 2010/10/30 22:46 mdawber 2010/10/30 22:46 mdawber 2010/10/30 18:34 mdawber 2010/10/30 18:33 mdawber 2010/10/28 14:06 mdawber 2010/10/27 01:03 mdawber 2010/10/27 01:02 mdawber 2010/10/27 01:02 mdawber 2010/10/25 00:40 mdawber created 2010/11/04 19:21 mdawber 2010/11/02 22:11 mdawber 2010/11/02 22:07 mdawber 2010/10/30 22:48 mdawber 2010/10/30 22:46 mdawber 2010/10/30 22:46 mdawber 2010/10/30 18:34 mdawber 2010/10/30 18:33 mdawber 2010/10/28 14:06 mdawber 2010/10/27 01:03 mdawber 2010/10/27 01:02 mdawber 2010/10/27 01:02 mdawber 2010/10/25 00:40 mdawber created Line 18: Line 18: {{wa_after_m2.png}} {{wa_after_m2.png}} + + + ===== 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) $\omega=\frac{v}{r}=\frac{1}{0.1}=10\mathrm{rad\,​s^{-1}}=10\times\frac{60}{2\pi}=95.5\mathrm{rpm}$ + + ===== 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 $\frac{1}{3}0.1\times9.8\times0.09375=0.031\mathrm{J}$ + + ===== 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 + + $v'​=\frac{(m-\frac{1}{12}M\frac{l^{2}}{r^{2}})}{(m+\frac{1}{12}M\frac{l^{2}}{r^{2}})}v$ ​ + + ===== Q3 solutions part 2  ===== + + $v'​=\frac{(m-\frac{1}{12}M\frac{l^{2}}{r^{2}})}{(m+\frac{1}{12}M\frac{l^{2}}{r^{2}})}v$ ​ + + 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 $10\mathrm{ms^{-1}}$ + + ===== 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) $F_{N}=T\cos\theta=126\cos(11.3^{0})=123.6\mathrm{N}$ + ===== Midterm 2 information ===== ===== Midterm 2 information =====