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Fall 2010 Midterm 2 Solutions

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

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

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

phy141/examprep/m2f10sols.txt · Last modified: 2012/10/22 20:45 by mdawber
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