# Fall 2011 Midterm 2 Solutions

## Fall 2011 Midterm 2 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 it 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

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

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

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

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

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