# Solutions to Fall 2015 Midterm 2

## Question 1 Solution (Average 20/30)

A catapult is designed to fire explosive rocks of mass $50\,\mathrm{kg}$ at a fixed angle of 45º above the horizontal. A knight begins to charge from a distance of 100 m towards the catapult at a constant speed of 10 m/s. As you found in your last midterm, in order to hit the knight the initial speed with which a rock should be fired is 25ms$^{-1}$. However, the rock explodes early, only 1 second after launch, breaking in to two pieces of equal mass. For all of this problem express vector quantities in terms of $\hat{i}$ and $\hat{j}$ based on the coordinate axis shown.

A. (6 points) What is the momentum and kinetic energy of the rock immediately after it has been launched?

$\vec{v}=25\times\cos45^{\circ}\hat{i}+25\times\sin45^{\circ}\hat{j}=17.67\hat{i}+17.67\hat{j}\,\mathrm{ms^{-1}}$

$\vec{p}=m\vec{v}=50\times17.67\hat{i}+50\times17.67\hat{j}=883.5\hat{i}+883.5\hat{j}\,\mathrm{kgms^{-1}}$

$KE=\frac{1}{2}mv^{2}=0.5\times50\times25^2=15625\,\mathrm{J}$

B. (6 points) What is the momentum and kinetic energy of the rock just before it explodes?

$v_{x}=v_{x0}=17.67\,\mathrm{ms^{-1}}$

$v_{y}=v_{y0}-gt=17.67-9.81*1=7.86\,\mathrm{ms^{-1}}$

$vec{v}=17.67\hat{i}+7.86\hat{j}\,\mathrm{ms^{-1}}$

$\vec{p}=m\vec{v}=50\times17.67\hat{i}+50\times7.86\hat{j}=883.5\hat{i}+393\hat{j}\,\mathrm{kgms^{-1}}$

$KE=\frac{1}{2}mv^{2}=0.5\times50\times(17.67^2+7.86^2)=9350\,\mathrm{J}$

C. (6 points) Immediately after the explosion, one piece of the rock is moving with velocity 10ms$^{-1}\,\hat{i}$. What is the velocity (as a vector) of the other piece of the rock immediately after the explosion?

Rock A has momentum $\vec{P_{A}}=250\hat{i}$

From conservation of momentum

$\vec{p_{initial}}=883.5\hat{i}+393\hat{j}=\vec{p_{A}}+\vec{p_{B}}$

$\vec{p_{B}}=(883.5-250)\hat{i}+393\hat{j}=633.5\hat{i}+393\hat{j}$

$\vec{v_{B}}=\frac{\vec{p_{B}}}{m_{b}}=\frac{633.5}{25}\hat{i}+\frac{393}{25}\hat{j}=25.34\hat{i}+15.72\hat{j}\,\mathrm{ms^{-1}}$

D. (6 points) What was the change in kinetic energy of the system during the explosion?

$KE_{final}=\frac{1}{2}\times25\times10^2+\frac{1}{2}\times25\times(25.34^2+15.72^2)=12365\,\mathrm{J}$

$\Delta KE=KE_{final}-KE_{initial}=12365-9350=3015\,\mathrm{J}$

E. (6 points) How long after the explosion takes place does each piece of the rock hit the ground?

First we need to find the height at which the explosion takes place

$y=17.67t-\frac{1}{2}gt^{2}=12.77m\,\mathrm{m}$

For rock A the equation for its vertical displacement after the explosion is

$y=12.77-\frac{1}{2}gt^{2}$

Solving for $y=0$ gives

$t_{A}=\sqrt{12.77/49}=1.6\mathrm{s}$

For rock B the equation for its vertical displacement after the explosion is

$y=12.77+15.72t-\frac{1}{2}gt^{2}$

Solving for $y=0$ gives

$t_{B}=\frac{15.72+\sqrt{15.72^{2}+4\times12.77\times4.9}}{2\times4.9}=3.88\mathrm{s}$

## Question 2 Solutions (Average: 23.7/35)

I want to raise a $0.5\,\mathrm{kg}$ object from the ground by pulling on a rope that passes over a solid uniform pulley which has mass $300\,\mathrm{g}$ and radius $15\,\mathrm{cm}$. The bearing on which the pulley turns can be treated as frictionless. The 0.5 kg object is initially at rest and I would like to lift it at a constant speed of $2\,\mathrm{ms^{-1}}$.

A. (3 points ) In order to have the weight be moving with a velocity of $2\,\mathrm{ms^{-1}}$, at time $t=0.5\,\mathrm{s}$ after I start pulling on the rope, what should the linear acceleration of the object be? (Assume uniform acceleration during this time period).

$a=\frac{v}{t}=\frac{2}{0.5}=4\,\mathrm{ms^{-2}}$

B. (3 points) What is the angular acceleration of the pulley while I am doing this?

$\alpha=\frac{a}{r}=\frac{4}{0.15}=26.67\,\mathrm{s^{-2}}$

D. (7 points) What is the magnitude of the force $F$ I need to apply to the rope for this to occur?

Considering the sum of the torques on the pulley

$\Sigma \tau=I\alpha$

$(F-T)r=\frac{1}{2}\times m\times r^2\frac{a}{r}$

$F-T=\frac{1}{2}\times0.3\times4$

From considering Newton's second Law applied to the object

$\Sigma F=ma$

$T-mg=ma$

$T=mg+ma$

So

$F=\frac{1}{2}\times0.3\times4+0.5\times(9.8+4)=7.5\,\mathrm{N}$

D. (7 points) How much work do I do during the $0.5\,\mathrm{s}$ that I am accelerating the object?

There are two approaches, either requires you to find out how much the object moved

$d=\frac{1}{2}at^{2}=\frac{1}{2}\times4\times0.5^2=0.5\,\mathrm{m}$

Now as this is the same distance the rope travels while the force $F$ is applied to it, the work done by $F$ is

$W=Fd=7.5\times0.5=3.75\,\mathrm{J}$

Alternatively you could compute the change in kinetic energy of both the pulley and the object and the change in the potential energy of the object. The sum of these is equal to the work done by $F$.

$W=\frac{1}{2}\times\frac{1}{2}\times0.3\times0.15^2\times(\frac{2}{0.15})^2+\frac{1}{2}\times0.5\times2^2+0.5\times9.8\times0.5=3.75\,\mathrm{J}$

E. (5 points) Once the object has reached a constant speed of $2\,\mathrm{ms^{-1}}$ what force should I apply to the rope maintain that constant speed.

If the pulley is not accelerated the tension in the rope attached to the object must be the same as that in the rope on the other side. As the object is not accelerating

$T=mg$

so

$F=mg=0.5\times9.8=4.9\,\mathrm{N}$

F. (5 points) What is the angular velocity (in rpm) of the pulley while I am raising the object with a constant speed of $2\,\mathrm{ms^{-1}}$?

$\omega=\frac{v}{r}=\frac{2}{0.15}=13.33\,\mathrm{s^{-1}}$

To convert to rpm

$\frac{13.33\times60}{2\pi}=127.3\,\mathrm{rpm}$

G. (5 points) What is the angular momentum of the pulley while I am raising the object with a constant speed of $2\,\mathrm{ms^{-1}}$? Give both the magnitude and the direction (ie. left, right, up, down, in to the page, out of the page).

$L=I\omega=\frac{1}{2}mr^{2}\omega=\frac{1}{2}\times0.3\times(0.15)^{2}\times13.33=.045\,\mathrm{kgm^{2}s^{-1}}$

Direction is out of page from right hand grip rule.

## Question 3 (Average: 25/35)

A uniform beam of mass $7\,\mathrm{kg}$ and length $80\,\mathrm{cm}$ is suspended from a rope, which is attached 20cm from the end of the beam and subject to a tension $T$. The rope makes an angle of $\theta$ with the horizontal. At its other end the beam is supported only by the normal force and the frictional force exerted by a wall (i.e. it is not attached to the wall by a hinge or nails or any other fastener). The beam makes a right angle with the wall.

A. (5 points) Add arrows to the diagram showing the locations and directions of the forces which act upon the beam. Label the arrows with an appropriate descriptive variable (ie. $F_{Fr}$, $F_{N}$, $mg$, $F_{T}$)

B. (10 points) Write equations for the sum of the horizontal forces acting on the beam, the sum of the vertical forces acting on the beam and the sum of the torques acting on the beam (calculate the torques around an axis of your choosing). (Note: Please do not just write zero three times….)

$\Sigma F_{x}=F_{N}-F_{T}\cos\theta=0$

$\Sigma F_{y}=F_{Fr}+F_{T}\sin\theta-mg=0$

Torques calculated around contact point with wall

$\Sigma \tau=F_{T}\sin\theta\times0.6-mg\times0.4=0$

C. (5 points) Find the tension in the rope when $\theta$ is $10^{o}$.

Using the torque equation

$F_{T}=\frac{7\times9.8\times0.4}{\sin10^{\circ}\times0.6}=263\,\mathrm{N}$

D. (5 points) Find the magnitude of the normal force exerted by the wall when $\theta$ is $10^{o}$.

From the equation for the horizontal forces

$F_{N}=F_{T}\cos10^{\circ}=263\cos10^{\circ}=259\,\mathrm{N}$

E. (10 points) If the coefficient of static friction of the wall is $\mu=0.2$, find the maximum value that $\theta$ can take for the beam to be in static equilibrium. (Hint: Look for a way to combine the 3 equations you wrote for part (b) so that they only contain the force due to tension and the angle $\theta$).

At the maximum $\theta$, $F_{Fr}=\mu F_{N}$ so the vertical force equation is

$\mu F_{N}+F_{T}\sin\theta-mg=0$

Using the horizontal force equation to replace $F_{N}$ with $F_{T}\cos\theta$

$\mu F_{T}\cos\theta+F_{T}\sin\theta-mg=0$

Now I want to get rid of $mg$ which I can do using the torque equation

$mg=F_{T}\sin\theta\frac{0.6}{0.4}=_{T}\sin\theta\frac{3}{2}$

Substituting this in, I get

$\mu F_{T}\cos\theta-\frac{1}{2}F_{T}\sin\theta=0$

$F_{T}\sin\theta=2\mu F_{T}\cos\theta$

$\frac{\sin\theta}{\cos\theta}=2\mu$

$\tan\theta=0.4$

$\theta=21.8^{\circ}$ 