User:Iamed18/sandbox

17.Mar.2013::Nicole
In the problem, you're given $$R_i=10^8m,T_i=30\text{ days}, R_f=10^4m$$, and in part (a) you identified that the conservation of angular momentum allowed you to find $$T_f=2.6\times10^{-2}s$$. That said, in part (b), finding the work can be done in a fashion pretty close to what you sent me in the email, but there are some details that need changing. Starting with

$$W=\frac{1}{2}\left[I_f\omega_f^2-I_i\omega_i^2\right]$$

is a good thing, but then one must be neat about what they substitute where insofar as details go. Recalling $$\omega=2\pi/T$$ and that the moment of inertia of a solid sphere is $$I=\frac{2}{5}MR^2$$, this becomes

$$W=\frac{4\pi^2}{5}M\left[\left(\frac{R_f}{T_f}\right)^2-\left(\frac{R_i}{T_i}\right)^2\right]$$,

which does give the correct numerical response of $$W=2.3\times10^{42}J$$.

Bernoulli's Equation of Ideal Stream Flow
Bernoulli's equation states that there exists a constant in a streamline flow that looks something like

$$\text{const.}=P+\rho gy+\frac{1}{2}\rho v^2$$

in which the second and third term both look strikingly like potential energy and kinentic energy, respectively. So, basically, we can have two points (we'll call them ``1 and ``2 for now) at which it must be true that

$$P_1+\rho gy_1+\frac{1}{2}\rho v_1^2=P_2+\rho gy_2+\frac{1}{2}\rho v_2^2$$.

Problem 87
Since the first two problems on today's discussion were covered, only the final problem will be done here.

Problem Statement

A water tank open to the atmosphere at the top has two small holes punched in its side, one above the other. The holes are $$5.00\text{cm}$$ and $$12.0\text{cm}$$ above the floor. How high does water stand in the tank if the two streams of water hit the floor at the same place?

What Does This Tell Us?

From the problem statement, there are a few things worth keeping track of:
 * The two holes in the tank are different heights, but their streams reach the same distance.
 * This is done by a common height of water in the tank
 * Not only will knowledge of fluids be needed to solve this problem, but some of the basic kinematics from the before-time.

Solution

The first thing to be done here is to relate the ejection velocities of the two holes; it's given that the two streams have the same range, so setting about that would be good. Firstly, starting with an initial velocity in $$x$$ only means that the initial velocity in $$y$$ will be zero, allowing

$$\Delta y=v_{iy}t+\frac{1}{2}at^2$$

to be solved

$$t=\sqrt{\frac{2h}{g}}.$$

The next thing to do is to use the $$x$$ range, $$R$$, to find our desired relationship (starting with $$v=R/t$$):

$$R_1=R_2$$

$$v_1t_1=v_2t_2$$

$$v_1\sqrt{\frac{2h_1}{g}}=v_2\sqrt{\frac{2h_2}{g}}$$

which can be rearranged to

$$v_1=v_2\sqrt{\frac{h_2}{h_1}},$$

where the two $$h$$ values are the heights of the two holes, respectively (labeled in the problem picture). Before applying Bernoulli's relationship, the last thing that needs fixing is the velocity of the tank water ($$v_3\approx 0$$), which may be assumed stationary relative to how fast the holes are expected to eject water. Hence, applying Bernoulli's relation at hole 1 yields

$$P_1+\rho gy_1+\frac{1}{2}\rho v_1^2=P_3+\rho gy_3+\frac{1}{2}\rho v_3^2$$

$$P_{\text{atm}}+\rho gy_1+\frac{1}{2}\rho v_1^2=P_{\text{atm}}+\rho gy_3+0$$

which can be solved for

$$v_1^2=2g(h_3-h_1)$$.

Similarly, doing the same at hole 2 yields

$$v_2^2=2g(h_3-h_2)$$.

Dividing these two gives

$$\frac{v_1^2}{v_2^2}=\frac{2g(h_3-h_1)}{2g(h_3-h_2)}=\frac{h_3-h_1}{h_3-h_2},$$

which, when using the above relationship between the two velocities $$v_1=v_2\sqrt{\frac{h_2}{h_1}},$$ reduces to

$$\frac{v_2^2(h_2/h_1)}{v_2^2}=\frac{h_2}{h_1}=\frac{h_3-h_1}{h_3-h_2},$$

in which only $$h_3$$ is unknown. This can be solved for the desired unknown as follows

$$\frac{h_2}{h_1}=\frac{h_3-h_1}{h_3-h_2}$$

$$(h_3-h_2)h_2=(h_3-h_1)h_1$$

$$h_3(h_2-h_1)=h_2^2-h_1^2$$

$$h_3=\frac{h_2^2-h_1^2}{h_2-h_1},$$

which, when number crunched, comes to $$h_3=17\text{cm}$$.

28.Nov.2012::Tomer
While it is quite true that the speed of a wave on a wire can be given by $$v=\lambda f$$, it also remains true that the speed can be written as $$v=\sqrt{F_T/\mu}$$. Using the two equations together, it is seen that

$$\lambda f=\sqrt{\frac{F_T}{\mu}}.$$

Now, your question as to why the frequency doesn't show up in the latter of the two is addressed as: the speed of the wave on the string, regardless of frequency/wavelength, is held constant by the tension in the string and the linear mass density. The use of it is to do the above equality, which relates the frequency to the wavelength, speed, and linear mass density in a convenient fashion. So, in short, frequency is absent from the latter expression because the speed of a wave on the string can be written in two ways: the first being the left hand of the above expression and the second being the right hand of the above expression.

08.Nov.2012::Useful Formulae and Discussion 21
Variable identification: $$F$$ is a force applied on a wire of cross-sectional area $$A$$ with original length $$L$$ which stretches by $$\Delta L$$. The material of the wire chooses its Young's Modulus $$Y$$ such that

$$\frac{F}{A}=Y\frac{\Delta L}{L}.$$

The strain on an object of thickness $$h$$ when a force causes the top to shear some distance $$\Delta x$$ is given by

$$\text{strain}=\frac{\Delta x}{h}.$$

Similarly, the stress on an object due to an applied force $$F$$ on an object with cross-sectional area $$A$$ goes as

$$\text{stress}=\frac{F}{A}.$$

The shear modulus is measured as

$$S=\frac{stress}{strain}.$$

First Problem
The magnitude of force on each wire can be found using the given $$\Delta L, L, Y,$$ and the diameter of the wire can be used to find the cross-sectional area $$A$$. The magnitude of the force then goes as

$$F=AY\frac{\Delta L}{L}=22N.$$

After obtaining the magnitude of 22N for the force on each wire, finding the net force on the tooth is done by adding the vector components of each wire together. You'll note that the $$x$$ direction forces will cancel each other out while the $$y$$ components will add together in the downward direction. If you find the $$y$$-component of $$F$$ and multiply it by 2, you get that the net force on the tooth is given by

$$F_{\text{net}}=2F\sin\theta,\text{ downward}=22N,\text{  downward.}$$

Second Problem
The stress, when rearranging the last formula given today, can be written as

$$\text{stress}=S\times\text{strain}=S\frac{\Delta x}{h}.$$

In the problem, it's given that a plate with thickness $$h=10km$$ is sheared $$\Delta x=5m$$. Using those numbers along with the provided shear modulus $$S$$ gives $$\text{stress}=7.5\times10^6 Pa.$$

Third Problem
This problem has two parts: while the two masses are in contact and when they're not. If we compress the spring by some initial amplitude $$A_i$$, the speed of the two masses as they come to the spring's equilibrium point can be found via energy conservation:

$$\frac{1}{2}kA_i^2=\frac{1}{2}(m_1+m_2)v_{12}^2\rightarrow v_{12}=A_i\sqrt{\frac{k}{m_1+m_2}}=0.5m/s.$$

Once the equilibrium point is passed, the two masses lose contact with each other as $$m_2$$ continues onward at $$v_{12}$$ while $$m_1$$ gets slowed down by the spring. It is now necessary to find the period of the spring-mass setup so that we can use this time as the time of travel for $$m_2$$.

$$T=\frac{1}{f}=\frac{2\pi}{\omega}=\frac{2\pi}{\sqrt{k/m_1}}=1.9sec.$$

Since we know that $$m_1$$ moves from equilibrium out to its amplitude for a fully stretched spring, we know then that the time this motion takes is $$T/4$$ since that describes only $$1/4$$ of a full oscillation (a full oscillator goes from equilibrium to amplitude, amplitude to equilibrium, equilibrium to negative amplitude, and negative amplitude back to equilibrium for the four-step oscillation). Hence the distance that mass two moves during this time can be found as

$$d_2=v_{12}(T/4).$$

As far as the distance of $$m_1$$ goes: we can use energy conservation to find the new amplitude of the system like so:

$$KE_i=PE_f$$

$$\frac{1}{2}m_1v_{12}^2=\frac{1}{2}kA_f^2$$

$$A_f=v_{12}\sqrt{\frac{m_1}{k}}$$

Hence, if we're interested in finding the distance between them at this point, we need only find the difference between the above found distances:

$$D=d_2-A_f=8.6cm.$$

30.Oct.2012::Sarah
So we're told that there exists a place between the Earth and Moon at which the forces of gravity from the Earth and Moon are equal. Let's say we write that out:

$$ F_{G,Earth}=F_{G,Moon}$$

$$\frac{GM_Em}{r_E^2}=\frac{GM_Mm}{r_M^2}$$

Where $$r_E$$ and $$r_M$$ represent the distance from this "balance point" to the Earth and Moon respectively, implying that $$r_E+r_M=R_{EM}$$, the total distance from the Earth to the Moon.

We're told that $$M_M=\frac{1}{81}M_E$$

$$\frac{GM_Em}{r_E^2}=\frac{G\left(\frac{1}{81}M_E\right)m}{r_M^2}$$

$$\frac{1}{r_E^2}=\frac{1}{81 r_M^2}$$

$$r_E=9r_M$$

Hence, the distance of this balance point from the Earth is 9 times the distance from the balance point to the moon, implying that it's 9/10ths of the way to the moon where the forces balance.

25.Oct.2012::Kelli
Pr 8.20



The problem asks you to find the magnitude of the torque on the rod from the force of gravity. If the rod has a length $$L$$, mass $$m$$, and is some angle $$\theta$$ above the horizontal, then we would write that as follows:

$$\tau=rF_{perp}=\frac{L}{2}mg\mbox{cos}(\theta)$$

where $$r=L/2$$ arises from the gravitational force acting on the center of mass (located half-way up the rod), $$mg$$ is the magnitude of the gravitational force, and the $$\mbox{cos}(\theta)$$ is used to pull out the component of the gravitational force that points along the perpendicular direction to the rod.

The picture illustrates what I mean by the angle.

12.Oct.2012::Lizzie
The problem, so far as I can tell, asks what work is done by gravity as a man walks up Mt. McKinley (elevation of $$h_{top}=6200$$m).

Let's start with a basic expression (the Work-Kinetic Energy Theorem):

$$W=\Delta KE$$

$$W=KE_f-KE_i$$

Now, since we know that the gravitational force is a conservative force, we know that while moving around in a gravitational field, we'll conserve energy.

$$KE_f+PE_f=KE_i+PE_i$$

Rearrange:

$$KE_f-KE_i=PE_i-PE_f=-\Delta PE$$

Notice that the expression here matches the one resulting from the work-kinetic energy theorem. This means:

$$-W=\Delta PE$$

$$W=mgh_i-mgh_f$$

$$W=-mgh_f$$, which checks out nicely.

Alternately, you could have looked on page 175 of your text instead of reading my miniature derivation!