For the Physics Concept[s] you need to indicate for the Hand-in
Homework, look through the file on Carmen→Modules named
"P1200 Hand_in_HW Concepts.pdf" and figure out the best ones
that match, and possibly make the list better since it is not very
comprehensive. In kinematics, the choices are "One-dimensional
kinematics", "Vectors", and "Projectile motion". [For example: In the
HiHW 2 (with the fish), I would have given "Two-dimensional
kinematics" and "Vectors". The fish is not a projectile and it is not
moving along a straight line.]
[click on the yellow box to expand it]
The numerical answer I get is $\fbox{$L_\mbox{H}=40.8\,\mbox{cm}$}$.
Your symbolic expression should involve quantities that
are given (or asked for) in the problem. For this problem, these symbols include:
\begin{align}
& L_\mbox{H} =\ \mbox{tuned length of Heavy string}\\[1ex]
& L_\mbox{L} =\ \mbox{given tuned length of Light string (equal to $50.0\,\mbox{cm}$)}\\[1ex]
& \frac{\mu_{\mbox{H}}}{\mu_{\mbox{L}}} = \ \mbox{ratio of mass densities of the strings (equal to 1.50)} \\[1ex]
\end{align}
It may also include other constants, like $g$ or $\pi$. Another thing you
can do is incorporate special values that are trivial or cancel out.
In this problem, the tensions are the same: Does the tension enter into the expression?
This is a perfect example to use ratio reasoning. To show in more detail how this works,
here is a more detailed solution.
The relationships for a standing wave on a clamped string are
\[ v = f\lambda = \sqrt{\frac{T}{\mu}} \]
For the fundamental standing wave,
\[ \lambda= 2L \]
Squaring the first and using the second (and arranging so it isn't a fraction),
\[ T = 4\mu f^2 L^2 \]
For two situations labeled L (for "light") and and H (for "heavy"), we can write,
\[ \frac{T_{\mbox{L}}}{T_{\mbox{H}}} = \frac{\mu_{\mbox{L}}\, f_{\mbox{L}}^2\, L_{\mbox{L}}^2}{\mu_{\mbox{H}}\, f_{\mbox{H}}^2\, L_{\mbox{H}}^2} \]
That formula looks really complicated and busy, but we can simplify it.
In this problem, the tensions are the same ($T_{\mbox{L}}=T_{\mbox{H}}$) and the
frequencies are the same ($f_{\mbox{L}}=f_{\mbox{H}}$). The mass densities are not equal,
and the lengths are not equal, so
\[ 1 = \frac{\mu_{\mbox{L}}\, L_{\mbox{L}}^2}{\mu_{\mbox{H}}\, L_{\mbox{H}}^2} \]
We are given one of the lengths and the ratio of mass densities, so solve for the unknown length,
\[ \fbox{$\displaystyle L_{\mbox{H}} = \sqrt{\frac{\mu_{\mbox{L}}}{\mu_{\mbox{H}}}}\cdot L_{\mbox{L}} $} \]
Numerically, the find the length of the heavier wire,
\[ L_{\mbox{H}} = \sqrt{\frac{1}{1.50}}\cdot (0.500) = 0.408\,\mbox{m} = 40.8\,\mbox{cm} \]
Intuition for musicians: The heavy strings can be given a higher pitch by making them shorter than the
lighter ones that are under similar tension. This fact is used, for example, to tune guitar strings relative
to each other.
The intended numerical answer is $\fbox{$L_\mbox{c}=0.80\,\mbox{m}$}$.
(The wrong answer that I got before was $\fbox{$L_\mbox{c}=0.75\,\mbox{m}$}$.)
Your symbolic expression should involve quantities that
are given (or asked for) in the problem. For this problem, these symbols include:
\begin{align}
& L_\mbox{o} =\ \mbox{tuned length of pipe open at both ends}\\[1ex]
& L_\mbox{c} =\ \mbox{given length of pipe open at one end and closed at the other end}\\[1ex]
& n_{\mbox{o}} =\ \mbox{which harmonic on the pipe with both ends open ($n_{\mbox{o}}=2$)} \\[1ex]
& n_{\mbox{c}} =\ \mbox{which harmonic on the pipe with one open end and one closed end ($n_{\mbox{c}}=3$)}
\end{align}
It may also include other constants, like $g$ or $\pi$. Another thing you
can do is incorporate special values that are trivial or cancel out.
The symbolic answer should include which harmonics you are trying to match up by
adjusting the length. On your first attempt, while working on this, you might substitute
the values in up-front, but try to incorporate those paramaters ($n_{\mbox{o}}$ and $n_{\mbox{c}}$) in your
submitted symbolic answer. [Correction: This hint might have been used, but it is not correct. $L_{\mbox{c}}=(2n_{\mbox{c}}-1)\lambda/4$ where $n_{\mbox{c}}=1,2,3,4...$. See below for the correct hint.]
In fact, the definition of harmonic means that
\[ L_{\mbox{c}} = \frac{n_{\mbox{c}}\lambda}{4}\qquad \mbox{where $n_{\mbox{c}}=1,3,5,7,...$ labels the harmonic for closed-on-one-end tube} \]
\[ L_{\mbox{o}} = \frac{n_{\mbox{o}}\lambda}{2}\qquad \mbox{where $n_{\mbox{o}}=1,2,3,4,...$ labels the harmonic for open tube} \]
Even that is not pedantically correct because the lowest frequency is called the "fundamental frequency", not the "first harmonic", although that is what it might be called.
If the frequency is the same for both tubes, so is the wavelength. Then,
\[ \lambda = \frac{4\,L_{\mbox{c}}}{n_{\mbox{c}}} = \frac{2\,L_{\mbox{o}}}{n_{\mbox{o}}} \]
Solving for $L_{\mbox{o}}$,
\[ L_{\mbox{o}} = \frac{2 n_{\mbox{o}}\,L_{\mbox{c}}}{n_{\mbox{c}}}
= \frac{2(2)\,L_{\mbox{c}}}{(3)}
= \frac43\,(0.60)
= 0.80\,\mbox{m}
\]
The numerical answer I get is $\fbox{$\rho_{\mbox{ore}}=2.7\,\mbox{gm/cm}^3=2700\,\mbox{kg/m}^3$}$.
Your symbolic expression should involve quantities that
are given (or asked for) in the problem. For this problem, these symbols include:
\begin{align}
& \rho_{\mbox{ore}} = \mbox{denisity of the ore sample} \\[1ex]
& W = \mbox{tension reading when ore is in air} \\[1ex]
& T = \mbox{tension reading when ore is fully immersed in water} \\[1ex]
& \rho_{\small\mbox{H$_2$O}} = \mbox{density of water}
\end{align}
It may also include other constants, like $g$ or $\pi$. Another thing you
can do is incorporate special values that are trivial or cancel out.
I get
\[ \rho_{\mbox{ore}} = \rho_{\small\mbox{H$_2$O}} \cdot\left[\frac{W}{W-T}\right] \]
The numerical answer I get is $\fbox{$r=4.8\,\mbox{cm}$}$.
Your symbolic expression should involve quantities that
are given (or asked for) in the problem. For this problem, these symbols include:
\begin{align}
& r = \mbox{radius of plunger, assumed to be round} \\[1ex]
& \Delta P = \mbox{pressure change} \\[1ex]
& \Delta x = \mbox{change in position of plunger} \\[1ex]
& k = \mbox{spring constant}
\end{align}
It may also include other constants, like $g$ or $\pi$. Another thing you
can do is incorporate special values that are trivial or cancel out.
The symbolic expression for the radius is
\[ r = \sqrt{\frac{k\cdot(\Delta x)}{\pi\cdot (\Delta P)}} \]
This answer is more straightforward than the previous one (HiHW 9a).
The numerical answer I get is $\fbox{$P=71.5\,\mbox{kW}$}$, which
seems like a lot of power.
Your symbolic expression should involve quantities that
are given (or asked for) in the problem. For this problem, these symbols include:
\begin{align}
& P = \mbox{power} = \cdots \\[1ex]
& M = 900\,\mbox{kg} \\[1ex]
& R = 0.45\,\mbox{m} \\[1ex]
& \alpha = -1.3\,\mbox{rad}/\mbox{s}^2 \\[1ex]
& (\Delta\theta) = 5.6\times 10^5\,\mbox{rad}
\end{align}
It may also include other constants, like $g$. Another thing you
can do is incorporate special values that are trivial, for example,
it if comes to rest, you can say $\omega_f=0\,\mbox{rad}/\mbox{s}$ and use that numerical fact
in your expression. Your symbolic expression must not include $(\Delta t)$ or $\omega_i$.
To make sure you are on track, I calculated a few intermediate quantities to
help you develop an intuition:
\begin{align}
& (\Delta t) = 15.5\,\mbox{minutes} \\[1ex]
& \omega_i = 1210\,\mbox{rad}/\mbox{s} = 11,500\,\mbox{RPM} \\[1ex]
& v_i = R\omega_i=543\,\mbox{m/s} = 1210\,\mbox{MPH} =\mbox{speed at edge of wheel}
\end{align}
That initial revolution rate is quite large, so it is a really good
design feature to place the entire spinning disk in a vacuum chamber
and use frictionless magnetic bearings: The speed on the rim is faster
than the speed of sound in air ($320\,\mbox{m/s}$). Another thing, before we
think about using this to replace foreign oil sources, realize that it
will be useful for only fifteen minutes. It also takes a lot of energy
to spin it up in the first place. Something like this could be used
to temporarily store electricity generated by solar or wind.
I'll split it into two parts, and then put it together.
First,
\[
E_i = \tfrac12\,I\omega^2 = \tfrac12\left(\tfrac12 MR^2\right)\left(-2\alpha(\Delta\theta)\right)\qquad\mbox{and}\qquad E_f = 0 \]
\[ (\Delta E) = E_f-E_i \]
Second, the time needed to stop,
\[ (0)=\omega_0 + \alpha\,(\Delta t)\qquad\Rightarrow\qquad (\Delta t ) = \frac{\omega_0-(0)}{-\alpha} = \sqrt{\frac{2\cdot (\Delta\theta)}{-\alpha}} \]
The $\alpha=-|\alpha|$ is negative. Putting things together,
\[
P = \frac{(\Delta E)}{(\Delta t)}
= \frac{-\tfrac12\left(\tfrac12 MR^2\right)\left(2|\alpha|(\Delta\theta)\right)}
{\displaystyle\sqrt{\frac{2\cdot (\Delta\theta)}{|\alpha|}}}
= - \frac{MR^2 \sqrt{|\alpha|^3\,(\Delta\theta)}} {2\sqrt{2}}
\]
The negative sign out front represent the fact that there is decrease in energy of the slowing flywheel,
so, if that is connected to a generator, it will output electric energy at a rate of $-P$.
I realize this is more complicated than usual.
The numerical answer I get is $\fbox{$m_2=250\,\mbox{gm}$}$.
You can imagine the two masses are connected by a massless but rigid wire
where you identify the balance point (using torques) as the center of mass position, OR
you can use the formula approach where
\[Mx_{\mbox{cm}}=\sum_{i=1}^N m_ix_i\]
for a set of $N$ masses (total mass $M=m_1+m_2+\cdots +m_N$) located along the $x\mbox{-axis}$.
In either case, you do need a diagram with coordinate axis.
The numerical answer I get is $\fbox{$v=4.0\,\mbox{m/s}$}$.
You can confirm that the total kinetic energy, before the
collision is $280\,\mbox{J}$, and after collision, it is a somewhat less than that.
This is neither a perfectly elastic (kinetic energy is conserved)
nor a perfectly inelastic (they stick together) collision. It is somewhere in between.
The initial speed of Tito before the collision with Jermaine is
\begin{align}
v_i &= \frac{M_{\small\mbox{Tito}}\sqrt{2gh_{\small\mbox{Tito}}}}{M_{\small\mbox{Jermaine}}}\\[2ex]
&= \frac{(45)\sqrt{2(9.8)(0.50)}}{(35)} = 4.0\,\mbox{m/s}
\end{align}
The numerical answer I get is $\fbox{$v=2.7\,\mbox{m/s}$}$.
To be sure you are using the correct angle, I get the radius of the uniform
circular motion to be $R=0.856\,\mbox{m}$.
The symbolic answer should be expressed in terms of
\begin{align}
& L = 1.3\,\mbox{m} \\[1ex]
& \theta = 48.8^\circ \\[1ex]
& g = 9.8\,\mbox{m/s}^2
\end{align}
I get the result (with a possible variant),
\[ v=\sqrt{\frac{gL\,\cos(\theta)}{\tan(\theta)}} =\sqrt{\frac{gL\,\cos^2(\theta)}{\sin(\theta)}} \]
The numerical answer I get is $\fbox{$\mu_k=0.45$}$. There are no units to $\mu_k$.
Note that the direction of the velocity is
not specified in the problem. You need that direction to determine the direction of
the kinetic friction. If you try both directions, you will discover that because
$ma\lt T$, the physically meaningful velocity direction is in the same
direction as the tension and the acceleration. The given numbers
result in a $\mu_k=0.45\gt 0$. You can think about what would happen if
$ma>T$, which would simply require different given numbers.
Either one of these would work:
\[ \mu_k = \frac{T - ma}{mg} = \frac{T}{mg} - \frac{a}{g} \]
The answer should have been expressed in terms of things that you are given, here
$T$, $a$, & $m$; and any constants, here $g$.
For other problems, you could use trivial givens, like if the initial
speed is zero, you could have used, $v_i=0$ in the symbolic expression. Here, we don't have that.
The numerical answer I get is $d=45\,\mbox{m}$.
In Representation, I will look for a picture of a car in the two situations, a free body diagram for the car, a motion diagram for the car, a coordinate axis, and a symbolization of the necessary physics
quantities, like $v$, $d$, etc. I suggest using "$(\Sigma F)$" or "$F_\mbox{net}$" to represent the total force
acting on the car.
To develop a
little intuition and learn a small driving lesson, first convert the speed to MPH and KPH,
\[ v=31\,\mbox{m/s} = 69\,\mbox{miles/hour} = 112\,\mbox{kilometer/hour} \]
In this problem, note that the deceleration is
\[ a = \frac{(\Sigma F)}{m} = 10.6\,\mbox{m/s}^2 = (1.1)\,g \]
which is a little bigger than 1-g. This is a rather significant braking,
and it is likely possible only under the very best road-tire conditions.
Yet, even traveling at this common highway speed limit, it still takes 45 meters (about half of
one football field) to stop, which will take just under three seconds (31/10.6=2.9). This also does not
take into account human reaction time.
How much room should you leave between you and the
driver in front of you on the highway? Is the old one-car-length for each 10-MPH of speed adequate?
\[ D = \frac{m\,v^2}{2\cdot(\Sigma F)} \]
I get an angle of "$\theta_f=52.4^\circ$ below the horizontal", or you can write, "$\theta_f=-52.4^\circ$ relative to
the positive x-axis".
The symbolic answer should express that $\theta_f$ (final theta) in
terms of the given symbols, $\Delta t$, $\Delta x$, and $g$. Note that
instead of simply $t$, I like
to use $\Delta t=t_f-t_i$ because it better represents the time interval analogous to the way $\Delta x=x_f-x_i$ is
the horizontal displacement (a distance interval).
Another caveat is that the symbolic
answer should not involve other symbols that may have been used to get
there, like the horizontal component of the velocity or the height.
I capitalize the arc-tangent because it actually becomes a
single-valued function, which is a little-known and little-used arcane
convention. (For example, $\arctan(0)={0,\pm 180^\circ,\pm 360^\circ,\cdots}$, but
$\mbox{Tan}^{-1}(0)=0$.) The minus sign out front says that the angle theta
$\theta_f$ points down below the horizontal.
Note that $\mbox{Tan}^{-1}(v_{y0}/v_{x0})$ is not enough for full credit.
I get $|v_f|=1.37\,\mbox{m/s}$ as the final speed of that fish. It is
the speed of the fish after that $(\Delta t)=0.25\,\mbox{s}$ interval
where it is accelerating.
Hint: the quantity $(\Delta\vec{v})=\vec{a}\cdot(\Delta t)$ is what
we must add to $\vec{v}_i$ to get $\vec{v}_f$. That hint is also
provided in the word problem: $(\Delta \vec{v})=\vec{v}_f - \vec{v}_i$.
Draw arrows representing these three things.
Note that I write $|\vec{v}_f|$ to mean the magnitude
of the final velocity $\vec{v}_f$.
Some people have trouble with compass directions. Symbolize the given
angle using $\theta$, and measure it relative to West going in a
counter-clockwise direction. This means West would be $\theta=0^\circ$,
South would be $\theta=90^\circ$, and exactly South-West would be $\theta=45^\circ$. This means we can use $\theta$ to symbolize how many degrees "South of West".
Your symbolic answer should express $v_w$ (velocity of wide receiver) in terms of
symbols (letters with subscripts) representing given values in the word problem.
If you discover something in your work, make sure you go back to earlier
steps and adjust things so everything is consistent.
Representation [0, 1, 2] In this section, we are looking for
diagrams, sketches, tabulation of givens and unknowns, etc. The
diagrams should be clear and representative of the problem at hand. A
graphic or pictorial or other visual representation of the
problem at hand is useful. Also important are relevant and properly
labeled axes, motion diagrams, free body or force diagrams, and so on.
Initial Equation(s) [0, 0.5, 1] List the relevant generic equations
needed to solve the problem.
When you first start the
problem, you might not know what you need, so when you are done, go
back and review this part. Mathematical relationships like those from trigonometry or the
quadratic equation are not needed here. The equations you choose to list
should line up with your Physics Concept(s), and also be equations
that you actually use.
Symbolic Answer [0, 1] This is the solution to the problem using
symbols representing the variables and parameters, rather than
numerical values. [However, if some things are specified to be zero,
you can simplify the solution using those values. For example, if the
final height of the ball is $y_f=0\,\mbox{m}$, you can use that fact in the symbolic answer.]
In this example, the symbolic answer is
\[ v_f = \frac{(\Delta y)}{(\Delta t)} - \tfrac12\,g\cdot(\Delta t) \]
If you solve a quadratic equation and come up with two mathematical
answers, but only one is physically relevant, you should identify
which one that is. For example, consider the contrived problem where
you must solve the following for $x$,
\[ Ax^2 + Bx + C = 0 \]
where you should get [mathematically],
\[ x_{\pm} = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \]
Suppose the one with the plus sign is the only interesting physical
one (which might not be obvious until you plug in numbers), then your
symbolic solution would be
\[ x = \frac{-B + \sqrt{B^2 - 4AC}}{2A} \]
It is usually useful to sketch $y=Ax^2+Bx+C$ and point out which root is relevant.
Units Check [0, 0.5, 1] If you have a symbolic answer (correct or
not), you should check it for units. If it doesn't check, you should
go back and try to find out why. (That is why it is called a units
check.) I like the notation that square brackets around a symbol
indicate its units, e.g., $[g]=\mbox{m/s}^2$. In the example problem, a
units check might look like:
\begin{align}
[v_f] & = \frac{[y]}{[t]} = [g]\cdot[t]\qquad ? \\[1ex]
\mbox{m/s} & = \frac{\mbox{m}}{\mbox{s}} = \frac{\mbox{m}}{\mbox{s}^2} \cdot \mbox{s} \qquad \checkmark
\end{align}
where I check that each term in the equation (symbolic solution) has the same units of m/s.
Neatness [-2, -1, 0] The neatness is relative. If everything is wrong but I can still read
what you have, it is "neat", but if I can't read what you have written, right or not, it is "not neat".
Correct Answer [N/A] This is the correct numerical answer. In this example, it is
$v_f=-9.7\,\mbox{m/s}$. This is not worth any points. Note the sign, significant figures,
and units are all relevant parts of the answer.
If the only thing you write on the paper (besides our names) is
\[ v_f = \frac{(-0.80)}{(1.9)} - \tfrac12\cdot (9.8)\cdot (1.9) = -9.7\,\mbox{m/s} \]
you will receive zero points total, even though it is the correct numerical
answer to the question. This is how Hand-in HomeWork differs from WebAssign homework problems.
A suggestion for plugging a number into the equation is to not
write the units, but instead enclose the numerical value in
parentheses. This will make the equation less cluttered and prepare it
for the calculator. Of course, you should properly round and restore
the units in the end.
To illustrate the confusion that may arise, consider this,
\[ v_f = \frac{-0.80\,\mbox{m}}{1.9\,\mbox{s}} - \tfrac12\times 9.8\,\mbox{m/s}^2\times 1.9\,\mbox{s} = -9.7\,\mbox{m/s} \]
Although perfectly correct, the extra symbols can lead to mistakes when entering the numbers into a
calculator. That expression is typeset, imagine the confusion if it is hand-written, especially if, for example,
"$m$" can represent mass and "$\mbox{m}$" can represent meters.
I might add that some teachers require that format, but here we can use the units check
result to make sure we have the correct units, and so we don't have to write them again here.
No FERPA-protected information (student names, grades, etc.) appears on this website.
This website does not use cookies, but it uses a little javascript mostly to render formulas, and even less php.
jones.3 (R Jones) Last Modified 23:59:45 18-Apr-2022