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\begin{center}
{\Large {Solution Derivations for Capa \#11\\[20pt]} }
\end{center}
1) A horizontal circular platform ($M = 128.1\ kg$, $r = 3.11\ m$) rotates
about a frictionless vertical axle. A student ($m = 68.3\ kg$) walks slowly
from the rim of the platform toward the center. The angular velocity $\omega$
of the system is $2.1\ rad/s$ when the student is at the rim. Find $\omega$
when the student is $2.23\ m$ from the center. Use ``rad/s'' as units.
\begin{quote}
$M$ of platform = Given\\
$R$ of platform = Given\\
$m$ of student = Given\\
$r_{i}$ of student = Given\\
$r_{f}$ of student = Given\\
$\omega _{i}$ = Given\\
$\omega _{f}$ = ?\\
This is a simple conservation of angular momentum problem.
\begin{eqnarray*}
L_{i} & = & L_{f}\\
I_{i}\omega _{i} & = & I_{f}\omega _{f}
\end{eqnarray*}
The moments of inertia are just the sum of moments of inertia of both the
disk and the student.
\begin{eqnarray*}
I_{i} & = & \frac{1}{2}MR^{2}+mr_{i}^{2}\\
I_{f} & = & \frac{1}{2}MR^{2}+mr_{f}^{2}
\end{eqnarray*}
Realizing that the student starts at the rim of the platform, the initial
radius of the student is equal to the radius of the platform, or $r_{i}=R$.
Plugging these new values in and solving for $\omega _{f}$,
\begin{eqnarray*}
\left( \frac{1}{2}MR^{2}+mR^{2}\right) \omega _{i} & = & \left( \frac{1}{2}
MR^{2}+mr_{f}^{2}\right) \omega _{f}\\
\frac{\left( \frac{1}{2}MR^{2}+mR^{2}\right) \omega _{i}}{\left( \frac{1}{2}
MR^{2}+mr_{f}^{2}\right) } & = & \omega _{f}
\end{eqnarray*}
No conversions should be necessary since $\omega _{i}$ is given in $rad/s$ and
you are asked to answer in rad/s.
\end{quote}
2) Two particles, each with mass $m = 5.8\ g$, are fastened to each other and
to a rotation axis as point $P$, by two thin rods, each with length
$L = 0.65\ m$ and a mass of $8.1\ g$, as shown. The combination rotates
around the rotation axis with an angular velocity of $13.6\ rad/s$. Find the
rotational inertia of the combination about $P$?
\begin{quote}
$m$ = Given in grams\\
$L$ = Given\\
$M$ = Given in grams\\
$\omega $ = Given\\
$I$ = ?\\
The moment of inertia of the system about the axis $P$ is the sum of the
moments of inertia for each rod and mass. Since the rods are equal length
and mass and since they are attached in a line, they can be treated as a
single rod of mass $2M$ and length $2L$.
\[
I=\frac{1}{3}\left( 2M\right) \left( 2L\right) ^{2}+mr_{1}^{2}+mr_{2}^{2}
\]
Since the masses are given in grams, first convert them to kilograms by
dividing by 1000. The radius to the first mass is the length of the first
rod. The radius of the second mass is the length of both rods.
\[
I=\frac{8}{3}ML^{2}+mL^{2}+m\left( 2L\right) ^{2}
\]
Units for moment of inertia are $kg \cdot m^{2}$.
\end{quote}
3) A large plate is balanced at its center and two students of equal mass
stand at its center. The plate is rotated on a frictionless pivot about an
axis through its center and perpendicular to its face. The students then
begin to walk out towards opposite edges. (Select T-True, F-False,
I-Increases, D-Decreases, S-Stays the same. ).
\begin{quote}
QUESTION:\\
A) The rate of rotation ... as the students walk outward.\\
B) The students produce a net torque on the plate.\\
C) When the students reach the edge and stop, the plate will have the same
angular speed as when they started.\\
D) The students do work in walking outward\\
E) The total angular momentum of the system ... as the students walk outward.\\
ANSWER:\\
A) D. As the mass of the students gets further away from the axis of
rotation, the rate slows down (remember the skater with arms out).\\
B) T. No explanation currently...\\
C) F. The plate slows down as they get further out. The rotation rate
does not matter if they are moving or not.\\
D) T. The students exert a force on the plate (and the plate on the
students) when walking out. Thus, that force over the distance walked
causes word to be down. Consequently the force of the plate on the
students is what slows the rotation rate.\\
E) S. Angular momentum is conserved.\\
\end{quote}
4) In the Figure, a block $m_1$ sits on a table. There {\em is} static
friction between the block and table. Block $m_2$ hangs from a knot, as
shown. Call the tension in the rope connecting the knot to $m_1$, ``T1''.
Call the tension in the rope connecting the knot to $m_2$, ``T2''. Call the
tension in the third rope (the one tipped up by an angle $\theta$, connecting
the knot to the wall, ``T3''. {\em The system is in equilibrium}. Which of
the following statements are true? (If A and E are true, and the others are
not, enter TFFFT)
\begin{quote}
QUESTION:\\
A) T3 is equal in magnitude to T1.\\
B) The net force on the knot is zero.\\
C) If the table was completely frictionless, the system as shown could not
be in equilibrium.\\
D) The tension T2 must equal $m_{2}g$.\\
E) The force of static friction on $m_{1}$ equals T1.\\
ANSWER:\\
A) F. Since T3 is directed an and angle $\theta $ above the horizontal,
it is not equal. T3's horizontal component, is, however.\\
B) T. Since the system is in equilibrium, all the forces sum to 0.\\
C) T. If there was no horizontal force (from friction) opposing T3's
horizontal component, the system would not be in equilibrium.\\
D) F. Since the mass is the only object connected to the string, it
must equal the weight.\\
E) T. Again, since the mass is the only object connected to the string,
it must equal to the force of friction.\\
\end{quote}
5) In the previous problem, block $m_1$ weighs $709\ N$. The coefficient of
static friction between the block and the table is $0.27$ and the angle
$\theta$ is $34.3^{\circ}$. Find the maximum weight of block $m_2$ for which
block $m_1$ will remain at rest. (Answer ins Newtons, N)
\begin{figure}[h]
\begin{center}
\includegraphics[width=4.0075in]{5a.png}
\end{center}
\end{figure}
\begin{quote}
$m_{1}$ = Given\\
$\mu _{s}$ = Given\\
$\theta $ = Given\\
From a free body diagram, we can see that
\begin{eqnarray*}
\sum F_{y} & = & -T_{2}+T_{3}\sin \theta\\
T_{3}\sin \theta & = & m_{2}g\\
T_{3} & = & \frac{m_{2}g}{\sin \theta }\\
\sum F_{x} & = & T_{3}\cos \theta -T_{1}\\
T_{1} & = & T_{3}\cos \theta =\frac{m_{2}g}{\sin \theta }\cos \theta\\
m_{2} & = & \frac{T_{1}\sin \theta }{g\cos \theta }=\frac{T_{1}}{g}\tan \theta
\end{eqnarray*}
Since $T_{1}$ is due to friction,
\begin{eqnarray*}
T_{1} & = & \mu _{s}N=\mu _{s}m_{1}g\\
m_{2} & = & \frac{\mu _{s}m_{1}g}{g}\tan \theta =\mu _{s}m_{1}\tan \theta
\end{eqnarray*}
\end{quote}
6) A physics student is sitting on a frictionless platform that is rotating
with an angular speed of $\omega_i = 3.8\ rad/s$. His arms are outstretched,
and he is holding a heavy weight in each hand. The moment of inertia of the
student, the extended weights, and the platform about the rotation axis is
$I_i = 8.5\ kg \cdot m^2$. When the student pulls the weights inward toward
his body, the moment of inertia decreases to $I_f = 5.0\ kg \cdot m^2$. What
is the resulting angular speed, $\omega_f$, of the platform?
\begin{quote}
$\omega _{i}$ = Given\\
$I_{i}$ = Given\\
$I_{f}$ = Given\\
$\omega _{f}$ = ?\\
This is another conservation of angular momentum problem.
\begin{eqnarray*}
L_{i} & = & L_{f}\\
I_{i}\omega _{i} & = & I_{f}\omega _{f}\\
\omega _{f} & = & \frac{I_{i}\omega _{i}}{I_{f}}
\end{eqnarray*}
This will give you an answer in rad/s.
\end{quote}
7) By how much is the kinetic energy of the system increased? (Think about
where this increase in energy comes from!)
\begin{quote}
Kinetic energy is related to the moment of inertia and the angular speed.
\[
KE=\frac{1}{2}I\omega ^{2}
\]
To find the increase in kinetic energy, subtract the initial from the final.
\[
\Delta KE=\frac{1}{2}I_{f}\omega _{f}^{2}-\frac{1}{2}I_{i}\omega _{i}^{2}=
\frac{1}{2}\left( I_{f}\omega _{f}^{2}-I_{i}\omega _{i}^{2}\right)
\]
\end{quote}
8) Compact discs and long-playing records are made from similar materials.
The former have a diameter of about $12\ cm$, and the latter, about $32\ cm$.
When in use, records spin at $33\frac{1}{3}\ rev/min$, and compact discs spin
at, say, $365\ rev/min$. Ignoring the holes in both objects and assuming that
a compact disc has half the thickness of a record and $0.90$ of its density,
what is the ratio of the angular momentum of a compact disc in use to that of
a record?
\begin{quote}
$R_{CD}$ = Given\\
$R_{LP}$ = Given\\
$\omega _{LP}$ = Given\\
$\omega _{CD}$ = Given\\
$h_{CD}$ = Thickness of the CD = Given in terms of the record (LP). Assume it
is a coefficient, $y$, times $h_{LP}$. That is, if your CD height is half of
your record height, $y=\frac{1}{2}$.\\
$d_{CD}$ = Given as a coefficient times $d_{LP}$. Assume it is a coefficient,
$x$, times $d_{LP}$. That is, if your CD density is 90\% of your record
density, $x=0.9$.\\
This problem deals with angular momentum.
\[
L=I\omega
\]
In this case, you are asked to ignore the holes in the CD and LP, so the
object is just like a solid disk. The moment of inertia for a disk is
given by $\frac{1}{2}mR^{2}$. So,
\[
L=\frac{1}{2}mR^{2}\omega
\]
You are asked to find the ratio of the angular of momentum of the CD to that
of a record
\[
\frac{L_{CD}}{L_{LP}}=\frac{\frac{1}{2}m_{CD}R_{CD}^{2}\omega _{CD}}{\frac{1
}{2}m_{LP}R_{LP}^{2}\omega _{LP}}=\frac{m_{CD}R_{CD}^{2}\omega _{CD}}{
m_{LP}R_{LP}^{2}\omega _{LP}}
\]
We know everything but the mass of the CD and record. Density is mass/volume.
So mass = density * volume. To find the volume of a CD, use the formula for a
right cylinder (i.e. $\pi R^{2}h$). The $h$ in this case will be
$y\ast h_{LP}$. So the volume of the CD is $\pi R_{CD}^{2}\ast y\ast h_{LP}$.
To find the mass of the CD, multiply the density given by the volume. The
given density is in terms of the density of the record ($x\ast d_{LP}$). So,
the mass of the CD is just
\begin{eqnarray*}
m_{CD} & = & v\ast d\\
m_{CD} & = & \pi R_{CD}^{2}\ast y\ast h_{LP}\ast x\ast d_{LP}
\end{eqnarray*}
The mass of the record can be found also by finding its volume ($\pi
R_{LP}^{2}h_{LP}$) and its density ($d_{LP}$). Don't worry that some of
these terms are unknown because they will cancel in the end. So,
\begin{eqnarray*}
m_{LP} & = & v\ast d\\
m_{LP} & = & \pi R_{LP}^{2}h_{LP}\ast d_{LP}
\end{eqnarray*}
Plugging these back into the ratio of angular momenta,
\[
\frac{L_{CD}}{L_{LP}}=\frac{m_{CD}R_{CD}^{2}\omega _{CD}}{
m_{LP}R_{LP}^{2}\omega _{LP}}=\frac{\pi R_{CD}^{2}\ast y\ast h_{LP}\ast
x\ast d_{LP}\ast R_{CD}^{2}\omega _{CD}}{\pi R_{LP}^{2}h_{LP}\ast d_{LP}\ast
R_{LP}^{2}\omega _{LP}}
\]
Look what cancels. $\pi ,$ $h_{LP},$ and $d_{LP}$. Simplifying,
\[
\frac{L_{CD}}{L_{LP}}=\frac{R_{CD}^{2}\ast y\ast x\ast R_{CD}^{2}\omega
_{CD}}{R_{LP}^{2}\ast R_{LP}^{2}\omega _{LP}}=\frac{R_{CD}^{4}\ast y\ast
x\ast \omega _{CD}}{R_{LP}^{4}\ast \omega _{LP}}
\]
This is your answer, and ratio's have no units.
\end{quote}
9) A uniform horizontal bar of length $L = 2\ m$ and weight $219\ N$ is pinned
to a vertical wall and supported by a thin wire that makes an angle of
$\theta = 24^{\circ}$ with the horizontal. A mass $M$, with a weight of
$311\ N$, can be moved anywhere along the bar. The wire can withstand a
maximum tension of $515\ N$. What is the maximum possible distance from the
wall at which mass $M$ can be placed before the wire breaks?
\begin{figure}[h]
\begin{center}
\includegraphics[width=3.4108in]{9a.png}
\end{center}
\end{figure}
\begin{quote}
$L$ = Given\\
$mg$ = Given (weight of rod)\\
$\theta $ = Given\\
$Mg$ = Given (weight of mass)\\
$T_{\max }$ = Given\\
$r$ = ? (radius to mass from axis $A$)\\
This system is obviously in equilibrium. In order for a system to be in
equilibrium, the sum of forces must be zero and the sum of torques must be
zero. However, we only need to be concerned with the torques for this part
of the problem. Looking at a free-body diagram, we can find the torques about
the axis $A$. The sum of torques is equal to the torque of the wire, the
torque of the rod, and the torque of the mass. Remember $\tau =rF\sin \theta$
\[
\tau =TL\sin \theta -\frac{1}{2}Lmg-rMg=0
\]
Solving for $r$,
\begin{eqnarray*}
TL\sin \theta -\frac{1}{2}Lmg & = & rMg\\
\frac{TL\sin \theta -\frac{1}{2}Lmg}{Mg} & = & r
\end{eqnarray*}
\end{quote}
10) With $M$ placed at this maximum distance what is the horizontal component
of the force exerted on the bar by the pin at $A$?
\begin{quote}
To find the horizontal component of the force, we can again look at the
free-body diagram. The only forces acting in the horizontal direction are
the component of the tension, and the force of the wall on the rod at point
$A$. So, they must be equal.
\[
F_{x}=T\cos \theta
\]
\end{quote}
11) Misako is about to do a push-up. Her center of gravity lies directly
above a point on the floor which is $d_1 = 1.04\ m$ from her feet and
$d_2 = 0.62\ m$ from her hands. If her mass is $55.0\ kg$, what is the force
exerted by the floor on her hands?
\begin{quote}
$d_{1}$ = Given\\
$d_{2}$ = Given\\
$m$ = Given\\
This problem involves torque once again. The person is pushing on the floor
in two different spots. She is exerting a torque on each spot on the ground.
Since she nor the ground is moving, the torque must be equal. In most cases,
the center of gravity will also be the center of mass. Call the force of her
hands (a distance of $d_{2}$ from the center of gravity) $F_{2}$. Call the
force of her feet (a distance of $d_{1}$ from her center of gravity) $F_{1}$.
Since she is pushing at right angles on the ground, and the force of her hands
is exerting an opposite torque due to the force of her feet so the total
torque can be calculated by
\[
\tau =F_{1}d_{1}-F_{2}d_{2}=0
\]
However, we must find $F_{1}$ and $F_{2}$ now. Realizing that her mass is
$m$, she is only exerting a force of $mg$ on the ground. Thus,
\[
F_{1}+F_{2}=mg
\]
Solving for $F_{1}$ in this equation will give $F_{2}$ in the second
equation (the force of her hands on the ground). This is also the force of
the ground on her hands since the ground pushes back with an equal and
opposite force.
\[
F_{1} = mg-F_{2}
\]
Plugging back into the torque equation,
\begin{eqnarray*}
0 & = & d_{1}\left( mg-F_{2}\right) -F_{2}d_{2}\\
0 & = & d_{1}mg-d_{1}F_{2}-F_{2}d_{2}\\
d_{1}F_{2}+d_{2}F_{2} & = & d_{1}mg\\
F_{2}\left( d_{1}+d_{2}\right) & = & d_{1}mg\\
F_{2} & = & \frac{d_{1}mg}{d_{1}+d_{2}}
\end{eqnarray*}
Force has units of Newtons (N).
\end{quote}
12) What is the force exerted by the floor on her feet?
\begin{quote}
This problem is simply an extension of the last. Since we know $F_{2}$,
you can use the equation $F_{1}+F_{2}=mg$ to find the other force (the force
of the ground on her feet). So,
\[
F_{1}=mg-F_{2}
\]
\end{quote}
13) The weightless horizontal bar in the figure below is in equilibrium.
Scale B reads $4.40\ kg$. The distances in the figure (which is not to scale)
are: D1 = $6.0\ cm$, D2 = $7.5\ cm$, and D3 = $6.0\ cm$. The mass of block X
is $0.98\ kg$ and the mass of block Y is $2.04\ kg$. Determine what the
reading on scale A must be.
\begin{quote}
$B$ = Given\\
$D_{1}$ = Given\\
$D_{2}$ = Given\\
$D_{3}$ = Given\\
$m_{x}$ = Given\\
$m_{y}$ = Given\\
$A$ = ?\\
This problem involves the torque. The torque on $A$ must equal the torque
on $B$ since the system is not moving. The torque on $B$ can be found by
summing the torque from mass $x$ and mass $y$. Since mass $z$ is on the
axis of rotation (at $B$), it produces no torque.
\[
\tau _{B}=(D_{2}+D_{3})m_{x}+D_{3}m_{y}
\]
The torque on $A$ can be calculated by referencing it from $B$. In other
words, what torque must be applied at $A$ to counteract the torques from
$m_{x}$ and $m_{y}$? This can by found by denoting the variable $m_{a}$ to
represent the mass (the scale reading) at $A$ to counteract the torque
calculated above. That is,
\[
m_{a}\left( D_{1}+D_{2}+D_{3}\right) =(D_{2}+D_{3})m_{x}+D_{3}m_{y}
\]
Thus, the scale reading at $A$ (that is, $m_{a}$)
\[
m_{a}=\frac{(D_{2}+D_{3})m_{x}+D_{3}m_{y}}{D_{1}+D_{2}+D_{3}}
\]
\end{quote}
\end{document}