From: W. Trevor King Date: Fri, 25 Mar 2011 15:43:45 +0000 (-0400) Subject: Add rest of winter 101 problems. X-Git-Url: http://git.tremily.us/?a=commitdiff_plain;h=82b1cfabdb6445aa6cc5c2c7941940745d9bfdae;p=course.git Add rest of winter 101 problems. --- diff --git a/latex/problems/Serway_and_Jewett_8/problem06.08.tex b/latex/problems/Serway_and_Jewett_8/problem06.08.tex new file mode 100644 index 0000000..42c3117 --- /dev/null +++ b/latex/problems/Serway_and_Jewett_8/problem06.08.tex @@ -0,0 +1,72 @@ +\begin{problem*}{6.8} +Consider a conical pendulum (Fig.~P6.8) with a bob of mass +$m=80.0\U{kg}$ on a string of length $L=10.0\U{m}$ that makes an angle +of $\theta=5.00\dg$ with the vertical. Determine \Part{a} the +horizontal and vertical components of the force exerted by the string +on the pendulum and \Part{b} the radial acceleration of the bob. +\begin{center} +\begin{asy} +import three; + +real u = 1cm; + +draw((0,0,2u)--(0,0,0), dashed); +draw((0,0,2u)--(0,u,0)); +draw(scale3(u)*unitcircle3); +dot((0,0,2u)); +dot((0,u,0)); +\end{asy} +\end{center} +\end{problem*} + +\begin{solution} +\Part{a} +Drawing a free body diagram for the bob +\begin{center} +\begin{asy} +import Mechanics; + +real u = 1cm; +real theta = 25; +real mg = 1.3u; + +draw((0,0)--(0, mg)); +Angle t = Angle(dir(90), (0,0), dir(90+theta), "$\theta$"); +t.draw(); +Vector T = Force((0,0), mag=mg/Cos(theta), dir=(90+theta), "$T$"); +T.draw(); +Vector G = Force((0,0), mag=mg, dir=(-90), "$mg$"); +G.draw(); +dot((0,0)); + +draw_ijhat((.5u,0), idir=0); +\end{asy} +\end{center} + +The bob does not move up or down, so the sum of forces in the vertical +direction must be zero. +\begin{align} + 0 &= \sum F_y = T_y - mg \\ + T\cos(\theta) = T_y &= mg + = 80.0\U{kg}\cdot9.80\U{m/s$^2$} = \ans{784\U{N}} \\ + T &= \frac{mg}{\cos(\theta)} \;, +\end{align} +where I've marked the vertical component of the string tension. The +horizontal (radial) component is given by +\begin{equation} + T_x = -T\sin(\theta) + = -\frac{mg\sin(\theta)}{\cos(\theta)} + = -mg\tan(\theta) = \ans{-68.6\U{N}} \;. +\end{equation} + +\Part{b} +The radial acceleration is given by Newton's second law +\begin{align} + F_r &= m a_r \\ + a_r &= \frac{F_r}{m} = \frac{T_x}{m} + = \frac{-mg\tan(\theta)}{m} + = -g\tan(\theta) + = -9.80\U{m/s$^2$}\cdot\tan(5.00\dg) + = \ans{0.857\U{m/s$^2$}} \;. +\end{align} +\end{solution} diff --git a/latex/problems/Serway_and_Jewett_8/problem06.11.tex b/latex/problems/Serway_and_Jewett_8/problem06.11.tex new file mode 100644 index 0000000..3b9fa0d --- /dev/null +++ b/latex/problems/Serway_and_Jewett_8/problem06.11.tex @@ -0,0 +1,70 @@ +\begin{problem*}{6.11} +A coin placed $30.0\U{cm}$ from the center of a rotating, horizontal +turntable slips when its speed is $50.0\U{cm/s}$. \Part{a} What force +causes the centripedal acceleration when the coin is stationary +relative to the turntable? \Part{b} What is the coefficient of static +friction between the coin and the turntable? +\end{problem*} + +\begin{solution} +\Part{a} +\begin{center} +\begin{asy} +import Mechanics; + +real u = 1cm; + +Mass m = Mass((u,0), "$m$"); +Vector F = Force(m.center, mag=0.7u, dir=180, "$F_f$"); + +draw(scale(u)*unitcircle); +Distance r = Distance((0,0), u*dir(45), "$r$"); +r.draw(); +F.draw(); +m.draw(); +\end{asy} +\end{center} + +The only centerward force on the coin is \ans{friction $F_f$}. + +\Part{b} +Drawing a free body diagram for the coin +\begin{center} +\begin{asy} +import Mechanics; + +real u = 1cm; +real theta = 25; +real mg = 1.3u; + +Vector F = Force((0,0), mag=0.7u, dir=180, "$F_f$"); +F.draw(); +Vector G = Force((0,0), mag=1u, dir=-90, "$mg$"); +G.draw(); +Vector N = Force((0,0), mag=1u, dir=90, "$N$"); +N.draw(); +dot((0,0)); + +draw_ijhat((.5u,0), idir=0); +\end{asy} +\end{center} + +Summing the forces in the vertical direction and noting that the coin +moves neither up nor down, +\begin{align} + 0 = \sum F_y &= N - mg \\ + N &= mg \;. +\end{align} + +The centerward acceleration needed to maintain the circular motion is +$a_r=v^2/r$. +\begin{align} + F_f &= m a_r = \mu_s N = \mu_s mg \\ + \mu_s &= \frac{m a_r}{mg} = \frac{a_r}{g} = \frac{v^2}{gr} + = \frac{(50.0\U{cm/s})^2}{980\U{cm/s$^2$}\cdot30.0\U{cm}} + = \ans{0.0850} \;. +\end{align} +Note that I converted $g$ into $\bareU{cm/s$^2$}$ so that all the +$\bareU{cm}$ would cancel out in the answer. You could also have +coverted everything to meters. +\end{solution} diff --git a/latex/problems/Serway_and_Jewett_8/problem06.16.tex b/latex/problems/Serway_and_Jewett_8/problem06.16.tex new file mode 100644 index 0000000..f0eedb1 --- /dev/null +++ b/latex/problems/Serway_and_Jewett_8/problem06.16.tex @@ -0,0 +1,44 @@ +\begin{problem*}{6.16} +A roller-coaster car (Fig. P6.16) has a mass of $500\U{kg}$ when fully +loaded with passengers. The path of the coaster from its intial point +in the figure to point $B$ involves only up-and-down motion (as seen +by the riders), with no motion to the left or right. \Part{a} If the +vehicle has a speed of $20.0\U{m/s}$ at point $A$, what is the force +exerted by the track on the car at this point? \Part{b} What is the +maximum speed the vehicle can have at point $B$ and still remain on +the track? Assume the roller-coaster tracks at points $A$ and $B$ are +pars of vertical circles of radius $r_1=10.0\U{m}$ and +$r_2=15.0\U{m}$, respectively. +\begin{center} +\begin{asy} +// TODO +draw(unitcircle); +\end{asy} +\end{center} +\end{problem*} + +\begin{solution} +\Part{a} +The roller coaster track exerts very little drag on the cars, so the +force must be almost entirely normal to the track. Because $A$ is at +the bottom of a circular portion, the force from the track will be +\ans{upwards}, turning the car up, and keeping it from falling through +the track towards the ground. To get a number... TODO +\begin{align} + N - mg &= mv^2 / r_1 \\ + N &= m (g + v^2/r_1) = \ans{24.9\U{kN}} +\end{align} + +\Part{b} +At $B$, the car must have a centerward (downward) acceleration of +$a_{rB}=v^2/r_2$ to remain on the track. Assuming the car does not +have wheels that wrap around the track, this downward acceleration +must come from gravity. At the \emph{maximum} speed, there will be no +normal force from the track itself, so the \emph{total} acceleration +will be the gravitational acceleration $g$. +\begin{align} + g &= a_rB = \frac{v^2}{r_2} \\ + v &= \sqrt{g r_2} = \sqrt{9.80\U{m/s$^2$}\cdot15.0\U{m}} + = \ans{12.1\U{m/s}} \;. +\end{align} +\end{solution} diff --git a/latex/problems/Serway_and_Jewett_8/problem06.21.tex b/latex/problems/Serway_and_Jewett_8/problem06.21.tex new file mode 100644 index 0000000..c4d1022 --- /dev/null +++ b/latex/problems/Serway_and_Jewett_8/problem06.21.tex @@ -0,0 +1,29 @@ +\begin{problem*}{6.21} +An object of mass $m=0.500\U{kg}$ is suspended from the ceiling of an +accelertating truck as shown in Figure P6.21. Taking +$a=3.00\U{m/s$^2$}$, find \Part{a} the angle $\theta$ that the string +makes with the vertical and \Part{b} the tension $T$ in the string. +\begin{center} +\begin{asy} +import Mechanics; + +real u = 1cm; +real a = 3.00; +real g = 9.80; +real L = 2u; +real theta = aTan(a/g); + +Mass m = Mass(L*dir(-90-theta), "$m$"); +draw((0,0)--m.center); +m.draw(); +draw((0,0)--(0, -0.5L), dashed); +Angle T = Angle((0,-1), (0,0), m.center, "$\theta$"); +T.draw(); +dot((0,0)); +// Vector A = Acceleration +\end{asy} +\end{center} +\end{problem*} + +\begin{solution} +\end{solution} diff --git a/latex/problems/Serway_and_Jewett_8/problem06.39.tex b/latex/problems/Serway_and_Jewett_8/problem06.39.tex new file mode 100644 index 0000000..afe642b --- /dev/null +++ b/latex/problems/Serway_and_Jewett_8/problem06.39.tex @@ -0,0 +1,12 @@ +\begin{problem*}{6.39} +A string under a tension of $50.0\U{N}$ is used to whirl a rock in a +horizontal circle of radius $2.50\U{m}$ at a speed of $20.4\U{m/s}$ on +a frictionless surface as shown in Figure P.39. As the string is +pulled in, the speed of the rock increases. When the string is +$1.00\U{m/s}$ long and the speed of the rock is $51.0\U{m/s}$, the +string breaks. What is the breaking strength, in newtons, of the +string? +\end{problem*} + +\begin{solution} +\end{solution} diff --git a/latex/problems/Serway_and_Jewett_8/problem06.61.tex b/latex/problems/Serway_and_Jewett_8/problem06.61.tex new file mode 100644 index 0000000..93483f3 --- /dev/null +++ b/latex/problems/Serway_and_Jewett_8/problem06.61.tex @@ -0,0 +1,11 @@ +\begin{problem*}{6.61} +A car rounds a banked curve as discussed in Example 6.4 and shown in +Figure 6.5. The radius of curvature of the road is $R$, the banking +angle is $\theta$, and the coefficient of static friction is +$\mu_s$. \Part{a} Determine the range of speeds the car can have +without slipping up or down the road. \Part{b} Find the minimum value +for $\mu_s$ such that the minimum speed is zero. +\end{problem*} + +\begin{solution} +\end{solution} diff --git a/latex/problems/Serway_and_Jewett_8/problem08.08.tex b/latex/problems/Serway_and_Jewett_8/problem08.08.tex new file mode 100644 index 0000000..7f993ac --- /dev/null +++ b/latex/problems/Serway_and_Jewett_8/problem08.08.tex @@ -0,0 +1,53 @@ +\begin{problem*}{8.8} +Two objects are connected by a light string passing over a light, +frictionless pulley as shown in Figure P8.7. The object of mass $m_1$ +is released from rest at a height $h$ above the table. Using the +isolated system model, \Part{a} determine the speed of $m_2$ just as +$m_1$ hits the table and \Part{b} find the maximum height above the +table to which $m_2$ rises. +% ceiling +% pulley +% | | +% | 1 +% | } h +% 2 } +%--table----- +\end{problem*} + +\begin{solution} +\Part{a} +Conserving energy, we just have to worry about gravitational potential +energy and kinetic energy (there are no springs, friction, external +forces, etc.). While they are attached by a taught string (i.e. while +$m_1$ is falling), the speeds of the two masses must match. Labeling +the release point $a$ and $m_1$ just hitting the floor $b$, we have +\begin{align} + E_a &= E_b \\ + K_{1a} + U_{1a} + K_{2a} + U_{2a} &= K_{1b} + U_{1b} + K_{2b} + U_{2b} \\ + 0 + m_1 gh + 0 + 0 + &= \frac{1}{2} m_1 v_b^2 + 0 + \frac{1}{2} m_2 v_b^2 + m_2 gh \\ + (m_1-m_2) gh &= \frac{1}{2} (m_1+m_2) v_b^2 \\ + v_b^2 &= 2gh\frac{m_1-m_2}{m_1+m_2} \\ + v_b &= \pm\sqrt{2gh\frac{m_1-m_2}{m_1+m_2}} + = \ans{\sqrt{2gh\frac{m_1-m_2}{m_1+m_2}}} \;, +\end{align} +where we dropped the $\pm$ because we only want the magnitude of the +velocity, not its direction. + +\Part{b} +After $m_1$ hits the table, the string goes slack as $m_2$ sails up in +a parabola $h(t)$ and peaks at some point $c$. Conserving energy (now +just for $m_2$, because $m_1$ just sits there on the table being +boring) +\begin{align} + E_{2b} &= E_{2c} \\ + K_{2b} + U_{2b} &= K_{2c} + U_{2c} \\ + \frac{1}{2} m_2 v_b^2 + m_2 gh &= 0 + m_2 g h_c \\ + h_c &= \frac{v_b^2}{2g} + h + = \frac{2gh\frac{m_1-m_2}{m_1+m_2}}{2g} + h + = h\p({\frac{m_1-m_2}{m_1+m_2} + 1}) \\ + &= h\p({\frac{m_1-m_2}{m_1+m_2}+\frac{m_1+m_2}{m_1+m_2}}) + = h\frac{m_1-m_2+m_1+m_2}{m_1+m_2} + = \ans{\frac{2m_1 h}{m_1+m_2}} +\end{align} +\end{solution} diff --git a/latex/problems/Serway_and_Jewett_8/problem08.11.tex b/latex/problems/Serway_and_Jewett_8/problem08.11.tex new file mode 100644 index 0000000..b66eabb --- /dev/null +++ b/latex/problems/Serway_and_Jewett_8/problem08.11.tex @@ -0,0 +1,38 @@ +\begin{problem*}{8.11} +The system shown in Figure P8.11 consists of a light, inextensible +cord, light, frictionless pulleys, and blocks of equal mass. Notice +that block $B$ is attached to one of the pulleys. The system is +initially held at rest so that the blocks are at the same height above +the ground. The blocks are then released. Find the speed of block +$A$ at the moment the vertical separation of the blocks is $h$. +% - pulley -- ceiling +% | | | +% | - pulley - +% | | +% A B +\end{problem*} + +\begin{solution} +Because the tension on both sides of a light pulley must match, block +$B$ has twice the upwards force (from the strings) as block $A$, so it +will rise as block $A$ drops. + +Block $A$ will move twice as fast and far as block $B$ (draw a few +snapshots or build a little model to prove that to yourself if you +need to). Therefore, when the blocks are $h$ appart, block $A$ will +be down $2h/3$ and block $B$ will be up $h/3$. Conserving energy, we +have +\begin{align} + E_i &= E_f \\ + K_{Ai} + U_{Ai} + K_{Bi} + U_{Bi} &= K_{Af} + U_{Af} + K_{Bf} + U_{Bf} \\ + 0 + 0 + 0 + 0 + &= \frac{1}{2} m v_{Af}^2 + mg\frac{-2h}{3} + \frac{1}{2} m \p({\frac{v_{Af}}{2}})^2 + mg\frac{h}{3} \\ + 0 &= v_{Af}^2 \p({\frac{1}{2} + \frac{1}{8}}) - g\frac{h}{3} \\ + v_{Af}^2 \frac{5}{8} &= \frac{gh}{3} \\ + v_{Af} &= \pm\sqrt{\frac{8gh}{15}} + = \ans{\sqrt{\frac{8gh}{15}}} +\end{align} +where we dropped the $\pm$ because we only want the magnitude of the +velocity, not its direction. +\end{solution} diff --git a/latex/problems/Serway_and_Jewett_8/problem08.15.tex b/latex/problems/Serway_and_Jewett_8/problem08.15.tex new file mode 100644 index 0000000..c6dd81e --- /dev/null +++ b/latex/problems/Serway_and_Jewett_8/problem08.15.tex @@ -0,0 +1,56 @@ +\begin{problem*}{8.15} +A block of mass $m=2.00\U{kg}$ is attached to a spring of force +constant $k=500\U{N/m}$ as shown in Figure P8.15. The block is pulled +to a position $x_i=5.00\U{cm}$ to the right of equilibrium and +released from rest. Find the speed the block has as it passes through +equilibrium if \Part{a} the horizontal surface is frictionless +and \Part{b} the coefficient of friction between the block and surface +is $\mu_k=0.350$. +% wall -- spring -- mass +\end{problem*} + +\begin{solution} +\Part{a} +Conserving energy, we just have to worry about kinetic and spring +potential energies (because the surface is horizontal, the +gravitational potential energy does not change). +\begin{align} + E_i &= E_f \\ + K_i + U_i &= K_f + U_f \\ + 0 + \frac{1}{2} k x_i^2 &= \frac{1}{2} m v_f^2 + 0 \\ + v_f^2 &= \frac{kx_i^2}{m} \\ + v_f &= \pm\sqrt{\frac{kx_i^2}{m}} + = \sqrt{\frac{kx_i^2}{m}} \\ + &= \sqrt{\frac{500\U{N/m}\cdot(5.00\E{-2}\U{m})^2}{2.00\U{kg}}} + = \ans{0.791\U{m/s}} +\end{align} +where we dropped the $\pm$ because we only want the magnitude of the +velocity, not its direction. + +\Part{b} +With friction, there is an additional +\begin{equation} + E_\text{int} = \vect{F}\cdot\vect{x} = \mu_k N x_i = \mu_k mg x_i +\end{equation} +going into internal energy (heat), so our conservation energy formula +looks like +\begin{align} + E_i &= E_f \\ + K_i + U_i &= K_f + U_f + E_\text{int} \\ + 0 + \frac{1}{2} k x_i^2 &= \frac{1}{2} m v_f^2 + 0 + \mu_k mg x_i \\ + v_f^2 &= \frac{kx_i^2}{m} - 2\mu_k g x_i\\ + v_f &= \pm\sqrt{\frac{kx_i^2}{m} - 2\mu_k g x_i} + = \sqrt{\frac{kx_i^2}{m} - 2\mu_k g x_i} \\ + &= \sqrt{\frac{500\U{N/m}\cdot(5.00\E{-2}\U{m})^2}{2.00\U{kg}} + - 2\cdot0.350\cdot9.80\U{m/s$^2$}\cdot5.00\E{-2}\U{m}} + = \ans{0.531\U{m/s}} \;. +\end{align} +To calculate the energy lost to friction, we assumed that the block +would slide all the way to the equilibrium position and not grind to a +halt somewhere in the middle. It's ok to assume that though, because +if we had guessed wrong and the block had stopped early, we would have +ended up with an imaginary velocity (from a negative sign inside the +square root). Because we got a positive number inside the square +root, we know the block did, in fact, make it to the equilibrium +position. +\end{solution} diff --git a/latex/problems/Serway_and_Jewett_8/problem08.16.tex b/latex/problems/Serway_and_Jewett_8/problem08.16.tex new file mode 100644 index 0000000..238fd1f --- /dev/null +++ b/latex/problems/Serway_and_Jewett_8/problem08.16.tex @@ -0,0 +1,67 @@ +\begin{problem*}{8.16} +A crate of mass $10.0\U{kg}$ is pulled up a rough incline with an +initial speed of $1.50\U{m/s}$. The pulling force is $100\U{N}$ +parallel to the incline, which makes an angle of $20.0\dg$ with the +horizontal. The coefficient of kinetic friction is $0.400$, and the +crate is pulled $5.00\U{m}$. \Part{a} How much work is done by the +gravitational force on the crate? \Part{b} Determine the increase in +internal energy of the crate-incline system owing to +friction. \Part{c} How much work is done by the $100\U{N}$ force on +the crate? \Part{d} What is the change in kinetic energy of the +crate? \Part{e} What is the speed of the crate after being pulled +$5.00\U{m}$? +\end{problem*} + +\begin{solution} +\Part{a} +The work done by gravity is +\begin{equation} + W_g = \vect{F}_g \cdot \vect{x} = mgx\cos(20.0\dg+90.0\dg) + = \ans{-168\U{J}} +\end{equation} +So the gravitationan potential energy increases by $168\U{J}$. + +You could also calculate the increase in gravitational potential +energy directly +\begin{equation} + \Delta U_g = mgh = mgx\sin(20.0\dg) = 168\U{J} +\end{equation} + +\Part{b} +The force of friction is always directly oposite the motion of the +block, so +\begin{equation} + \Delta E_\text{int} = -W_f = -\mu_k Nx\cos(180\dg) + = \mu_k Nx = \mu_k mg\cos(20.0\dg) = \ans{184\U{J}} +\end{equation} + +\Part{c} +The external force is parallel to the motion, so +\begin{equation} + W_F = Fx\cos(0\dg) = Fx = 500\U{J} +\end{equation} + +\Part{d} +Conserving energy +\begin{align} + K_i + U_{gi} + W_F &= K_f + U_{gf} + \Delta E_\text{int} \\ + \Delta K &= K_f - K_i = W_F - (U_{gf}-U_{gi} - \Delta E_\text{int} \\ + &= W_F - \Delta U_g - \Delta E_\text{int} \\ + &= (500 - 168 - 184)\U{J} + = \ans{148\U{J}} +\end{align} + +\Part{e} +\begin{align} + \Delta K &= K_f - K_i + = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2 + = \frac{m}{2}(v_f^2 - v_i^2) \\ + v_f^2 - v_i^2 &= \frac{2}{m}\Delta K \\ + v_F^2 &= \frac{2}{m}\Delta K + v_i^2 \\ + v_f &= \pm\sqrt{\frac{2}{m}\Delta K + v_i^2} + = \sqrt{\frac{2}{m}\Delta K + v_i^2} + = \ans{5.65\U{m/s}} \;, +\end{align} +where we dropped the $\pm$ because we only want the magnitude of the +velocity, not its direction. +\end{solution} diff --git a/latex/problems/Serway_and_Jewett_8/problem08.22.tex b/latex/problems/Serway_and_Jewett_8/problem08.22.tex new file mode 100644 index 0000000..df0f2a8 --- /dev/null +++ b/latex/problems/Serway_and_Jewett_8/problem08.22.tex @@ -0,0 +1,34 @@ +\begin{problem*}{8.22} +The coefficient of friction between the block of mass $m_1=3.00\U{kg}$ +and the surface in Figure P8.22 is $\mu_s=0.400$. The system starts +from rest. What is the speed of the bal of mass $m_2=5.00\U{kg}$ when +it has fallen a distance $h=1.50\U{m}$? +% m1-block-on-table -- pulley -- hanging m2 +\end{problem*} + +\begin{solution} +Because they are linked by a taut string, the blocks will move at the +same speed. Therefore, $m_1$ drags a distance $h$ across the table, +loosing +\begin{equation} + \Delta E_\text{int} = -W_f = -\mu_k mgh\cos(180\dg) = \mu_k mgh +\end{equation} +to friction. + +Conserving energy +\begin{align} + E_i &= E_f \\ + K_{1i} + U_{1i} + K_{2i} + U{2i} + &= K_{1f} + U_{1f} + K_{2f} + U{2f} + E_\text{int} \\ + 0 + 0 + 0 + 0 + &= \frac{1}{2} m_1 v_f^2 + 0 + + \frac{1}{2} m_2 v_f^2 - m_2 gh + \mu_k m_1 gh \\ + (m_2-\mu_k m_1) gh &= \frac{m_1+m_2}{2} v_f^2 \\ + v_F^2 &= 2gh\frac{m_2-\mu_k m_1}{m_1+m_2} \\ + v_f &= \pm\sqrt{2gh\frac{m_2-\mu_k m_1}{m_1+m_2}} + = \sqrt{2gh\frac{m_2-\mu_k m_1}{m_1+m_2}} + = \ans{3.74\U{m/s}} +\end{align} +where we dropped the $\pm$ because we only want the magnitude of the +velocity, not its direction. +\end{solution} diff --git a/latex/problems/Serway_and_Jewett_8/problem08.28.tex b/latex/problems/Serway_and_Jewett_8/problem08.28.tex new file mode 100644 index 0000000..8b0086b --- /dev/null +++ b/latex/problems/Serway_and_Jewett_8/problem08.28.tex @@ -0,0 +1,22 @@ +\begin{problem*}{8.28} +A certain rain cloud at an altitude of $1.75\U{km}$ contains +$3.20\E{7}\U{kg}$ of water vapor. How long would it take a +$2.70\U{kW}$ pump to raise the same amount of water from the Earth's +surface to the cloud's position? +\end{problem*} + +\begin{solution} +The gravitational potential energy of the cloud is +\begin{equation} + U_g = mgh = 3.20\E{7}\U{kg}\cdot9.80\U{m/s$^2$}\cdot1.75\U{km} + = 549\U{GJ} +\end{equation} + +The pump would take +\begin{align} + P &= \deriv{t}{E} \\ + \Delta t &= \frac{E}{P} = \frac{549\U{GJ}}{2.70\U{kW}} + = \ans{203\U{Ms} = 6.44\U{years}} +\end{align} +to raise the water. +\end{solution} diff --git a/latex/problems/Serway_and_Jewett_8/problem08.35.tex b/latex/problems/Serway_and_Jewett_8/problem08.35.tex new file mode 100644 index 0000000..a3e06a5 --- /dev/null +++ b/latex/problems/Serway_and_Jewett_8/problem08.35.tex @@ -0,0 +1,29 @@ +\begin{problem*}{8.35} +When an automobile moves with a constant speed down a highway, most of +the power developed in the engine is used to compensate for the energy +transformations due to friction forces exerted on the car by the air +and the road. If the power developed by the engine is $175\U{hp}$, +estimate the total friction force acting on the car when it is moving +at a speed of $29\U{m/s}$. One horsepower equals $746\U{W}$. +\end{problem*} + +\begin{solution} +The energy going into friction in a distance $x$ is +\begin{equation} + E_\text{int} = -W_f = -F_f x\cos(180\dg) = F_f x \;. +\end{equation} + +The power going into friction is thus +\begin{equation} + P_f = \deriv{t}{E_\text{int}} = F_f \deriv{t}{x} = F_f v \;. +\end{equation} + +If the power lost to friction matches the power generated by the +engine, then the force of friction is given by +\begin{align} + P_e &= P_f = F_f v \\ + F_f &= \frac{P_e}{v} + = \frac{175\U{hp}\cdot\frac{746\U{W}}{1\U{hp}}}{29\U{m/s}} + = \ans{4.50\U{kN}} \;. +\end{align} +\end{solution} diff --git a/latex/problems/Serway_and_Jewett_8/problem09.09.tex b/latex/problems/Serway_and_Jewett_8/problem09.09.tex new file mode 100644 index 0000000..c36e4b2 --- /dev/null +++ b/latex/problems/Serway_and_Jewett_8/problem09.09.tex @@ -0,0 +1,18 @@ +\begin{problem*}{9.9} +Two blocks of mass $m$ and $3m$ are placed on a frictionless, +horizontal surface. A light spring is attached to the more massive +block, and the blocks are pushed together with the spring between them +(Fig.~P9.9). A cord initially holding the blocks together is burned; +after that happens, the block if mass $3m$ moves to the right with a +speed of $2.00\U{m/s}$. \Part{a} What is the velocity of the block of +mass $m$? \Part{b} Find the system's original elastic potential +energy, taking $m=0.350\U{kg}$. \Part{c} Is the original energy in +the spring or in the cord? \Part{d} Explain your answer +to \Part{c}. \Part{e} Is the momentum of the system conserved in the +bursting-apart process? Explain how that is possible +considering \Part{f} there are large forces acting and \Part{g} there +is no motion beforehand and plenty of motion afterward. +\end{problem*} + +\begin{solution} +\end{solution} diff --git a/latex/problems/Serway_and_Jewett_8/problem09.28.tex b/latex/problems/Serway_and_Jewett_8/problem09.28.tex new file mode 100644 index 0000000..5a6d3ca --- /dev/null +++ b/latex/problems/Serway_and_Jewett_8/problem09.28.tex @@ -0,0 +1,13 @@ +\begin{problem*}{9.28} +Two automobiles of equal mass approach an intersection. One vehicle +is traveling with speed $13.0\U{m/s}$ toward the east, and the other +is traveling north with speed $v_2$. Neither driver sees the other. +The vehicles collide in the intersection and stick together, leaving +parallel skid marks at an angle of $55.0\dg$ north of east. The speed +limit for both roads is $35\U{mi/h}$, and the driver of the +northward-moving vehicle claims he was within the speed limit when the +collision occured. Is he telling the truth? Explain your reasoning. +\end{problem*} + +\begin{solution} +\end{solution} diff --git a/latex/problems/Serway_and_Jewett_8/problem09.55.tex b/latex/problems/Serway_and_Jewett_8/problem09.55.tex new file mode 100644 index 0000000..e1fb00b --- /dev/null +++ b/latex/problems/Serway_and_Jewett_8/problem09.55.tex @@ -0,0 +1,10 @@ +\begin{problem*}{9.55} +A $3.00\U{kg}$ steel ball strikes a wall with a speed of $10.0\U{m/s}$ +at an angle of $\theta=60.0\dg$ with the surface. It bounces off with +the same speed and angle (Fig.~P9.55). If the ball is in contact with +the wall for $0.200\U{s}$, what is the average force exerted by the +wall on the ball? +\end{problem*} + +\begin{solution} +\end{solution} diff --git a/latex/problems/Serway_and_Jewett_8/problem09.56.tex b/latex/problems/Serway_and_Jewett_8/problem09.56.tex new file mode 100644 index 0000000..818e052 --- /dev/null +++ b/latex/problems/Serway_and_Jewett_8/problem09.56.tex @@ -0,0 +1,19 @@ +\begin{problem*}{9.56} +Figure P9.56 shows three points in the operation of the ballistic +pendulum discussed in Example 9.6 (and shown in Fig.~9.9b). The +projectile approaches the pendulum in Figure P9.56a. Figure P9.56b +shows the situation just after the projectile is captured in the +pendulum. In Figure P9.56c, the pendulum arm has swung upward and +come to rest at a height $h$ above its initial position. \Part{a} +Prove that the ratio of kinetic energy of the projectile-pendulum +system immediately after the collision to the kenetic energy +immediately before is $m_1/(m_1+m_2)$. \Part{b} What is the ratio of +the momentum of the system immediately after the collision to the +momentum immediately before? \Part{c} A student believes that such a +large decrease in mechanical energy must be accompanied by at least a +small decrease in momentum. How would you convince this student of +the truth? +\end{problem*} + +\begin{solution} +\end{solution} diff --git a/latex/problems/Serway_and_Jewett_8/problem09.65.tex b/latex/problems/Serway_and_Jewett_8/problem09.65.tex new file mode 100644 index 0000000..e068c5c --- /dev/null +++ b/latex/problems/Serway_and_Jewett_8/problem09.65.tex @@ -0,0 +1,10 @@ +\begin{problem*}{9.65} +A bullet of mass $m$ is fired into a block of mass $M$ initially at +rest at the edge of a frictionless table of height $h$ (Fig.~P9.65). +The bullet remains in the block and after impact the block lands a +distance $d$ from the bottom of the table. Determine the initial +speed of the bullet. +\end{problem*} + +\begin{solution} +\end{solution} diff --git a/latex/problems/Serway_and_Jewett_8/problem09.73.tex b/latex/problems/Serway_and_Jewett_8/problem09.73.tex new file mode 100644 index 0000000..6a4730e --- /dev/null +++ b/latex/problems/Serway_and_Jewett_8/problem09.73.tex @@ -0,0 +1,14 @@ +\begin{problem*}{9.73} +A $5.00\U{g}$ bullet moving with an initial speed $v_i=400\U{ms/s}$ is +fired in to and passes through a $1.00\U{kg}$ block as shown in Figure +P7.73. The block, intially at rest on a frictionless, horizontal +surface, is connected to a spring with force constant $900\U{N/m}$. +The block moves $d=5.00\U{cm}$ to the right after impact before being +brought to rest by the spring. Find \Part{a} the speed at which the +bullet emerges from the block and \Part{b} the amount of the initial +kinetic energy of the bullet that is converted into internal energy in +the bullet-block system during the collision. +\end{problem*} + +\begin{solution} +\end{solution} diff --git a/latex/problems/Serway_and_Jewett_8/problem10.18.tex b/latex/problems/Serway_and_Jewett_8/problem10.18.tex new file mode 100644 index 0000000..70abc97 --- /dev/null +++ b/latex/problems/Serway_and_Jewett_8/problem10.18.tex @@ -0,0 +1,41 @@ +\begin{problem*}{10.18} +A car accelerates uniformly from rest and reaches a speed of +$22.0\U{m/s}$ in $9.00\U{s}$. Assuming the diameter of a tire is +$58.0\U{cm}$, \Part{a} find the number of revolutions the tire makes +during this motion, assuming that no slipping occurs. \Part{b} What +is the final angular speed of a tire in revolutions per second? +\end{problem*} + +\begin{solution} +\Part{a} +From our uniform acceleration work, we know $v = at + v_0$, so +\begin{align} + v - v_0 &= at \\ + a &= \frac{v - v_0}{t} +\end{align} +The distance traveled during this time is +\begin{equation} + \Delta x = x - x_0 + = \frac{1}{2} a t^2 + v_0 t + = \frac{v-v_0}{2} t + v_0 t + = \frac{v+v_0}{2} t + = \frac{vt}{2} +\end{equation} +where we use the fact that the car starts from rest ($v_0=0$) in the +last step. + +With each rotation the tire covers a distance of $2\pi r = \pi d$, so +the total number of rotations is +\begin{equation} + \text{rot} = \Delta x \frac{1\U{rotation}}{\pi d} + = \frac{vt}{2\pi r}\U{rotations} + = \ans{54.3\U{rotations}} +\end{equation} + +\Part{b} +Taking the time derivative of $\dd x=r\dd\theta$, we have +\begin{align} + v = \deriv{t}{x} &= r \deriv{t}{\theta} = r\omega \\ + \omega &= \frac{v}{r} = = \frac{2v}{d} = 75.9\U{rad/s} = \ans{12.1\U{rev/s}} +\end{align} +\end{solution} diff --git a/latex/problems/Serway_and_Jewett_8/problem10.35.tex b/latex/problems/Serway_and_Jewett_8/problem10.35.tex new file mode 100644 index 0000000..7993788 --- /dev/null +++ b/latex/problems/Serway_and_Jewett_8/problem10.35.tex @@ -0,0 +1,19 @@ +\begin{problem*}{10.35} +Find the net torque on the wheel in Figure P10.35 about the axle +through $O$, taking $a=10.0\U{cm}$ and $b=25.0\U{cm}$. +% 10.0N at a point b N of O, pulling E +% 9.00N at a point b E of O, pulling S +% 12.0N at a point a ~NW of O, pulling tangentially 30dg S of W. +\end{problem*} + +\begin{solution} +Taking the positive direction to be counter clockwise and summing the +torque $\tau = \vect{r}\times\vect{F}$: +\begin{equation} + \sum \tau = 12.0\U{N}\cdot a - 10.0\U{N}\cdot b - 9.00\U{N}\cdot b + = -3.55\U{J} +\end{equation} +We don't care about the $30\dg$ angle (where the force is applied). +What matters is the distance from the axle to the force ($a$) and the +angle between the radius and force ($90\dg$). +\end{solution} diff --git a/latex/problems/Serway_and_Jewett_8/problem10.44.tex b/latex/problems/Serway_and_Jewett_8/problem10.44.tex new file mode 100644 index 0000000..42170e6 --- /dev/null +++ b/latex/problems/Serway_and_Jewett_8/problem10.44.tex @@ -0,0 +1,54 @@ +\begin{problem*}{10.44} +Consider the system shown in Figure P10.44 with $m_1=20.0\U{kg}$, +$m_2=12.5\U{kg}$, $R=0.200\U{m}$, and the mass of the pulley +$M=5.00\U{kg}$. Object $m_2$ is resting on the floor, and object +$m_1$ is $4.00\U{m}$ above the floor when it is released from rest. +The pulley axis is frictionless. The cord is light, does not stretch, +and does not slip on the pulley. \Part{a} Calculate the time interval +required for $m_1$ to hit the floot. \Part{b} How would your answer +change if the pulley were massless? +\end{problem*} + +\begin{solution} +\Part{a} +Let the positive direction to be the direction of motion (down for +$m_1$, up for $m_2$, and counter clockwise for the pulley). Balancing +forces and torques on each object we have +\begin{align} + \sum F_1 &= m_1 g - T_1 = m_1 a \\ + \sum F_2 &= T_2 - m_2 g = m_2 a \\ + \sum \tau &= T_1 R - T_2 R = I\alpha + = \p({\frac{1}{2}MR^2})\cdot\frac{a}{R} + = \frac{MRa}{2} \;, +\end{align} +where we used the moment of interia for a solid cylinder +$I=\frac{1}{2}MR^2$ and the relationship between linear and angular +acceleration $a = r\alpha$. Now we have three equations for our three +unknowns ($a$, $T_1$, and $T_2$) and we can solve for acceleration. +\begin{align} + T_1 &= m_1 (g-a) \\ + T_2 &= m_2 (g+a) \\ + T_1 - T_2 &= \frac{Ma}{2} \\ + m_1 g - m_1 a - m_2 g - m_2 a &= \frac{Ma}{2} \\ + (m_1 - m_2)g - (m_1 + m_2)a &= \frac{M}{2}a \\ + (m_1 - m_2)g &= a \p({\frac{M}{2} + m_1 + m_2}) \\ + a &= g\frac{m_1 - m_2}{\frac{M}{2} + m_1 + m_2} \;. +\end{align} + +The time-to-floor is the a constant acceleration problem. +\begin{align} + h &= \frac{1}{2} a t^2 \\ + t^2 &= \frac{2h}{a} \\ + t &= \sqrt{\frac{2h}{a}} + = \sqrt{\frac{2h}{g}\frac{\frac{M}{2} + m_1 + m_2}{m_1 - m_2}} + = \ans{1.95\U{s}} \;. +\end{align} + +\Part{b} +If the mass of the pulley was zero, +\begin{equation} + t = \sqrt{\frac{2h}{g}\frac{m_1 + m_2}{m_1 - m_2}} + = \ans{1.88\U{s}} \;, +\end{equation} +which is slightly faster than the original time. +\end{solution} diff --git a/latex/problems/Serway_and_Jewett_8/problem10.51.tex b/latex/problems/Serway_and_Jewett_8/problem10.51.tex new file mode 100644 index 0000000..ae5dde8 --- /dev/null +++ b/latex/problems/Serway_and_Jewett_8/problem10.51.tex @@ -0,0 +1,58 @@ +\begin{problem*}{10.51} +An object with a mass $m=5.10\U{kg}$ is attached to the free end of a +light string wrapped around a reel of radius $R=0.250\U{m}$ and mass +$M=3.00\U{kg}$. The reel is a solid disk, free to rotate in a +vertical plane about the horiizontal axis passing through its center +as shown in Figure P10.51. The suspended object is released from rest +$6.00\U{m}$ above the floor. Determine \Part{a} the tension in the +string, \Part{b} the acceleration of the object, and \Part{c} the +speed with which the object hits the floor. \Part{d} Verify your +answer to \Part{c} by using the isolated system (energy) model. +\end{problem*} + +\begin{solution} +\Part{a} +Summing the forces and tourques, +\begin{align} + \sum F &= mg - T = ma \\ + \sum \tau &= RT = I\alpha + = \p({\frac{1}{2}MR^2})\cdot\frac{a}{R} + = \frac{MRa}{2} \;. +\end{align} +Solving for tension +\begin{align} + a &= g - \frac{T}{m} \\ + RT &= \frac{MRa}{2} \\ + 2T &= Ma \\ + 2T &= Mg - \frac{M}{m}T \\ + T \p({2 + \frac{M}{m}}) &= Mg \\ + T &= \frac{Mg}{2 + \frac{M}{m}} = \ans{11.4\U{N}} \;. +\end{align} + +\Part{b} +Plugging back in for $a$, +\begin{equation} + a = g - \frac{T}{m} = 9.80\U{m/s$^2$} - \frac{11.4\U{N}}{5.10\U{kg}} + = \ans{7.57\U{m/s$^2$}} +\end{equation} + +\Part{c} +Finding the floor-hitting speed is a constant acceleration problem +\begin{align} + v^2 &= v_0^2 + 2a\Delta x \\ + v &= \sqrt{2a\Delta x} = \ans{9.53\U{m/s}} +\end{align} + +\Part{d} +Conserving energy, the initial gravitational energy is converted into +linear and angular kinetic energies. +\begin{align} + E_i = mgh &= E_f = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 + = \frac{1}{2}v^2 + \frac{1}{2} \p({\frac{1}{2}MR^2}) \p({\frac{v}{R}})^2 + = \frac{m}{2}mv^2 + \frac{M}{4} v^2 + = \frac{2m + M}{4}v^2 \\ + v^2 &= \frac{4mgh}{2m+M} \\ + v &= \sqrt{\frac{4mgh}{2m+M}} = \ans{9.53\U{m/s$^2$}} \;, +\end{align} +which is the same speed we got in \Part{c}. +\end{solution} diff --git a/latex/problems/Serway_and_Jewett_8/problem10.55.tex b/latex/problems/Serway_and_Jewett_8/problem10.55.tex new file mode 100644 index 0000000..b9ad38c --- /dev/null +++ b/latex/problems/Serway_and_Jewett_8/problem10.55.tex @@ -0,0 +1,27 @@ +\begin{problem*}{10.55} +A cylinder of mass $10.0\U{kg}$ rolls without slipping on a horizontal +surface. At a certain instant, its center of mass has a speed of +$10.0\U{m/s}$. Determine \Part{a} the translational kinetic energy of +its center of mass, \Part{b} the rotational kinetic energy about its +center of mass, and \Part{c} its total energy. +\end{problem*} + +\begin{solution} +\Part{a} +\begin{equation} + K_t = \frac{1}{2}mv^2 = \ans{500\U{J}} +\end{equation} + +\Part{b} +\begin{equation} + K_r = \frac{1}{2}I\omega^2 + = \frac{1}{2}\p({\frac{1}{2}MR^2})\p({\frac{v}{R}})^2 + = \frac{1}{4}Mv^2 + = \ans{250\U{J}} +\end{equation} + +\Part{c} +\begin{equation} + E = K_t + K_r = \ans{750\U{J}} +\end{equation} +\end{solution} diff --git a/latex/problems/Serway_and_Jewett_8/problem10.59.tex b/latex/problems/Serway_and_Jewett_8/problem10.59.tex new file mode 100644 index 0000000..ce7e0f3 --- /dev/null +++ b/latex/problems/Serway_and_Jewett_8/problem10.59.tex @@ -0,0 +1,43 @@ +\begin{problem*}{10.59} +A uniform solid disk and a uniform hoop are placed side by side at the +top of an incline of height $h$. \Part{a} If they are released from +rest and roll without slipping, which object reaches the bottom +first? \Part{b} Verify your answer by calculating their speeds when +they reach the bottom in terms of $h$. +\end{problem*} + +\begin{solution} +\Part{a} +The disk reaches the bottom first, because it has a lower moment of +inertia, so there is less energy wasted on rotational kinetic energy +and more left over for translational kinetic energy. + +\Part{b} +Conserving energy +\begin{align} + E_i = mgh &= E_f = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 + = \frac{1}{2}mv^2 + \frac{1}{2}I\p({\frac{v}{r}})^2 + = \frac{1}{2}\p({m+\frac{I}{r^2}})v^2 \\ + v^2 &= \frac{2mgh}{m+\frac{I}{r^2}} \\ + v &= \sqrt{\frac{2mgh}{m+\frac{I}{r^2}}} \;. +\end{align} + +For a disk, $I=\frac{1}{2}mr^2$, so +\begin{equation} + v = \sqrt{\frac{2mgh}{m+\frac{1}{2}m}} + = \sqrt{\frac{2gh}{\frac{3}{2}}} + = \ans{\sqrt{\frac{4gh}{3}}} \;. +\end{equation} + +For a hoop, $I=mr^2$, so +\begin{equation} + v = \sqrt{\frac{2mgh}{m+m}} + = \sqrt{\frac{2gh}{2}} + = \ans{\sqrt{gh}} \;. +\end{equation} + +As we claimed in \Part{a}, +\begin{equation} + v_d = \sqrt{\frac{4gh}{3}} > \sqrt{gh} = v_h \;. +\end{equation} +\end{solution} diff --git a/latex/problems/Serway_and_Jewett_8/problem11.03.tex b/latex/problems/Serway_and_Jewett_8/problem11.03.tex new file mode 100644 index 0000000..dda182e --- /dev/null +++ b/latex/problems/Serway_and_Jewett_8/problem11.03.tex @@ -0,0 +1,30 @@ +\begin{problem*}{11.3} +Two vectors are given by $\vect{A}=\ihat+2\jhat$ and +$\vect{B}=-2\ihat+3\jhat$. Find \Part{a} $\vect{A}\times\vect{B}$ +and \Part{b} the angle between $\vect{A}$ and $\vect{B}$. +\end{problem*} + +\begin{solution} +\Part{a} +Using the component form for the cross product +\begin{align} + \vect{A}\times\vect{B} + = \p|{\begin{matrix} + \ihat & \jhat & \khat \\ + A_x & A_y & A_z \\ + B_x & B_y & B_z + \end{matrix}}| + = \ihat(A_yB_z - A_zBy) - \jhat(A_xB_z-A_zB_x) + \khat(A_xB_y-A_yB_x) + = (1\cdot3-(-2)\cdot2)\khat + = 7\khat +\end{align} + +\Part{b} +Using the $\sin$ form for the cross product +\begin{align} + |\vect{A}\times\vect{B}| &= |\vect{A}|\cdot|\vect{B}|\cdot\sin\theta \\ + \theta &= \arcsin\p({\frac{|\vect{A}\times\vect{B}|}{AB}}) + = \arcsin\p({\frac{7}{\sqrt{1^2+2^2}\cdot\sqrt{2^2+3^2}}}) + = \ans{60.3\dg} +\end{align} +\end{solution} diff --git a/latex/problems/Serway_and_Jewett_8/problem11.05.tex b/latex/problems/Serway_and_Jewett_8/problem11.05.tex new file mode 100644 index 0000000..b0b3ed3 --- /dev/null +++ b/latex/problems/Serway_and_Jewett_8/problem11.05.tex @@ -0,0 +1,26 @@ +\begin{problem*}{11.5} +Caclulate the net torque (magnitude and direction) on the beam in +Figure~P11.5 about \Part{a} an axis through $O$ perpendicular to the +page and \Part{b} an axis through $C$ perpendicular to the page. +\end{problem*} + +\begin{solution} +\Part{a} +\begin{equation} + \sum \tau = 0\U{m}\cdot30\U{N} + + 2.0\U{m}\cdot25\U{N}\sin(90\dg-30\dg) + + 4.0\U{m}\cdot10\U{N}\sin(-20\dg) + = \ans{29.6\U{J}} \;. +\end{equation} +Because we used the usual right handed cross product, the positive +torque will cause counter clockwise rotation. + +\Part{b} +\begin{equation} + \sum \tau = 2\U{m}\cdot30\U{N}\sin(45\dg) + + 0\U{m}\cdot25\U{N} + + 2.0\U{m}\cdot10\U{N}\sin(-20\dg) + = \ans{35.6\U{J}} \;. +\end{equation} +As in \Part{a}, positive torques are counter clockwise. +\end{solution} diff --git a/latex/problems/Serway_and_Jewett_8/problem11.16.tex b/latex/problems/Serway_and_Jewett_8/problem11.16.tex new file mode 100644 index 0000000..30e01c6 --- /dev/null +++ b/latex/problems/Serway_and_Jewett_8/problem11.16.tex @@ -0,0 +1,40 @@ +\begin{problem*}{11.16} +A conical pendulum consists of a bob of mass $m$ in motion in a +circular path in a horizontal plane as shown in Figure~P11.16. During +the motion, the supporing wire of length $l$ maintains a constant +angle $\theta$ with the vertical. Show that the magnitude of the +angular momentum of the bob about the vertical dashed line is +\begin{equation} + L = \p({\frac{m^2gl^3\sin^4\theta}{\cos\theta}})^{\frac{1}{2}} +\end{equation} +\end{problem*} + +\begin{solution} +Digging back to Problem~6.8, we have the radial acceleration of a +conical pendulum bob +\begin{equation} + a_r = -g\tan(\theta) \;. +\end{equation} +This gives a tangential speed of +\begin{align} + a_r &= -\frac{v^2}{r} \\ + v^2 &= -a_r r \\ + v &= \sqrt{-a_r r} +\end{align} +The angular speed is then +\begin{equation} + \omega = \frac{v}{r} = \frac{\sqrt{-a_r r}}{r} = \sqrt{\frac{-a_r}{r}} \;. +\end{equation} +And the angular momentum is +\begin{align} + L &= I\omega + = mr^2 \cdot \sqrt{\frac{-a_r}{r}} + = \sqrt{\frac{-m^2r^4a_r}{r}} + = \sqrt{-m^2r^3a_r} \\ + &= \sqrt{m^2[l\sin(\theta)]^3 [g\tan(\theta)]} + = \sqrt{m^2l^3\sin^3(\theta)g\tan(\theta)} + = \sqrt{\frac{m^2gl^3\sin^3(\theta)\sin(theta)}{\cos(\theta)}} + = \sqrt{\frac{m^2gl^3\sin^4(\theta)}{\cos(\theta)}} \;, +\end{align} +which is what we set out to show. +\end{solution} diff --git a/latex/problems/Serway_and_Jewett_8/problem11.30.tex b/latex/problems/Serway_and_Jewett_8/problem11.30.tex new file mode 100644 index 0000000..62f3e3e --- /dev/null +++ b/latex/problems/Serway_and_Jewett_8/problem11.30.tex @@ -0,0 +1,26 @@ +\begin{problem*}{11.30} +A disk with moment of inertia $I_1$ rotates about a frictionless, +vertical axle with angular speed $\omega$. A second disk, this one +having a moment of intertia $I_2$ and intitially not rotating, drops +onto the first disk (Fig.~P11.30). Because of friction between the +surfaces, the two eventually reach the same angular speed +$\omega_f$. \Part{a} Calculate $\omega_f$. \Part{b} Calculate the +ratio of the final to the initial rotational energy. +\end{problem*} + +\begin{solution} +\Part{a} +Conserving angular momentum +\begin{align} + L_i = I_1\omega &= L_f = (I_1+I_2)\omega_f \\ + \omega_f &= \ans{\frac{I_1}{I_1+I_2} \omega} \;. +\end{align} + +\Part{b} +\begin{equation} + \frac{K_f}{K_i} + = \frac{\frac{1}{2}(I_1+I_2)\omega_F^2}{\frac{1}{2}I_1\omega^2} + = \frac{(I_1+I_2)\p({\frac{I_1}{I_1+I_2} \omega})^2}{I_1\omega^2} + = \ans{\frac{I_1}{I_1+I_2}} \;. +\end{equation} +\end{solution} diff --git a/latex/problems/Serway_and_Jewett_8/problem11.39.tex b/latex/problems/Serway_and_Jewett_8/problem11.39.tex new file mode 100644 index 0000000..b358497 --- /dev/null +++ b/latex/problems/Serway_and_Jewett_8/problem11.39.tex @@ -0,0 +1,41 @@ +\begin{problem*}{11.39} +A wad of sticky clay with mass $m$ and velocity $\vect{v}_i$ is fired +at a solid cylinder of mass $M$ and radius $R$ (Fig.~P11.39). The +cylinder is initially at rest and is mounted on a fixed horizontal +axle that runs through its center of mass. The line of motion of the +projectile is perpendiuclar to the axle and at a distance $d