--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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}
--- /dev/null
+\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 $d<R$ from
+the center. \Part{a} Find the angular speed of the system just after
+the clay strikes and sticks to the surface of the cylinder. \Part{b}
+Is the mechanical energy of the clay--cylinder system constant in this
+process? Explain your answer. \Part{c} Is the momentum of the
+clay--cylinder system constant in this process? Explain your answer.
+\end{problem*}
+
+\begin{solution}
+\Part{a}
+Conserving angular momentum
+\begin{align}
+ L_i = (md^2)\cdot\p({\frac{v_i}{d}})
+ &= L_f = \p({mR^2+\frac{1}{2}MR^2})\cdot\omega \\
+ \omega &= \frac{mdv_i}{mR^2 + \frac{1}{2}MR^2}
+ = \ans{\frac{2mv_id}{(2m+M)R^2}} \;.
+\end{align}
+
+\Part{b}
+Mechanical energy is not constant, because some kinetic energy is lost
+to internal energy during the collision.
+\begin{equation}
+ \frac{K_f}{K_i}
+ = \frac{\frac{1}{2}(mR^2+\frac{1}{2}MR^2)\omega^2}{\frac{1}{2}mv_i^2}
+ = \frac{(2m_M)R^2\p({\frac{2mv_id}{(2m+M)R^2}})^2}{2mv_i^2}
+ = \frac{2md^2}{(2m+M)R^2} \;.
+\end{equation}
+
+\Part{c}
+\ans{Angular momentum is conserved} because there are no external
+torques (we used this in \Part{a}). \ans{Linear momentum is not
+conserved} because the axle exerts a backwards force on the cylinder
+as the clay strikes. The axle cannot exert any net vertical force
+though, because the vertical component of the clay--cylinder momentum
+is zero both before and after the collision.
+\end{solution}
projects / course.git / commitdiff