diff --git a/concepts/em/u11/e11-5-kirchhoff.tex b/concepts/em/u11/e11-5-kirchhoff.tex index 42082d7..85861b7 100644 --- a/concepts/em/u11/e11-5-kirchhoff.tex +++ b/concepts/em/u11/e11-5-kirchhoff.tex @@ -138,7 +138,6 @@ Substitute equation (1) into equation (2): \[ 7.0\,I_1 + 3.0\,I_2 = 12. \] -\label{eq:a} Substitute equation (1) into equation (3): \[ @@ -158,9 +157,8 @@ so \[ I_1 = 2.0 - 3.0\,I_2. \] -\label{eq:b} -Substitute equation (\ref{eq:b}) into equation (\ref{eq:a}): +Substitute equation (3) into equation (2): \[ 7.0\,(2.0 - 3.0\,I_2) + 3.0\,I_2 = 12, \] @@ -177,7 +175,7 @@ Substitute equation (\ref{eq:b}) into equation (\ref{eq:a}): I_2 = \frac{2.0}{18.0} = \frac{1}{9}\,\mathrm{A}. \] -From equation (\ref{eq:b}): +From equation (3): \[ I_1 = 2.0 - 3.0\left(\frac{1}{9}\right) = 2.0 - \frac{1}{3} = \frac{6}{3} - \frac{1}{3} = \frac{5}{3}\,\mathrm{A}. \] diff --git a/concepts/em/u12/e12-3-force-on-current.tex b/concepts/em/u12/e12-3-force-on-current.tex index ca941be..7a894ee 100644 --- a/concepts/em/u12/e12-3-force-on-current.tex +++ b/concepts/em/u12/e12-3-force-on-current.tex @@ -111,9 +111,10 @@ U = -\mu B\,\cos\phi = -\vec{\mu}\cdot\vec{B}. \nt{The SI unit of magnetic dipole moment $\vec{\mu}$ is ampere-square meter, $\mathrm{A\cdot m^2}$. This is equivalent to joules per tesla, $\mathrm{J/T}$.} -\ex{Illustrative example}{A coil with $N=50$ turns, each of area $A = 8.0\,\mathrm{cm^2} = 8.0\times 10^{-4}\,\mathrm{m^2}$, carries current $I = 2.0\,\mathrm{A}$ in a field $B = 0.40\,\mathrm{T}$. The maximum torque occurs at $\phi = 90^\circ$: +\ex{Illustrative example}{A coil with $N=50$ turns, each of area $A = 8.0\,\mathrm{cm^2} = 8.0\times 10^{-4}\,\mathrm{m^2}$, carries current $I = 2.0\,\mathrm{A}$ in a field $B = 0.40\,\mathrm{T}$. The maximum torque occurs at $\phi = 90^\circ$:} \[ -\tau_{\text{max}} = N\,I\,A\,B = (50)(2.0\,\mathrm{A})(8.0\times 10^{-4}\,\mathrm{m^2})(0.40\,\mathrm{T}) = 3.2\times 10^{-2}\,\mathrm{N\cdot m}.} +\tau_{\text{max}} = N\,I\,A\,B = (50)(2.0\,\mathrm{A})(8.0\times 10^{-4}\,\mathrm{m^2})(0.40\,\mathrm{T}) = 3.2\times 10^{-2}\,\mathrm{N\cdot m}. +\] \qs{Worked example}{A rectangular wire loop has width $a=0.10\,\mathrm{m}$ (horizontal side) and height $b=0.050\,\mathrm{m}$ (vertical side). The loop has $N=100$ turns and carries current $I=2.0\,\mathrm{A}$ in the direction shown: clockwise when viewed from the front (from the $+x$-axis toward the $yz$-plane). The loop sits in a uniform magnetic field $\vec{B} = (0.60\,\mathrm{T})\,\hat{\imath}$ pointing to the right. The normal vector $\hat{n}$ points along the $-\hat{k}$ direction (into the page) by the right-hand rule for clockwise current. diff --git a/concepts/mechanics/u6/m6-4-rolling.tex b/concepts/mechanics/u6/m6-4-rolling.tex index 380b867..8d194e3 100644 --- a/concepts/mechanics/u6/m6-4-rolling.tex +++ b/concepts/mechanics/u6/m6-4-rolling.tex @@ -32,48 +32,13 @@ Thus the wheel's center moves at $2.4\,\mathrm{m/s}$, while the top point moves \mprop{Rolling kinematics, energy, and incline dynamics}{Consider a rigid body of mass $M$, radius $R$, and moment of inertia $I_{\mathrm{cm}}$ about its center of mass. Let $s_{\mathrm{cm}}$, $v_{\mathrm{cm}}$, and $a_{t,\mathrm{cm}}$ denote the center-of-mass distance, speed, and tangential acceleration, and let $\theta$, $\omega$, and $\alpha$ denote the corresponding angular variables. Let $K_i$ and $U_i$ denote the initial kinetic and potential energies, and let $K_f$ and $U_f$ denote the final kinetic and potential energies. -\begin{enumerate}[label=\textbf{\arabic*.}] -\item For pure rolling, -\[ -s_{\mathrm{cm}}=R\theta, -\qquad -v_{\mathrm{cm}}=R\omega, -\qquad -a_{t,\mathrm{cm}}=R\alpha. -\] +1. For pure rolling, $s_{\mathrm{cm}}=R\theta$, $v_{\mathrm{cm}}=R\omega$, $a_{t,\mathrm{cm}}=R\alpha$. -\item The total kinetic energy is the sum of translational and rotational parts: -\[ -K=\tfrac12 Mv_{\mathrm{cm}}^2+\tfrac12 I_{\mathrm{cm}}\omega^2. -\] -Using $v_{\mathrm{cm}}=R\omega$, this may also be written as -\[ -K=\tfrac12\left(M+\frac{I_{\mathrm{cm}}}{R^2}\right)v_{\mathrm{cm}}^2. -\] +2. The total kinetic energy is the sum of translational and rotational parts: $K=\tfrac12 Mv_{\mathrm{cm}}^2+\tfrac12 I_{\mathrm{cm}}\omega^2$. Using $v_{\mathrm{cm}}=R\omega$, this may also be written as $K=\tfrac12\left(M+\frac{I_{\mathrm{cm}}}{R^2}\right)v_{\mathrm{cm}}^2$. -\item If no nonconservative force removes mechanical energy from the system, then for rolling without slipping, -\[ -K_i+U_i=K_f+U_f. -\] -For a vertical drop of magnitude $h$ from rest, -\[ -Mgh=\tfrac12 Mv_{\mathrm{cm}}^2+\tfrac12 I_{\mathrm{cm}}\omega^2. -\] +3. If no nonconservative force removes mechanical energy from the system, then for rolling without slipping, $K_i+U_i=K_f+U_f$. For a vertical drop of magnitude $h$ from rest, $Mgh=\tfrac12 Mv_{\mathrm{cm}}^2+\tfrac12 I_{\mathrm{cm}}\omega^2$. -\item For a body rolling without slipping down an incline of angle $\beta$, choose positive down the incline. Let $a_{\mathrm{cm}}$ denote the center-of-mass acceleration magnitude along the incline. If $f_s$ denotes the magnitude of the static friction force, then -\[ -Mg\sin\beta-f_s=Ma_{\mathrm{cm}}, -\qquad -f_sR=I_{\mathrm{cm}}\alpha, -\qquad -a_{\mathrm{cm}}=R\alpha. -\] -Therefore, -\[ -a_{\mathrm{cm}}=\frac{g\sin\beta}{1+I_{\mathrm{cm}}/(MR^2)}, -\qquad -f_s=\frac{I_{\mathrm{cm}}}{R^2}a_{\mathrm{cm}}. -\] +4. For a body rolling without slipping down an incline of angle $\beta$, choose positive down the incline. Let $a_{\mathrm{cm}}$ denote the center-of-mass acceleration magnitude along the incline. If $f_s$ denotes the magnitude of the static friction force, then $Mg\sin\beta-f_s=Ma_{\mathrm{cm}}$, $f_sR=I_{\mathrm{cm}}\alpha$, $a_{\mathrm{cm}}=R\alpha$. Therefore, $a_{\mathrm{cm}}=\frac{g\sin\beta}{1+I_{\mathrm{cm}}/(MR^2)}$, $f_s=\frac{I_{\mathrm{cm}}}{R^2}a_{\mathrm{cm}}$. For an object accelerating down the incline, the static friction force on the object points up the incline.} \qs{Worked example}{A solid cylinder of mass $M=2.0\,\mathrm{kg}$ and radius $R=0.20\,\mathrm{m}$ is released from rest on an incline that makes an angle $\beta=30^\circ$ with the horizontal. The cylinder rolls without slipping through a vertical drop $h=0.75\,\mathrm{m}$. Let $g=9.8\,\mathrm{m/s^2}$.