checkpoint 1

concepts/mechanics/u6/m6-1-rotational-energy.tex

@@ -0,0 +1,152 @@
+\subsection{Rotational Kinetic Energy and Work by Torque}
+
+This subsection introduces the energy description of fixed-axis rotation. In AP mechanics, the key idea is that a net torque acting through an angular displacement changes a rigid body's rotational kinetic energy in the same way that a net force acting through a displacement changes translational kinetic energy.
+
+\dfn{Rotational kinetic energy and incremental work by torque}{Consider a rigid body rotating about a fixed axis with unit vector $\hat{k}$. Let $I$ denote the body's moment of inertia about that axis, let $\vec{\omega}=\omega\hat{k}$ denote its angular velocity, and let $\omega=|\vec{\omega}|$ denote the angular speed. The \emph{rotational kinetic energy} of the rigid body is
+\[
+K_{\mathrm{rot}}=\tfrac12 I\omega^2.
+\]
+
+Now let the body undergo an infinitesimal angular displacement $d\theta$ about the same axis, with the positive direction chosen consistently with $\hat{k}$. Let $\vec{\tau}_{\mathrm{net}}=\tau_{\mathrm{net}}\hat{k}$ denote the net external torque about that axis. Then the incremental work done by the net torque is
+\[
+dW=\tau_{\mathrm{net}}\,d\theta.
+\]
+If $\tau_{\mathrm{net}}$ is constant over a finite angular displacement $\Delta\theta$, then
+\[
+W=\tau_{\mathrm{net}}\Delta\theta.
+\]
+The SI unit of both rotational kinetic energy and work is the joule, where $1\,\mathrm{J}=1\,\mathrm{N\cdot m}$.}
+
+\thm{Rotational work-energy relation for fixed-axis motion}{Consider a rigid body rotating about a fixed axis with moment of inertia $I$. Let $\omega_i$ and $\omega_f$ denote the initial and final angular speeds, let $\theta_i$ and $\theta_f$ denote the corresponding angular positions, and let $\tau_{\mathrm{net}}$ denote the signed net external torque about the axis. Then the net rotational work from the initial state to the final state is
+\[
+W_{\mathrm{net}}=\int_{\theta_i}^{\theta_f} \tau_{\mathrm{net}}\,d\theta,
+\]
+and the rotational work-energy theorem is
+\[
+W_{\mathrm{net}}=\Delta K_{\mathrm{rot}}=K_{\mathrm{rot},f}-K_{\mathrm{rot},i}
+=\tfrac12 I\omega_f^2-\tfrac12 I\omega_i^2.
+\]
+Thus a positive net torque doing positive work increases rotational kinetic energy, while negative net work decreases it.}
+
+\nt{This result is the rotational analog of the translational work-energy theorem $W_{\mathrm{net}}=\Delta K$ with $K=\tfrac12 mv^2$. The correspondence is
+\[
+\vec{F}_{\mathrm{net}}\leftrightarrow \vec{\tau}_{\mathrm{net}},
+\qquad
+x\leftrightarrow \theta,
+\qquad
+v\leftrightarrow \omega,
+\qquad
+\tfrac12 mv^2\leftrightarrow \tfrac12 I\omega^2.
+\]
+In this subsection, assume a rigid body rotating about one fixed axis so that every part of the body shares the same angular displacement $\theta$ and angular speed $\omega$, and so that $I$ stays constant about that axis. The sign convention for $\tau_{\mathrm{net}}$ must match the sign convention for $\theta$.}
+
+\pf{Why $K_{\mathrm{rot}}=\tfrac12 I\omega^2$ and $W_{\mathrm{net}}=\Delta K_{\mathrm{rot}}$}{Model the rigid body as particles labeled by an index $i$. Let particle $i$ have mass $m_i$ and perpendicular distance $r_{\perp,i}$ from the fixed axis. Because the body is rigid, each particle has speed
+\[
+v_i=r_{\perp,i}\omega.
+\]
+Therefore the total kinetic energy is
+\[
+K_{\mathrm{rot}}=\sum_i \tfrac12 m_i v_i^2
+=\sum_i \tfrac12 m_i(r_{\perp,i}\omega)^2
+=\tfrac12 \omega^2\sum_i m_i r_{\perp,i}^2.
+\]
+Since
+\[
+I=\sum_i m_i r_{\perp,i}^2,
+\]
+it follows that
+\[
+K_{\mathrm{rot}}=\tfrac12 I\omega^2.
+\]
+
+For the work-energy relation, start with the fixed-axis rotational form of Newton's second law,
+\[
+\tau_{\mathrm{net}}=I\alpha,
+\]
+where $\alpha=d\omega/dt$. Multiply both sides by $d\theta$:
+\[
+\tau_{\mathrm{net}}\,d\theta=I\alpha\,d\theta.
+\]
+Using $\omega=d\theta/dt$, we have $d\theta=\omega\,dt$, so
+\[
+I\alpha\,d\theta=I\frac{d\omega}{dt}(\omega\,dt)=I\omega\,d\omega.
+\]
+Thus
+\[
+dW=\tau_{\mathrm{net}}\,d\theta=I\omega\,d\omega=d\!\left(\tfrac12 I\omega^2\right)=dK_{\mathrm{rot}}.
+\]
+Integrating from the initial state to the final state gives
+\[
+W_{\mathrm{net}}=\Delta K_{\mathrm{rot}}.
+\]}
+
+\qs{Worked example}{A wheel of radius $R=0.25\,\mathrm{m}$ rotates about a frictionless fixed axle. Its moment of inertia about the axle is $I=1.0\,\mathrm{kg\cdot m^2}$. A light string is wrapped around the rim, and a student pulls tangentially on the string with constant force magnitude $F=12\,\mathrm{N}$. The string does not slip, and a length $s=2.0\,\mathrm{m}$ of string unwinds. Initially the wheel is already spinning in the same direction as the applied torque with angular speed $\omega_i=4.0\,\mathrm{rad/s}$.
+
+Find:
+\begin{enumerate}[label=(\alph*)]
+\item the torque magnitude $\tau$ applied by the string,
+\item the angular displacement $\Delta\theta$ during the pull,
+\item the work done on the wheel by the torque, and
+\item the final angular speed $\omega_f$.
+\end{enumerate}}
+
+\sol Because the pull is tangential to the rim, the lever arm equals the radius $R$. Therefore the applied torque magnitude is
+\[
+\tau=RF=(0.25\,\mathrm{m})(12\,\mathrm{N})=3.0\,\mathrm{N\cdot m}.
+\]
+
+Since the string does not slip, the unwound length equals the arc length at the rim:
+\[
+s=R\Delta\theta.
+\]
+Hence
+\[
+\Delta\theta=\frac{s}{R}=\frac{2.0\,\mathrm{m}}{0.25\,\mathrm{m}}=8.0\,\mathrm{rad}.
+\]
+
+The torque is constant and acts in the direction of rotation, so the work done on the wheel is
+\[
+W=\tau\Delta\theta=(3.0\,\mathrm{N\cdot m})(8.0\,\mathrm{rad})=24\,\mathrm{J}.
+\]
+This also agrees with $W=Fs=(12\,\mathrm{N})(2.0\,\mathrm{m})=24\,\mathrm{J}$.
+
+Now apply the rotational work-energy theorem:
+\[
+W=\Delta K_{\mathrm{rot}}=\tfrac12 I\omega_f^2-\tfrac12 I\omega_i^2.
+\]
+Substitute the known values:
+\[
+24=\tfrac12 (1.0)\omega_f^2-\tfrac12 (1.0)(4.0)^2.
+\]
+Since
+\[
+\tfrac12 (1.0)(4.0)^2=8,
+\]
+we get
+\[
+24=\tfrac12 \omega_f^2-8.
+\]
+Add $8$ to both sides:
+\[
+32=\tfrac12 \omega_f^2.
+\]
+Multiply by $2$:
+\[
+\omega_f^2=64.
+\]
+Therefore,
+\[
+\omega_f=8.0\,\mathrm{rad/s}.
+\]
+
+So the results are
+\[
+\tau=3.0\,\mathrm{N\cdot m},
+\qquad
+\Delta\theta=8.0\,\mathrm{rad},
+\qquad
+W=24\,\mathrm{J},
+\qquad
+\omega_f=8.0\,\mathrm{rad/s}.
+\]
+The key idea is that the applied torque does positive work, so the wheel's rotational kinetic energy increases.