Differences

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--- 183_notes:examples:a_meter_stick_on_the_ice [2014/11/16 21:20] – created pwirving
+++ 183_notes:examples:a_meter_stick_on_the_ice [2014/11/20 16:17] – pwirving
@@ Line 6: / Line 6: @@
 === Facts ===
+Mass of meter stick 300g
+Pull at end of meter stick at right angles to the stick: 6N
+Remember a meter stick is a meter long
@@ Line 13: / Line 16: @@
 === Lacking ===
+Rate of change of the center-of-mass speed $v_{CM}$?
+Rate of change of the angular speed $\omega$?
 === Approximations & Assumptions ===
+No friction due to ice
@@ Line 23: / Line 29: @@
 === Representations ===
+System: Stick
-{{course_planning:projects:mi3e_11-006.jpg?400}}
+Surroundings: Your hand (pulling); ice (negligible effect)
+{{183_projects:mi3e_11-050.jpg?300}}
+$\frac{d\vec{P}}{dt}$ = $\vec{F}_{net}$
+$\frac{d\vec{L}_{rot}}{dt}$ = $\vec{\tau}_{net,CM}$
+$\tau = r_{A}Fsin \theta$
+${\vec{L}_{rot}} = I \vec{\omega}$
@@ Line 31: / Line 48: @@
 === Solution ===
-Direction: At both locations, the direction of the translational angular momentum of the comet is in the -z direction (into the computer); determined by using the right-hand rule.
+We can use the momentum principle to find the rate of change of the center of mass speed $v_{cm}$
+We know that the change in momentum over change in time is equal to $\vec{F}_{net}$
-At location 1:
+$d\vec{P}/dt = d(m\vec{v}_{CM})/dt = \vec{F}_{net}$
-$\mid\vec{L}_{trans,Sun}\mid$ = $(8.77$ x $10^{10}m)(2.2$ x $10^{14}kg)(5.46$ x $10^4m/s)sin 90^{\circ}$
+We are given $\vec{F}_{net}$ and the mass of the meter stick so we can find $v_{CM}$.
-$= 1.1$ x $10^{30}$ $kg \cdot m^2/s$
+$dv_{CM}/dt = (6N)/(0.3kg) = 20m/s^2$
-$\vec{L}_{trans,Sun}$ = $\langle{0, 0, -1.1 x 10^30}\rangle$ $kg \cdot m^2/s$
+Similarly we know that the Angular Momentum Principle about center of mass states that the change in Rotational Angular Momentum divided by the change in time is equal to the net torque about the center of mass of the meter stick.
+$d\vec{L}_{rot}/dt = \vec{\tau}_{net,CM}$
-At location 2:
+We know from the right hand rule that the component is into the screen (-z direction) and that $d\vec{L}_{rot}/dt = I \frac{d\omega}{dt}$
-$\mid\vec{L}_{trans,Sun}\mid$ = $(1.19$ x $10^{12}m)(2.2$ x $10^{14}kg)(1.32$ x $10^4m/s)sin 17.81^{\circ}$
+$Id\omega/dt = (0.5m)(6N)sin90^{\circ} = 3N \cdot m$
-$= 1.1$ x $10^{30}$ $kg \cdot m^2/s$
+$d\omega/dt = (3N \cdot m)/[(0.3 kg \cdot m^2)/12] = 120 radians/s^2$
-$\vec{L}_{trans,Sun}$ = $\langle{0, 0, -1.1 x 10^30}\rangle$ $kg \cdot m^2/s$
+In vector terms, $d\vec{\omega}/dt$ points into the page, corresponding to the fact that the angular velocity points into the page and is increasing.