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--- 183_notes:examples:a_meter_stick_on_the_ice [2014/11/20 16:16] – pwirving
+++ 183_notes:examples:a_meter_stick_on_the_ice [2014/11/20 16:28] (current) – pwirving
@@ Line 24: / Line 24: @@
 No friction due to ice
+Assume the meter stick is a uniform rod
@@ Line 62: / Line 64: @@
 $d\vec{L}_{rot}/dt = \vec{\tau}_{net,CM}$
-We know from the right hand rule that the component is into the screen (-z direction) and that $d\vec{L}_{rot}/dt = I d\frac{\vec{\omega}}{dt}$
+We know from the right hand rule that the component is into the screen (-z direction) and that $d\vec{L}_{rot}/dt = I \frac{d\omega}{dt}$
+We also know that $\tau = r_{A}Fsin \theta$ and that $F=6N$ and $\theta=90^{\circ}$. We also know that $r$ is the distance from where the force is applied to the center of mass which is $.5m$. We input all these variables to the right hand side of our equation and compute $Id\omega/dt$:
 $Id\omega/dt = (0.5m)(6N)sin90^{\circ} = 3N \cdot m$
+Divide by the moment of inertia which around the center of mass of a uniform rod of mass M and length L is $\frac{ML^{2}}{12}$ and L is 1m in this situation. Solve for $d\omega/dt$
 $d\omega/dt = (3N \cdot m)/[(0.3 kg \cdot m^2)/12] = 120 radians/s^2$
 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.