183_notes:examples:a_meter_stick_on_the_ice

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183_notes:examples:a_meter_stick_on_the_ice [2014/11/20 16:00] – pwirving183_notes:examples:a_meter_stick_on_the_ice [2014/11/20 16:28] (current) – pwirving
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 No friction due to ice No friction due to ice
 +
 +Assume the meter stick is a uniform rod
  
  
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 $\tau = r_{A}Fsin \theta$ $\tau = r_{A}Fsin \theta$
 +
 +${\vec{L}_{rot}} = I \vec{\omega}$
  
  
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 === Solution === === Solution ===
  
-From the momentum principle:+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 $\vec{F}_{net}$
  
 $d\vec{P}/dt = d(m\vec{v}_{CM})/dt = \vec{F}_{net}$ $d\vec{P}/dt = d(m\vec{v}_{CM})/dt = \vec{F}_{net}$
 +
 +We are given $\vec{F}_{net}$ and the mass of the meter stick so we can find $v_{CM}$.
  
 $dv_{CM}/dt = (6N)/(0.3kg) = 20m/s^2$ $dv_{CM}/dt = (6N)/(0.3kg) = 20m/s^2$
  
-Angular Momentum Principle about center of mass:+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.
  
 $d\vec{L}_{rot}/dt = \vec{\tau}_{net,CM}$ $d\vec{L}_{rot}/dt = \vec{\tau}_{net,CM}$
  
-Component into screen (-z direction):+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 $Id\omega/dt$:
  
 $Id\omega/dt = (0.5m)(6N)sin90^{\circ} = 3N \cdot m$ $Id\omega/dt = (0.5m)(6N)sin90^{\circ} = 3N \cdot m$
 +
 +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 $d\omega/dt$
  
 $d\omega/dt = (3N \cdot m)/[(0.3 kg \cdot m^2)/12] = 120 radians/s^2$ $d\omega/dt = (3N \cdot m)/[(0.3 kg \cdot m^2)/12] = 120 radians/s^2$
  
 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. 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.
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  • by pwirving