183_notes:examples:a_meter_stick_on_the_ice

Differences

This shows you the differences between two versions of the page.

Link to this comparison view

Both sides previous revision Previous revision
183_notes:examples:a_meter_stick_on_the_ice [2014/11/20 16:28] – pwirving183_notes:examples:a_meter_stick_on_the_ice [2014/11/20 16:28] (current) – pwirving
Line 70: Line 70:
 $Id\omega/dt = (0.5m)(6N)sin90^{\circ} = 3N \cdot m$ $Id\omega/dt = (0.5m)(6N)sin90^{\circ} = 3N \cdot m$
  
-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 $d\omega/dt$+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.
  • 183_notes/examples/a_meter_stick_on_the_ice.1416500893.txt.gz
  • Last modified: 2014/11/20 16:28
  • by pwirving