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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] – pwirving | ||
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No friction due to ice | No friction due to ice | ||
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+ | Assume the meter stick is a uniform rod | ||
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$d\vec{L}_{rot}/ | $d\vec{L}_{rot}/ | ||
- | We know from the right hand rule that the component is into the screen (-z direction) and that $d\vec{L}_{rot}/ | + | We know from the right hand rule that the component is into the screen (-z direction) and that $d\vec{L}_{rot}/ |
+ | |||
+ | 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/ | ||
$Id\omega/ | $Id\omega/ | ||
+ | |||
+ | 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/ | ||
$d\omega/dt = (3N \cdot m)/[(0.3 kg \cdot m^2)/12] = 120 radians/ | $d\omega/dt = (3N \cdot m)/[(0.3 kg \cdot m^2)/12] = 120 radians/ | ||
In vector terms, $d\vec{\omega}/ | In vector terms, $d\vec{\omega}/ |