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183_notes:examples:a_meter_stick_on_the_ice [2014/11/20 15:59] – pwirving | 183_notes:examples:a_meter_stick_on_the_ice [2014/11/20 16:10] – pwirving | ||
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Pull at end of meter stick at right angles to the stick: 6N | Pull at end of meter stick at right angles to the stick: 6N | ||
+ | |||
+ | Remember a meter stick is a meter long | ||
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$\frac{d\vec{L}_{rot}}{dt}$ = $\vec{\tau}_{net, | $\frac{d\vec{L}_{rot}}{dt}$ = $\vec{\tau}_{net, | ||
- | $\tau = r_{A}Fsin\teta$ | + | $\tau = r_{A}Fsin \theta$ |
+ | |||
+ | ${d\vec{L}_{rot}} = I \vec{\omega}$ | ||
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=== Solution === | === Solution === | ||
- | From the momentum principle: | + | We can use the momentum principle |
+ | |||
+ | We know that the change in momentum over change in time is equal to $\vec{F}_{net}$ | ||
$d\vec{P}/ | $d\vec{P}/ | ||
+ | |||
+ | We are given $\vec{F}_{net}$ and the mass of the meter stick so we can find $v_{CM}$. | ||
$dv_{CM}/dt = (6N)/ | $dv_{CM}/dt = (6N)/ | ||
- | 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}/ | $d\vec{L}_{rot}/ |