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183_notes:examples:a_meter_stick_on_the_ice [2014/11/16 21:20] – created pwirving | 183_notes:examples:a_meter_stick_on_the_ice [2014/11/20 16:22] – pwirving | ||
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=== Facts === | === Facts === | ||
+ | Mass of meter stick 300g | ||
+ | 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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=== Lacking === | === Lacking === | ||
+ | Rate of change of the center-of-mass speed $v_{CM}$? | ||
+ | Rate of change of the angular speed $\omega$? | ||
=== Approximations & Assumptions === | === Approximations & Assumptions === | ||
+ | No friction due to ice | ||
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=== Representations === | === Representations === | ||
+ | System: Stick | ||
- | {{course_planning: | + | Surroundings: |
+ | |||
+ | {{183_projects:mi3e_11-050.jpg?300}} | ||
+ | |||
+ | $\frac{d\vec{P}}{dt}$ = $\vec{F}_{net}$ | ||
+ | |||
+ | $\frac{d\vec{L}_{rot}}{dt}$ = $\vec{\tau}_{net, | ||
+ | |||
+ | $\tau = r_{A}Fsin \theta$ | ||
+ | |||
+ | ${\vec{L}_{rot}} = I \vec{\omega}$ | ||
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=== Solution === | === Solution === | ||
- | Direction: At both locations, | + | We can use the momentum principle to find the rate of change |
+ | |||
+ | We know that the change | ||
+ | |||
+ | $d\vec{P}/ | ||
- | At location 1: | + | We are given $\vec{F}_{net}$ and the mass of the meter stick so we can find $v_{CM}$. |
- | $\mid\vec{L}_{trans, | + | $dv_{CM}/dt = (6N)/(0.3kg) = 20m/s^2$ |
- | $= 1.1$ x $10^{30}$ $kg \cdot m^2/s$ | + | 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. |
- | $\vec{L}_{trans,Sun}$ = $\langle{0, 0, -1.1 x 10^30}\rangle$ $kg \cdot m^2/s$ | + | $d\vec{L}_{rot}/dt = \vec{\tau}_{net,CM}$ |
+ | We know from the right hand rule that the component is into the screen (-z direction) and that $d\vec{L}_{rot}/ | ||
- | At location 2: | + | 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. |
- | $\mid\vec{L}_{trans, | + | $Id\omega/ |
- | $= 1.1$ x $10^{30}$ $kg \cdot m^2/s$ | + | $d\omega/ |
- | $\vec{L}_{trans, | + | In vector terms, |