Differences
This shows you the differences between two versions of the page.
| Both sides previous revision Previous revision Next revision | Previous revision | ||
| 183_notes:examples:rotational_kinetic_energy_and_work [2014/11/05 21:06] – pwirving | 183_notes:examples:rotational_kinetic_energy_and_work [2014/11/14 07:01] (current) – pwirving | ||
|---|---|---|---|
| Line 1: | Line 1: | ||
| - | ===== Example: | + | ===== Example: |
| In the figure which is in the representations section you observe that a wheel is mounted on a stationary axel, which is nearly frictionless so that the wheel turns freely. The wheel has an inner ring with mass 5 kg and radius 10 cm and an outer ring with mass 2 kg and radius 25 cm; the spokes have negligible mass. A string with negligible mass is wrapped around the outer ring and you pull on it, increasing the rotational speed of the wheel. During the time that the wheel' | In the figure which is in the representations section you observe that a wheel is mounted on a stationary axel, which is nearly frictionless so that the wheel turns freely. The wheel has an inner ring with mass 5 kg and radius 10 cm and an outer ring with mass 2 kg and radius 25 cm; the spokes have negligible mass. A string with negligible mass is wrapped around the outer ring and you pull on it, increasing the rotational speed of the wheel. During the time that the wheel' | ||
| Line 44: | Line 44: | ||
| $E_{f} = E{i} + W$ | $E_{f} = E{i} + W$ | ||
| + | |||
| + | Substitute in equation for rotational energy for energy initial and energy final. | ||
| $\frac{1}{2}I\omega^{2}_{f} = \frac{1}{2}I\omega^{2}_{i} + W$ | $\frac{1}{2}I\omega^{2}_{f} = \frac{1}{2}I\omega^{2}_{i} + W$ | ||
| + | |||
| + | You are trying to find work so rearrange the equation to isolate W. | ||
| $W = \frac{1}{2}I(\omega^{2}_{f} - \omega^{2}_{i})$ | $W = \frac{1}{2}I(\omega^{2}_{f} - \omega^{2}_{i})$ | ||
| Line 52: | Line 56: | ||
| $I = (m_{1}r^{2}_{\perp1}$ + $m_{2}r^{2}_{\perp2}$ + $\cdot \cdot \cdot)_{inner}$ + $(m_{1}r^{2}_{\perp1}$ + $m_{2}r^{2}_{\perp2})_{outer}$ + $\cdot \cdot \cdot$ | $I = (m_{1}r^{2}_{\perp1}$ + $m_{2}r^{2}_{\perp2}$ + $\cdot \cdot \cdot)_{inner}$ + $(m_{1}r^{2}_{\perp1}$ + $m_{2}r^{2}_{\perp2})_{outer}$ + $\cdot \cdot \cdot$ | ||
| + | |||
| + | The total inertia is then the inertia inner + the inertia outer. | ||
| $I = I_{inner} + I_{outer}$ | $I = I_{inner} + I_{outer}$ | ||
| Line 58: | Line 64: | ||
| $I = M_{inner}R^{2}_{inner} + M_{outer}R^{2}_{outer}$ | $I = M_{inner}R^{2}_{inner} + M_{outer}R^{2}_{outer}$ | ||
| + | |||
| + | Substitute the corresponding values for the variables: | ||
| $I = (5kg)(.1m)^2$ + $(2 kg)(.25 m)^2$ = $(0.050 + 0.125) kg \cdot m^2 = 0.175 kg \cdot m^2$ | $I = (5kg)(.1m)^2$ + $(2 kg)(.25 m)^2$ = $(0.050 + 0.125) kg \cdot m^2 = 0.175 kg \cdot m^2$ | ||
| Line 69: | Line 77: | ||
| You, the Earth, and the axle will exert forces on the system. How much work does the Earth do? Zero, because the center of mass of the wheel doesn' | You, the Earth, and the axle will exert forces on the system. How much work does the Earth do? Zero, because the center of mass of the wheel doesn' | ||
| - | $W = \frac{1}{2}(0.175 kg \cdot m^{2})(44.0^2 | + | $W = \frac{1}{2}(0.175 kg \cdot m^{2})(44.0^2 |