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| 183_notes:examples:a_yo-yo [2014/11/06 02:40] – pwirving | 183_notes:examples:a_yo-yo [2014/11/06 02:59] (current) – pwirving | ||
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| {{course_planning: | {{course_planning: | ||
| - | $\int_i^f \vec{F}_{net} \cdot d\vec{r}_{cm}$ | + | $\Delta K_{trans}$ = $\int_i^f \vec{F}_{net} \cdot d\vec{r}_{cm}$ |
| + | |||
| + | $\Delta E_{sys}$ = $W_{surr}$ | ||
| === Solution === | === Solution === | ||
| Line 68: | Line 70: | ||
| a: | a: | ||
| - | From the Energy Principle (point particle only has $K_{trans}$): | + | From the Energy Principle ( when dealing with a point particle |
| + | |||
| + | $\Delta K_{trans}$ = $\int_i^f \vec{F}_{net} \cdot d\vec{r}_{cm}$ | ||
| + | |||
| + | Substituting in for the forces acting on the yo-yo for $F_{net}$ and the change in position in the y direction for the centre of mass for $d\vec{r}_{cm}$ we get: | ||
| $\Delta K_{trans} = (F - mg)\Delta y_{CM}$ | $\Delta K_{trans} = (F - mg)\Delta y_{CM}$ | ||
| - | $\Delta y_{CM} = -h (from\; digram)$ | + | As indicated in diagram in the b section of the representation: |
| - | $\Delta K_{trans} = (F - mg)(-h) = (mg - F)h$ | + | $\Delta y_{CM} = -h$ |
| + | |||
| + | Substitute in $-h$ for $y_{CM}$ | ||
| + | |||
| + | $\Delta K_{trans} = (F - mg)(-h)$ | ||
| + | |||
| + | Multiply across by a minus and you get an equation for $\Delta K_{trans}$ that looks like: | ||
| + | |||
| + | $\Delta K_{trans} | ||
| b: | b: | ||
| + | |||
| + | From the energy principle we know: | ||
| + | |||
| + | $\Delta E_{sys}$ = $W_{surr}$ | ||
| + | |||
| + | In this case we know that the change in energy in the system is due to the work done by the hand and the work done by the Earth. | ||
| $\Delta E_{sys} = W_{hand} + W_{Earth}$ | $\Delta E_{sys} = W_{hand} + W_{Earth}$ | ||
| + | |||
| + | Because we are dealing with the real system in this scenario the change in energy is equal to the change in translational kinetic energy + the change in rotational kinetic energy. | ||
| + | |||
| + | $\Delta K_{trans} + \Delta K_{rot} = W_{hand} + W_{Earth}$ | ||
| + | |||
| + | Substitute in the work represented by force by distance for both the hand and the Earth. | ||
| $\Delta K_{trans} + \Delta K_{rot} = Fd + (-mg)(-h)$ | $\Delta K_{trans} + \Delta K_{rot} = Fd + (-mg)(-h)$ | ||
| - | $\Delta K_{trans} = (mg - F)h$ (From part (a)) | + | From part (a) of the problem we can substitute in $(mg - F)h$ for $\Delta K_{trans}$ as the translational kinetic energy will be the same. |
| + | |||
| + | $\Delta K_{trans} = (mg - F)h$ | ||
| + | |||
| + | Substituting this into our equation leaves us with: | ||
| $(mg - F)h + \Delta K_{rot} = Fd + mgh$ | $(mg - F)h + \Delta K_{rot} = Fd + mgh$ | ||
| + | |||
| + | Solve for change in rotational kinetic energy: | ||
| $\Delta K_{rot} = F(d + h)$ | $\Delta K_{rot} = F(d + h)$ | ||