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--- 183_notes:examples:a_yo-yo [2014/11/06 02:41] – pwirving
+++ 183_notes:examples:a_yo-yo [2014/11/06 02:59] (current) – pwirving
@@ Line 63: / Line 63: @@
  $\Delta K_{trans}$  =  $\int_i^f \vec{F}_{net} \cdot d\vec{r}_{cm}$
+ $\Delta E_{sys}$  =  $W_{surr}$
 === Solution ===
@@ Line 68: / Line 70: @@
 a:
-From the Energy Principle (point particle only has  $K_{trans}$ ):
+From the Energy Principle ( when dealing with a point particle it only has  $K_{trans}$ ):
+ $\Delta K_{trans}$  =  $\int_i^f \vec{F}_{net} \cdot d\vec{r}_{cm}$
+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:
  $\Delta K_{trans} = (F - mg)\Delta y_{CM}$
- $\Delta y_{CM} = -h (from\; digram)$
+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:
+$\Delta K_{trans} = (mg - F)h$
 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.
  $\Delta E_{sys} = W_{hand} + W_{Earth}$
+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.
  $\Delta K_{trans} + \Delta K_{rot} = Fd + (-mg)(-h)$
- $\Delta K_{trans} = (mg - F)h$  (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:
  $(mg - F)h + \Delta K_{rot} = Fd + mgh$
+Solve for change in rotational kinetic energy:
  $\Delta K_{rot} = F(d + h)$