183_notes:examples:a_yo-yo

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183_notes:examples:a_yo-yo [2014/10/31 16:29] – pwirving183_notes:examples:a_yo-yo [2014/11/06 02:59] (current) – pwirving
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 === Assumptions and Approximations === === Assumptions and Approximations ===
 +
 +You are able to maintain constant force when pulling up on yo-yo
 +
 +Assume no slipping of string around the axle. Spindle turns the same amount as string that has unravelled
 +
 +No wobble included
 +
 +String has no mass
 +
  
 === Lacking === === Lacking ===
 +
 +Change in translational kinetic energy of the yo-yo
 +
 +Change in the rotational kinetic energy of the yo-yo
  
 === Representations === === Representations ===
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 Surroundings: Earth and hand Surroundings: Earth and hand
 +
 +{{course_planning:course_notes:mi3e_09-034.jpg?100|}}
  
 b:  b: 
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 Surroundings: Earth and hand Surroundings: Earth and hand
 +
 +{{course_planning:course_notes:mi3e_09-035.jpg?100|}}
 +
 +$\Delta K_{trans}$ = $\int_i^f \vec{F}_{net} \cdot d\vec{r}_{cm}$
 +
 +$\Delta E_{sys}$ = $W_{surr}$
  
 === Solution === === Solution ===
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 a:  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:
  
 $\Delta K_{trans} = (F - mg)\Delta y_{CM}$ $\Delta K_{trans} = (F - mg)\Delta y_{CM}$
  
-$\Delta y_{CM} = -h (from\; digram)$+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:  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:
  
 $\Delta K_{rot} = F(d + h)$ $\Delta K_{rot} = F(d + h)$
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  • Last modified: 2014/10/31 16:29
  • by pwirving