183_notes:examples:a_yo-yo

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183_notes:examples:a_yo-yo [2014/10/31 16:22] – created 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|}}
 +
 +ΔKtrans = ∫fi→Fnet⋅d→rcm
 +
 +ΔEsys = Wsurr
  
 === Solution === === Solution ===
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 a:  a: 
  
-From the Energy Principle (point particle only has Ktrans):+From the Energy Principle ( when dealing with a point particle it only has Ktrans): 
 + 
 +ΔKtrans = ∫fi→Fnet⋅d→rcm 
 + 
 +Substituting in for the forces acting on the yo-yo for Fnet and the change in position in the y direction for the centre of mass for d→rcm we get: 
 + 
 +ΔKtrans=(F−mg)ΔyCM 
 + 
 +As indicated in diagram in the b section of the representation: 
 + 
 +ΔyCM=−h  
 + 
 +Substitute in −h for yCM 
 + 
 +ΔKtrans=(F−mg)(−h) 
 + 
 +Multiply across by a minus and you get an equation for ΔKtrans that looks like:   
 + 
 +ΔKtrans=(mg−F)h 
 + 
 + 
 +b:  
 + 
 +From the energy principle we know: 
 + 
 +ΔEsys = Wsurr 
 + 
 +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. 
 + 
 +ΔEsys=Whand+WEarth 
 + 
 +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. 
 + 
 +ΔKtrans+ΔKrot=Whand+WEarth 
 + 
 +Substitute in the work represented by force by distance for both the hand and the Earth. 
 + 
 +ΔKtrans+ΔKrot=Fd+(−mg)(−h) 
 + 
 +From part (a) of the problem we can substitute in (mg−F)h for ΔKtrans as the translational kinetic energy will be the same. 
 + 
 +ΔKtrans=(mg−F)h  
 + 
 +Substituting this into our equation leaves us with: 
 + 
 +(mg−F)h+ΔKrot=Fd+mgh 
 + 
 +Solve for change in rotational kinetic energy: 
 + 
 +ΔKrot=F(d+h)
  
-\deltaKtrans=(F−mg)\deltayCM 
  
  
  
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