Tension/polar co ordinates

  • Thread starter keelejody
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A particle P of mass m is suspended from a xed point O by a light inextensible string of
length a. The particle is subject to uniform gravity and moves in a vertical plane through
O with the string taut

Assuming the standard expression for the components of acceleration in polar coordinates, show from rst principles that

((d^ϴ)/dt^2) +(g/a)sinϴ=0

where where ϴ is the angle between the string and the downwards vertical, and g is the acceleration due to gravity. Give an expression for the tension in the string.

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ok so i thought maybe write about acceleration is in polar co-ordinates and postion but that wasnt smart because obviously they wont sum to one

maybe solve the second order diff but wouldnt i need (g/a)sinϴ in terms of t?

maybe take the (g/a)sinϴ over and intergrate but sides

so i got v=ϴ dot= (gt/a)cosϴ

and x =ϴ=(gt^2)sinϴ

but i actually dont have a clue how to tackle this......
 

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