How Does Tension Affect Particle Movement in Polar Coordinates?

[keelejody]

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 won't sum to one

maybe solve the second order diff but wouldn't 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 don't have a clue how to tackle this...

 

[NateD]

any help?Answer:The acceleration of the particle can be expressed in terms of the polar coordinates (r, ϴ) as a = (d²r/dt², r(d²ϴ/dt²)) Since the particle is constrained to move in a vertical plane, the radial component of the acceleration is zero, i.e., (d²r/dt²) = 0. The angular component of the acceleration is given by: (d²ϴ/dt²) = - (g/a)sinϴ The tension in the string is given by T = m(d²r/dt²) + ma(d²ϴ/dt²) = ma(d²ϴ/dt²) = -mgsinϴ