Force, Given Parametric Polar Coordinates

## Question:

The path of motion in the horizontal plane of a 5lb particle is described in terms of polar coordinates as r=2*t+10 where r is the distance from the origin of the reference frame to the particle, and q=5*t2-6*t where q is the angle from the x axis of the reference frame to the distance line. Time (t) is in seconds. Determine the unbalanced force acting on the particle when t=2s.

The rate of change of the position vector with respect to time is the velocity vector and its rate of change with respect to time is the acceleration vector. Position p=2*t+10ft at an angle of 5*t2-6*t radians. Putting this in rectangular coordinates x and y we have
px=(2*t+10)*cos(5*t2-6*t) and
py=(2*t+10)*sin(5*t2-6*t)

The rate of change of p with respect to time, dp/dt is a vector whose components are the derivatives of px and py. To avoid a lot of calculus let's make up a table symmetrical about t=2.0 and calculate the values of px and py for each time in the table. Then from the finite differences between px and py, calculate the rate of change of px and py, we call them px' and py'. Then from the finite differences between px' and py', calculate the rate of change of px' and py', we call them px'' and py''.

 t 1.999 2 2.001 5t2-6t 7.986005 8 8.014005 cos(5t2-6t) -0.1316402 -0.1455 -0.1593413 sin(5t2-6t) 0.99129757 0.98935825 0.98722356 2t+10 13.998 14 14.002 px -1.8426991 -2.0370005 -2.2310965 py 13.8761833 13.8510155 13.8231043 px' -194.30139 -194.09605 py' -25.167888 -27.911159 px'' 205.346831 py'' -2743.2711
Magnitude of the acceleration is sqrt(px''2+py''2) = 2750.94591 f/s/s
Mass is 5lb/32.2f/s/s = 0.1552795 slugs
Force is mass times acceleration = 426.396616N

I did this on a spreadsheet so I could try different delta t values. Decreasing delta t by a factor of 1000 only changed force in the fifth significant digit.

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