Force, Given Parametric Polar Coordinates


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

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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