Calculate the Radial & Angular Components of Force for Particle Trajectory

[mopar969]

A particle with mass 10 kg starts from rest at r=10 and theta=0 radians, following a trajectory given by r=10-2t, theta = 0.2t. Find the radial and angular components Fr and Ftheta of the force, as functions of time, that will cause this motion.

Please help me get started on this problem what is my first move?

 

[ehild]

Write up the position vector of the particle,
determine the acceleration: it is the second derivative of the position vector with respect time. Then apply Newton second law to get force from acceleration.
You can work either in a polar coordinate system or convert r(t) and theta(t) to the Descartes coordinates, x and y.

ehild

 

[mopar969]

I am a little confused as to what you mean by the position vector isn't it the equation given?

 

[Paulie323]

the position vector i believe is the displacement which would involve an equation relating position and time and the derivative of that equation would leave you with the equation of or acceleration.

 

[ehild]

I am a little confused as to what you mean by the position vector isn't it the equation given?

You need the components of acceleration. What have you learnt, how can you calculate the components of acceleration in polar coordinates?

ehild

 

[mopar969]

Sorry about that I was not thinking what you meant. Here is wwhat I have done:
Vector r= r r(hat)
Velocity = r(dot) r(hat)+r r(hat dot)
I found what r(hat dot) is and plugged it into the equation
So now,
velocity = r(dot) r(hat)+r theta(dot) theta(hat)
acceleration = vector r (r(double dot) -r theta(dot squared))+theta(hat) (r theta(double dot)+2 r(dot) theta(dot)

I know that r= 10-2t
r(dot) = -2 and r(double dot)=0

So, all I need to do now is plugg them into the acceleration equation right?

 

[ehild]

Correct! :) (if you meant vector r as r(hat))

ehild

 

[mopar969]

Her is where I am then what:

acceleration= 10 rhat -2trhat+(-1.12)

Now what do I need to do to find fr and f theta?

 

[ehild]

You have derived correctly the expression for acceleration. The force is mass times the acceleration. If you multiple a vector by a number, all components have to be multiplied. The force, just like the acceleration has both radial and angular components, which are m times those of the acceleration. So the components of the force are:

radial: Fr = [r(double dot) -r theta(dot squared)] * m

angular: Ftheta = [r theta(double dot)+2 r(dot) theta(dot)] * m

Just plug in everything and simplify. ehild

 

[mopar969]

radial: -4+0.8t
angular: -8

This is what I got when I simplified everything please check my answers. Also, how come there is no t value in the second answer? Did I do something wrong or what does this mean?

 

[DaveC426913]

10kg makes for one heckuva particle, btw.

 

[mopar969]

So, are my answers correct?

 

[ehild]

You solved the problem, it is correct. This is a motion along a spiral, with constant angular velocity. The period of one turn is constant, while the radius decreases. Therefore the linear speed decreases, at a constant rate. ehild