Uniform Circular Motion
A particle moves along a circular path over a horizontal xy coordinate system, at constant speed. At time t(1)=4.00s, it is at point (5.00m, 6.00m) with a velocity (3.00 m/s)j and acceleration in the position x direction. At time t(2) = 10.0s, it has a velocity of (-3.00 m/s)i and acceleration in the positive y direction. What are the x and y coordinates of the center of the circular path if t(2)-t(1) is less than one period?
I have tried a bunch of stuff on this problem and I keep getting the wrong answer....help please.
Ima agree with iceman here, I think you solved for the angular velocity wrong top.
I think maybe we can use a relation relating the constant velocity around the circular track to the distance traveled to find r
we know velocity of a particle in uniform circular motion, v = omega*r... we have omega, we have v... vooooooila make sure your units are rad/s
I would suggest this method of solving the problem:
I'm not sure you can make the assumption that this person is experienced w/ calculus.... it may or may not be the case
To Elite, if you can follow the calculus...
Just note that be broke the velocity vector into x and y components (dx/dt) and (dy/dt). Same way you solve for the magnitude of any right triangle you would take sqrt(dx/dt^2 + dy/dt^2) sorry for not using the script.
Assuming you don't know calculus, I would
1) Solve for omega = (change in angle)/(change in time) = 3pi/2 / 6 = pi/4 rad/s
2) Note that v = omega*r => r = v/omega = 3/pi/4
Same solution, different method depending on your background
Thanks guys. Don't know where my head was at. (Doh)
Where did you guys get these equations exactly?
Originally Posted by icemanfan