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'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
if the particle is moving clockwise
if the particle is moving counterclockwise.
We know from the direction of acceleration, the given fact that between t = 4 and t = 10 we do not have a full cycle, and the directions of velocity at t = 4 and t = 10 that we have moved 3/4 of a revolution clockwise. In radians, a full revolution is radians, so 3/4 of a revolution is radians. This is equal to the total change in angle, which is = (rate of change in angle per time)x(time taken), or in this case