The magnitude of acceleration is constant, and the direction is towards the center.
So you are given two values of acceleration in directions which intersect at the center of rotation.
mag acceleration= sqrt(36+16)=sqrt(52)
The next step is to find the displacement between these two times.
theta=Mag acceleration * time^2 * 1/2
theta= sqrt52 * 9*2 radians
so ang velocity= w= theta/time=
= you do it.
Finally,
a=w*r or r=a/w= sqrt52/w and you have it.
check my thinking
At time t1 = 2.00 s, the acceleration of a particle in counterclockwise circular motion is (6.00 m/s2) i^+ (4.00 m/s2) j^. It moves at constant speed. At time t2 = 5.00 s, its acceleration is (4.00 m/s2) i^+ (-6.00 m/s2) j^. What is the radius of the path taken by the particle?
1 answer
1 answer