At time t1 = 2.00 s, the acceleration of a particle in counterclockwise circular motion is (9.00i + 8.00j) m/s^2. It moves at constant speed. At time t2 = 3.00s (3/4 of a revolution later), it's acceleration is (8.00i - 9.00j) m/s^2. Find the radius of the path taken by the particle.
see below.
To find the radius of the path taken by the particle, we can use the relationship between acceleration and radius in circular motion.
The acceleration in circular motion is given by the equation:
a = -ω^2 * r
where a is the acceleration, ω (omega) is the angular velocity, and r is the radius of the circular path.
In counterclockwise motion, the acceleration vector is given as (9.00i + 8.00j) m/s^2. Since the particle moves at a constant speed, the magnitude of the acceleration vector remains constant at all points on the circular path. Thus, the magnitude of the acceleration vector at time t1 is:
|a1| = √[(9.00)^2 + (8.00)^2] m/s^2
Similarly, at time t2, the magnitude of the acceleration vector is:
|a2| = √[(8.00)^2 + (-9.00)^2] m/s^2
Now, the change in angle between t1 and t2 is 3/4 of a revolution. Since a full revolution is 2π radians, the change in angle can be calculated as:
Δθ = (3/4) * 2π = (3/2)π radians
With this information, we can set up the following equation using the relationship between acceleration and angular velocity:
|a2| = ω^2 * r
Substituting the values we have:
√[(8.00)^2 + (-9.00)^2] = ω^2 * r
Solving for ω^2, we get:
ω^2 = √[(8.00)^2 + (-9.00)^2] / r
Now, we can solve for r by rearranging the equation:
r = √[(8.00)^2 + (-9.00)^2] / ω^2
Once you calculate the value of ω^2, you can substitute it into this equation to find the radius r.