Question

At time t1 = 2.00 s, the acceleration of a particle in counterclockwise circular motion is (7.00 m/s2)i hat + (6.00 m/s2)j. It moves at constant speed. At time t2 = 5.00 s, its acceleration is (6.00 m/s2)i hat + (−7.00 m/s2)j. What is the radius of the path taken by the particle if t2 − t1 is less than one period?

Answer 1

Answer:

r = 3519.55 m

Explanation:

We know that the acceleration of a particle in a circular motion is directed towards the center of the circumference and has magnitude:

F = rω^2

Where r is the radius of the circumference and ω is the angular velocity.

From the two acceleration vectors we find that their magnitude is

√(7^2+6^2) = √85

Therefore:

√85 m/s^2= rω^2

Now we need to calculate the angular velocity to obtain the radius. Since t2-t1 = 3s is less than one period we can be sure that the angular velocity is equals to the angle traveled between this time divided by 3 s.

The angle with respect to the x-axis  for the particle at t1 and t2 is:

$\theta 1 =\cos ^{-1}\left(\frac{7}{\sqrt{85}}\right)\\\theta 2 =\cos ^{-1}\left(\frac{6}{\sqrt{85}}\right)$

Therefore, the angular velocity ω is (in radians per second):

$\omega = \frac{\theta2 - \theta1}{3 s} = 0.0511813 \frac{1}{s}$

Therefore:

r = √85 / (0.0511813)^2 = 3519.55 m