Uniform circular motion particle problem

[EngnrMatt]

Homework Statement

At t1 = 3.00 s, the acceleration of a particle moving at constant speed in counterclockwise circular motion is

a₁=(6m/s²)i+(5m/s²)j

At t2 = 4.00 s (less than one period later), the acceleration is

a₂=(5m/s²)i-(6m/s²)j

The period is more than 1.00 s. What is the radius of the circle?

Homework Equations

T= (2∏r)/v
a=v²/r

The Attempt at a Solution

I drew out the vectors, but that's all I could do with the problem. Maybe I'm supposed to integrate to find velocity or something? I just really don't understand where to start. If I knew that, I could probably finish it up. But there's so much information given in the book for circular motion, I don't know what to do with it.

 

[ehild]

What angle the acceleration vectors do enclose? What can be the total time period? What is the magnitude of the acceleration? How is the acceleration related to the angular speed and radius of circle?ehild

 

[EngnrMatt]

I don't know whether I'm supposed to find the angles of the individual vectors or average the acceleration over the given time period then find that one angle. And once that's done, I don't know what do with that information either. I wish I had a better professor, all he does is prove equations during class. Never goes over any examples.

 

[ehild]

Try to answer my questions. That will lead to the solution. It is uniform circular motion. The acceleration is centripetal. The acceleration vectors are along the radii. What is the angular displacement of the particle from t1 to t2?


ehild

 

[Nickie]

As a scientist, it is important to approach a problem like this systematically and use the relevant equations to find a solution. In this case, we can use the equation a=v^2/r to find the radius of the circle.

First, we need to find the magnitude of the acceleration at t1 and t2. We can do this by using the Pythagorean theorem to combine the x and y components of the acceleration vectors.

At t1: a1 = √((6m/s^2)^2 + (5m/s^2)^2) = √(36 + 25) = √61 m/s^2
At t2: a2 = √((5m/s^2)^2 + (-6m/s^2)^2) = √(25 + 36) = √61 m/s^2

Next, we can use the equation a=v^2/r to find the radius of the circle. We know that the speed is constant, so we can set the magnitudes of the accelerations at t1 and t2 equal to each other.

a1 = a2
√61 = v^2/r

We also know that the period (T) is more than 1.00 s, so we can use the equation T=(2∏r)/v to find the radius.

T=(2∏r)/v
r = (Tv)/(2∏)

Substituting this into our previous equation, we get:

√61 = v^2/[(2∏)(Tv)/(2∏)]
√61 = v

Now, we can use the given information about the period to solve for the velocity.

T = (2∏r)/v
1.00 s < T < ∞
1.00 s < (2∏r)/v < ∞
1.00 s < (2∏r)/√61 < ∞

We can see that the value for (2∏r)/√61 must be greater than 1.00 s, so we can set this as a lower bound for the velocity.

1.00 s < (2∏r)/√61

Rearranging this equation, we get:

√61 < (2∏r)/1.00 s

Now, we can use the given information about the accelerations at t1