Circular Motion with time-dependent radial acceleration

[TwoSeventyOne]

Homework Statement

A particle is traveling on a circle with a radius R. The particle's radial acceleration is given as:
[itex] a_r=At^4 [/itex]
At time [itex] t=0 [/itex] the particle is at [itex] (R,0) [/itex].

A. Find the distance that the particle has traveled as a function of time [itex] S(t) [/itex].
B. Display the particle's acceleration in polar coordinates.
C. Display the particle's acceleration in Cartesian coordinates.

Homework Equations

##a_r=\frac {v^2} {R}##
##x=rcos\theta, y=rsin\theta##

The Attempt at a Solution

I assumed at first that since the particle is moving in a uniform circular motion, I can use
[itex] a_r=\frac {v^2} {R} [/itex]
[itex] At^4=\frac {v^2} {R} [/itex]
[itex] v=\sqrt{RAt^4} [/itex]
Then I treated [itex] v [/itex] as [itex] \frac {ds} {dt} [/itex] and got
[itex] S(t)=\sqrt{RA}*\int_0^t t'^2 \, dt' = \frac{\sqrt{RA}t^3} {3} [/itex]

But then I realized that the radial acceleration changes according to time. Wouldn't that affect the solution? Also, I'm confused as to converting the acceleration from polar to Cartesian. I'm not sure if the same rules apply to acceleration and exactly what are my ##\hat r## and ##\hat \theta##.

 

[Orodruin]

There is nothing stating that the motion is uniform. In fact, the changing radial acceleration tells you that it is not. However, ##a = v^2/r## holds for any circular motion - otherwise the particle would not be moving in a circle.

 

[TwoSeventyOne]

I see. Does that mean that there is necessarily tangential acceleration as well? If so, any hint on finding it?

 

[Orodruin]

Yes, there must be a tangential acceleration. You have already found the velocity, so finding the acceleration is a simple matter of differentiating.

 

[HalfLostAndSearching]

I would first clarify the problem statement with my professor or colleagues to ensure that my assumptions are correct. It seems that the problem is asking for the position, acceleration, and velocity of a particle moving in a circle with a time-dependent radial acceleration.

To find the distance traveled as a function of time, I would use the formula for displacement in circular motion:
S(t) = Rθ(t), where θ(t) is the angle swept by the particle at time t. To find θ(t), I would need to integrate the angular velocity, ω(t), which can be found by integrating the radial acceleration, a_r(t):
ω(t) = ∫a_r(t) dt = At^5/5 + C
Using the initial condition, ω(0) = 0, we can solve for C and get ω(t) = At^5/5.
Thus, θ(t) = ∫ω(t) dt = At^6/30 + C
Using the initial condition, θ(0) = 0, we can solve for C and get θ(t) = At^6/30.
Finally, plugging this into the equation for distance, we get:
S(t) = RAt^6/30

To display the particle's acceleration in polar coordinates, we can use the formula a_r = ω^2R, where ω is the angular velocity. Using the values we found earlier, we get:
a_r(t) = (At^5/5)^2 * R = A^2Rt^10/25

To display the particle's acceleration in Cartesian coordinates, we can use the equations x = Rcosθ and y = Rsinθ, where θ = At^6/30. Differentiating these equations with respect to time, we get:
vx = -Rsin(At^6/30) * (A^6t^5)/30
vy = Rcos(At^6/30) * (A^6t^5)/30
Thus, the acceleration in Cartesian coordinates is:
ax = -Rsin(At^6/30) * (A^6t^5)/30
ay = Rcos(At^6/30) * (A^6t^5)/30

In summary, to solve this problem, we needed to