Constant radial and angular velocity; find magnitude of acceleration.

• oddjobmj
In summary, a particle moving in a plane with constant radial velocity and constant angular velocity was given specific values for its variables. The speed of the particle was determined to be 8.026 m/s and the magnitude of its acceleration was found to be 22.43 m/s^2, with a radial component of 16.93 m/s^2 and a tangential component of 14.72 m/s^2. The formula for centripetal acceleration, a=v^2/r, was used, as well as the concept of expressing velocity and acceleration in polar coordinates. The particle was found to be moving along a spiral of Archimedes.
oddjobmj

Homework Statement

(a) A particle moves in a plane with constant radial velocity [(r)\dot] and constant angular velocity [θ\dot]. When the particle is at distance r from the origin, determine the speed of the particle.
Data: [(r)\dot] = 3.2 m/s; [θ\dot] = 2.3 rad/s; r = 3.2 m.

(b) When the particle is at distance r from the origin, determine the magnitude of the acceleration.

a=v^2/r

The Attempt at a Solution

This question was posted on these forums before which I found using the search function. The answer to part (a) was derived after the help from that thread. The result is 8.026 m/s.

I am not sure about part (b), however. In my search I have found many suggesting that a=v^2/r. Why is this the case? Also, I haven't been able to get that to work.

What I tried is this:

Find the circumference of the circle with radius 3.2m which is 20.1062 m. Find the period of the particle which is 2*pi/2.3 which is 2.73182 seconds. Using that I was thinking I could find the angular velocity in m/s which I calculated to be 7.447 m/s. If I use that in the equation a=v^2/r I get a=7.447^2/3.2=16.93 m/s^2 which is incorrect.

What am I missing? Also, where does a=v^2/r come from? Thanks in advance for the help!

This is a problem that apparently assumes that you are familiar with expressing velocity and acceleration in polar coordinates. See for example this link.

The expression ##v^2/r## is the formula for "centripetal acceleration" for a particle moving on a circle. But, in your problem the particle is not moving on a circle. In your problem the particle will have two components of acceleration (radial and tangential) as shown near the bottom of the page in the link above.

Your calculation of 16.93 m/s2 does in fact happen to give the correct answer for the magnitude of the radial component of acceleration for this problem. (It would not give the correct radial acceleration if ##\ddot{r} \neq 0##.) To find the total acceleration, you will also need to find the tangential component of acceleration.

Thank you for your help TSny.

I am confused about how to find r, r(dot), etc. As you suggest it is not moving in a circle but since the radial and angular velocity are constant it is hard for me to imagine the radius changing. I might just be tired but I can't even seem to remember what r dot and theta dot are. I was under the impression that these were just derivatives but I thought r was constant so the derivative would be 0?

I have no idea how to find the tangential component of acceleration. The link you provided suggests a formula in two parts for acceleration; one is radial and the other tangential. As you noted I have solved for the radial portion but can't seem to figure out how to solve the other portion. If I understood how to find rdot and theta dot I would probably be better suited to take the next step.

Is it really just: 2*3.2*2.3+3.2*0=14.72 for the tangential portion of acceleration? i.e. 2*r(dot)*theta(dot)+r*theta'

EDIT: Apparently that worked. If you take the above value of 14.72 and find the hypotenuse formed with the other component of acceleration you get 22.43 m/s^2 which is correct.

Last edited:
oddjobmj said:
Thank you for your help TSny.

I am confused about how to find r, r(dot), etc. As you suggest it is not moving in a circle but since the radial and angular velocity are constant it is hard for me to imagine the radius changing. I might just be tired but I can't even seem to remember what r dot and theta dot are. I was under the impression that these were just derivatives but I thought r was constant so the derivative would be 0?

Yes, ##\dot{r}## and ##\dot{\theta}## are just the time derivatives of ##r## and ##\theta##. As ##\theta## increases, so does ##r##. The particle is moving along a spiral of Archimedes

Is it really just: 2*3.2*2.3+3.2*0=14.72 for the tangential portion of acceleration? [STRIKE] i.e. 2*r(dot)*theta(dot)+r*theta'[/STRIKE]

EDIT: Apparently that worked. If you take the above value of 14.72 and find the hypotenuse formed with the other component of acceleration you get 22.43 m/s^2 which is correct.

That's it. Good work.

Dear student,

Thank you for your question. Your attempt at finding the magnitude of acceleration is on the right track, but there are a few things that need to be clarified.

Firstly, the equation a=v^2/r is derived from the centripetal acceleration formula, which is a=v^2/r. This formula is used to calculate the acceleration of an object moving in a circular path. In this case, the particle is moving in a circular path with a constant speed, so the magnitude of its acceleration can be calculated using this formula.

Now, let's take a closer look at your attempt. You calculated the angular velocity in m/s to be 7.447 m/s, but this is incorrect. The angular velocity is given in radians per second, so it should not have units of m/s. Instead, you should use the angular velocity directly in the equation a=v^2/r. This will give you the correct answer.

Also, the period of the particle is not 2*pi/2.3 seconds. The period is the time it takes for the particle to complete one full revolution, which in this case is 2*pi/2.3 seconds. However, we are only interested in the acceleration at a specific point in time, when the particle is at a distance r from the origin. So, the period is not relevant in this case.

Therefore, using the correct angular velocity of 2.3 rad/s and the radius of 3.2 m, we can calculate the acceleration as a=(2.3)^2*3.2=16.768 m/s^2. This is the correct magnitude of acceleration at a distance of 3.2 m from the origin.

I hope this helps to clarify things for you. Keep up the good work in your studies!

Best,

1. What is constant radial and angular velocity?

Constant radial and angular velocity refers to an object moving in a circular path at a constant speed. The radial velocity is the speed at which the object moves towards or away from the center of the circle, while the angular velocity is the speed at which the object rotates around the center.

2. How do you calculate the magnitude of acceleration in a circular motion?

The magnitude of acceleration in a circular motion can be calculated using the formula a = v^2/r, where v is the tangential velocity and r is the radius of the circle. This formula can also be written as a = ω^2r, where ω is the angular velocity. It is important to note that the acceleration in a circular motion is always directed towards the center of the circle.

3. Can the magnitude of acceleration change in a circular motion?

Yes, the magnitude of acceleration can change in a circular motion if the speed or direction of the object changes. For example, if the object speeds up or slows down, the magnitude of acceleration will change accordingly. Additionally, if the object changes direction or moves in a non-uniform circular motion, the magnitude of acceleration will also change.

4. How does centripetal acceleration relate to constant radial and angular velocity?

Centripetal acceleration is the acceleration that keeps an object moving in a circular path. In the case of constant radial and angular velocity, the centripetal acceleration is equal to the magnitude of acceleration and is directed towards the center of the circle. This acceleration is necessary to maintain the object's motion in a circular path.

5. Can the magnitude of acceleration be negative in a circular motion?

Yes, the magnitude of acceleration can be negative in a circular motion. This occurs when the object is slowing down, and the acceleration is directed in the opposite direction of the object's motion. However, the magnitude of acceleration can never be negative if the object is moving at a constant speed, as there is no change in the object's velocity or direction.

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