# Intro to Uniform Circular Motion/Centripetal Acceleration

1. Nov 8, 2015

### a lone fishy

Hi everyone,

1. The problem statement, all variables and given/known data

2. Relevant equations

1) centripetal acceleration = v^2 / R
2) centripetal acceleration = 4 pi^2 R / T^2
3) average velocity = displacement / time

3. The attempt at a solution

For #1a I first found the displacement of A to B. And then divided that by the speed to get time. However I feel like im doing it wrong. Perhaps we need to find the centripetal acceleration first using equation 1 and then use equation 2 to find time. But then that would give the time of going around the entire circle. A bit torn on what do for 1a

For #1b I'm not sure where to put the angle. We just need to find the change in velocity and then divide it by the time we found in 1a. I just can't seem to pin point which angle I need to use and where I have to put it.

For #2a I used equation 1 to find the radius. And then used equation 2 to find the time. Since we're only dealing with a quarter of a circle, I divided the time by 4. I believe I did #2a correctly

For #2b I found the the change of velocity and then divided it by the time we found in 2a. I don't think I found the change of velocity correctly

For #3a I first found the radius with equation 2. And then I found the displacement with pythagorean theory. Now that we have displacement, I divided it by the given time to get the average velocity

For #3bI got a little confused. To find the change in velocity, we need to do v2 - v1 however we don't know any velocities so I just went with speed which I believe is incorrect. I used equation 1 to find speed and then found the change in velocity using speed.

I attached my solution attemps below. Thank You !!

2. Nov 8, 2015

### BvU

Hello Fishy, welcome to PF !

You''ll get used to the culture around here, especially when you have read the guidelines -- and follow them a bit more.

So three threads for three exercises, and a minimum of photographed own work (only the pictures, not the workings)

Furthermore, if you don't tell what you've done and what you got, helping isn't made easy for us. E.g. in 1a): what was your reasoning, what did yuo calculate and what was the result ? What you wrote first looks quite reasonable, so why not tell us ? -- Unless you mean the handwritten stuff, but there you calculate the distance from A to B over a straight line, which can't be right indeed: the car follows the circular trajectory !

 Oh, and exercise looks a bit garbled to me; did teacher mix up two different exercises ? plane and car ?

Last edited: Nov 8, 2015
3. Nov 8, 2015

### a lone fishy

Ah I see my mistake sorry!

Ill delete this thread and make a new one !

Also right above the pictures at the bottom, I have written down what I thought was correct for each question

Edit: Can't find the delete thread button if there is one

4. Nov 8, 2015

### BvU

No need to delete the thread - let's continue with what there is now, otherwise it'll be a lot of work for the helper too !

5. Nov 8, 2015

### a lone fishy

Ok.

For 1.a I believe I can find time once I get distance and speed. We have speed given however Im not sure how to find the distance.
Can we do the following?:

We know that distance is 2πR so therefore we can find the time of the entire circle. We know the entire circle is 360 degrees however we just want 68 degrees of the circle. Therefore could we just do 360 divided by 68? And then divide the total time by that value?

It holds some logic to me but Im not sure if that's the correct way of doing it.

Thanks

6. Nov 8, 2015

### BvU

Length of an arc of 68$^\circ$ on a circle with radius 50 m should not pose a problem ! If the whole circle is $2\pi R$ that $2\pi$ is the angle. If the angle is $\alpha$ the arc is $\alpha R$ ! (with $\alpha$ in radians !!!)

for b) you want to use the definition of average acceleration: $\ \vec a_{\rm \,avg} \equiv {\Delta \vec v \over \Delta t}\$.

Delta v is the change in v, so $\vec v_2 - \vec v_1$. It has a magnitude and a direction.

7. Nov 8, 2015

### a lone fishy

Ah I never thought about doing that.

I did: 68° × π/180 × 50 = 59.34m

59.34/18 = 3.3 s for the time

However for 1b Im still unsure on what to do. We need velocity but we only have speed

8. Nov 9, 2015

### BvU

Speed is the same, only: the direction has changed !
In your left drawing I see on the left $v_2$ and on the right $-v_1$. Now add them up, as you do in your right drawing. For the ? you know two magnitudes (18 m/s) and one angle (68$^\circ$). Bob's your uncle !

9. Nov 9, 2015

### a lone fishy

I still don't understand where to put the angle. Are we even supposed to use the 68 degree angle

10. Nov 9, 2015

### BvU

Can you describe the velocity vector at A in Cartesian coordinates ?
Let x+ = East and y+ = North

11. Nov 10, 2015

### a lone fishy

nevermind i got it solved now. thx