# Centripetal acceleration derivation

Discussion in 'Physics' started by logearav, Sep 30, 2011.

1. ### logearav Thread Starter Member

Aug 19, 2011
248
0
Revered members,
Please go through my attachment, which describes the derivation of centripetal acceleration.
Change in velocity along horizontal direction = vcosdθ - v
Change in velocity along vertical direction = v sindθ - 0
change in velocity means final velocity - initial velocity
why V is taken as initial velocity in horizontal direction and 0 is taken as initial velocity in vertical direction?

File size:
7.9 KB
Views:
39
File size:
141.2 KB
Views:
37
File size:
139.2 KB
Views:
47
2. ### BillO Distinguished Member

Nov 24, 2008
985
136
Simply put, because of the chosen initial conditions. Look at the diagram in your first attachment.

3. ### logearav Thread Starter Member

Aug 19, 2011
248
0
Thanks BillO. But i can't figure out what are the initial conditions. Velocity acts tangentially at the points A and B. Thats all i could figure it out.

4. ### BillO Distinguished Member

Nov 24, 2008
985
136
It's where the angle is measured from. They chose to start measuring the angle from point A.

In the attached drawing we have a particle p traveling around the shown circle with velocity v. we can break v down into horizontal and vertical components vx and vy.

Using angle a;

vx=-vsin(a) and vy=vcos(a)

Using angle b:

vx=-vcos(b) and vy=vsin(b)

So sin(b)=cos(a) and cos(b)=sin(a).

Can you see what I mean now?

File size:
5.3 KB
Views:
71
5. ### logearav Thread Starter Member

Aug 19, 2011
248
0
i feel embarrassed to say that i don't understand. My poor basics is the reason for that. I apologize for wasting the time of you, by my stupid questions.
I interpreted in this way, change in velocity in horizontal direction is
vcosdθ - vcos(0) so vcosdθ - v
Similarly change in velocity in vertical direction is vsindθ - vsin(0) which
vsindθ - 0.
I don't know whether i am correct.

6. ### BillO Distinguished Member

Nov 24, 2008
985
136
logerav, there is no such thing as stupid questions, there are only just stupid answers. If you are not getting my answers, it is my failing, not yours.

Give me a little time to think about how I should help with this.

7. ### logearav Thread Starter Member

Aug 19, 2011
248
0
BillO, thanks for your constant support and encouragement. Regarding your diagram in
post #4, i tried to resolve the velocity into its horizontal and vertical components, both for angle 'a' and angle 'b'. Please tell if i proceeded right.
Change in velocity in horizontal direction = vcos(b) - vsin(a)
Change in velocity in vertical direction = vsin(b) - vcos(a).
Now what is the value for the angle 'a' and 'b'?

File size:
244.8 KB
Views:
33
8. ### BillO Distinguished Member

Nov 24, 2008
985
136
Hmmm...

Angles a and b are continuously changing as p moves around the circle. They also follow the relationship;

a + b = pi/2

Now, p is moving around the circle in a counter clockwise direction with angular velocity v.

So, if we want to decompose v down into its x and y components we must choose an angle.

If we choose angle a to do our decomposition, then:

The component of the velocity in the x direction = -vsin(a)
The component of the velocity in the y direction = vcos(a)

This is all we need. You had asked why cos and/or why sin were chosen, and I told you it was because of the angle that was chosen. So to demonstrate this, if, instead of angle a, we happen to choose angle b, then:

The component of the velocity in the x direction = -vcos(b)
The component of the velocity in the y direction = vsin(b)

You do not need both a and b. Just pick one or the other. Either one is fine, it's just that the functions you will use will change as noted above.

9. ### logearav Thread Starter Member

Aug 19, 2011
248
0

Thanks a lot BillO. But, why its negative i.e., -vsin(a) and -vcos(b)?

10. ### BillO Distinguished Member

Nov 24, 2008
985
136
So as to make the direction along the x axis correct. As you can see, when p gets to the top (a=pi/2 or b=0) it is moving directly in the negative direction parallel to the x axis.