# Deduction of centripetal force

1. Feb 9, 2010

### Misr

I have questions about the deduction of law of centripetal force

http://img42.imageshack.us/img42/5841/centripetalforce.gif [Broken]
Why do we assume that the change in direction of velocity of the object is V(t+dt)
And similarly the change in direction of radius?

Also what is the triangle of velocity ? and how to draw it?
http://img203.imageshack.us/img203/4669/triangleofvelocity2.gif [Broken]
Do we use only the magnitude of these directions to express the sides triangle of velocity bec. The triangle sides can't be negative while the directions could be +ve or negative?

Thanks

Last edited by a moderator: May 4, 2017
2. Feb 10, 2010

### Staff: Mentor

The velocity (for uniform circular motion) is always tangent to the circle. V(t+dt) is the velocity at time t+dt, not the change in velocity. The change in velocity is V(t+dt) - V(t).

You already have a drawing of the triangle of velocity. It represents the vector addition: V(t+dt) = V(t) + ΔV. The "trick" is realizing that the angle between the two velocity vectors is θ.

When drawing a vector, the length of the arrow represents the magnitude, which is always positive. The direction of the vector is represented by the direction of the arrow.

Read this derivation: http://hyperphysics.phy-astr.gsu.edu/HBASE/cf.html#cf2"

Last edited by a moderator: Apr 24, 2017
3. Feb 10, 2010

### Misr

Thanks this really helped me but actually i can't understand this point

You already have a drawing of the triangle of velocity. It represents the vector addition: V(t+dt) = V(t) + ΔV. The "trick" is realizing that the angle between the two velocity vectors is θ.

----
I have another question
can we prove that the triangle of velocity in the second circle is similar to the triangular cab (ABC) in the first circle?

Thanks again

4. Feb 11, 2010

### Staff: Mentor

Two points:
(1) The angle between the two velocity vectors is θ. To understand that, realize that the velocity vector is always perpendicular to the radius. Thus when the radius moves by an angle θ, so must the velocity vector.
(2) The triangles are similar. In the first circle you have an isosceles triangle with angle θ formed by the radii; In the second circle you have an isosceles triangle with angle θ formed by the velocity vectors. Two isosceles triangles with the same angle are similar.

Another point: For a small change in angle dθ, the length AB (in the first diagram) will equal rdθ.

5. Feb 11, 2010

### Misr

mmm actually i can't imagine can u draw an illustration?
i'll put a proof later to make sure

Thanks so much

Last edited: Feb 11, 2010
6. Feb 11, 2010

### Misr

bec ΔV = V(t+dt) - V(t)
therefore V(t+dt) = V(t) + ΔV ?Right?

7. Feb 12, 2010

### Staff: Mentor

The velocity vector is always a fixed angle (90°) with respect to the radius. If the radius points in the +x direction, the velocity vector points in the +y direction. When the radius has moved to be 30° (for example) above the x-axis, where must the velocity vector point?
Right.

8. Feb 12, 2010

### Misr

Thanks so much
I want to ask if this works ?

http://img215.imageshack.us/img215/3401/proof3f.jpg [Broken]
http://img693.imageshack.us/img693/8039/proofe.jpg [Broken]
http://img705.imageshack.us/img705/5425/proof2t.jpg [Broken]

I'm very sorry 4 the handwriting but i'm in a hurry

Thanks again

Last edited by a moderator: May 4, 2017
9. Feb 12, 2010

### Misr

Also i made an animated picture to demonstrate ur words
http://img30.imageshack.us/img30/1725/centripetalanimated.gif [Broken]
Is my understanding right?

I want to make sure if my two posts are right

Last edited by a moderator: May 4, 2017
10. Feb 12, 2010

### Staff: Mentor

Your animation looks fine to me.

11. Feb 13, 2010

### Misr

Good
I think i got it now
but i have another way and also i want to make sure if its right ?

[url]http://img24.imageshack.us/i/prooffz.jpg/[/url]

IS this right

Thanks so much

12. Feb 14, 2010

### Staff: Mentor

I'm afraid that I don't understand what you are doing in that image. Could you define your triangles and angles, and then describe in words what you are trying to prove?

13. Feb 14, 2010

### Misr

I'm very sorry its my fault I should have explained more
Actually I don't really need to put this proof but I'm just eager to know if its right or wrong
http://img246.imageshack.us/img246/745/centripetalforce2.jpg [Broken]
This pink line is an extension to the velocity vector
And I drew a green radius to help me in the proof
I don't want to prove any thing new I just want to prove that the angle between the velocity vectors is equal to the angle between the two radii .
I understand what u said but I want to make sure if this method also works .

Since the velocity vector is at point b is parallel to ax
Therefore m(<MXY)=m(MXA)=m(<b) = 90 degrees

Therefore XA = XY (According to a theorem in geometry )

And MX is a common side in the two trianles MXY and MXA
Therefore triangle MXY is congruent to triangle MXA

Therefore m(<2)=m(<3)

Since m(<1) = 1/2 m(<AMY) (angle of tangency and centeral angle subtended by the same arc)

Therefore m(<1) = m(<2)
I hope its obvious now
Thanks

Last edited by a moderator: May 4, 2017
14. Feb 14, 2010

### Staff: Mentor

Looks good to me. You can also say that m(<1) = m(<2) since they are both complementary to angle MAX.

15. Feb 15, 2010

### Misr

Yeah this way is better and much more easier
this is a very interesting discussion
nothing is well-illustrated in my txt book

thanks so so much this really helped

16. Feb 15, 2010

### ideasrule

As an aside, there's a much easier derivation of this.

$$\vec v \cdot \vec r =C$$

Derive both sides:

$$\vec a \cdot \vec r + \vec v \cdot \vec v =0$$

Get rid of the vectors:

$$-ar + v^2 = 0$$

17. Feb 15, 2010

### Staff: Mentor

Nice. Combine that with a similar treatment of $\vec v \times \vec r = C$ to show that the acceleration is radial.

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