# Velocity & Accelerationvectors Help Needed B4 Tomorrow Please!

1. Feb 8, 2006

### Lisa...

NOTE: bold characters are vectors

Could somebody give me some clues in order to solve this problem?
A particle moves in a circle that is centered at the origin. The particle has position r and angular velocity w

a) Show that its velocity is v = w x r
b) Show that its centripetal acceleration is a =w x v = w x (w x r)

I know r= x i + y j + z k , but from now I don't know how to finish solving the problem....

2. Feb 8, 2006

### vaishakh

A - How is velocity defined in circular motion? (Hint l(arc) = R*theta)
A implies B.

3. Feb 8, 2006

### Lisa...

I know how to solve this for scalars, but with vectors I must use the cross product and how do I need to solve this problem that way around?

4. Feb 8, 2006

### vaishakh

You don't know what is cross product?
mod(A cross B) = modA*modB*sintheta where theta is the angle between the vectors. The cross product has the direction mutually perpendicular to the plane containing the vectors given by the right hand rule(keep you hand on vector A and then curl your fingers toward B, Now you thumb denotes the direction of resultant).
So by this definition, i cross j = k, j cross k = i and k cross i = j. Cross product is not commutative. That is any A cross B = -(B cross A) since the magnitude remains the same.

5. Feb 8, 2006

### vaishakh

You don't know what is cross product?
mod(A cross B) = modA*modB*sintheta where theta is the angle between the vectors. The cross product has the direction mutually perpendicular to the plane containing the vectors given by the right hand rule(keep you hand on vector A and then curl your fingers toward B, Now you thumb denotes the direction of resultant).
So by this definition, i cross j = k, j cross k = i and k cross i = j. Cross product is not commutative. That is any A cross B = -(B cross A) since the magnitude remains the same.

6. Feb 8, 2006

### Lisa...

I know what the cross product is, I just don't know how to USE it in THIS problem in order to get to v= w (crossproduct) r. I figured:

w= (d(theta)/dt) * k
r= xi + yj +zk

w (crossproduct) r = (d(theta)/dt) * xj - (d(theta)/dt) * yi

But why would that equal the velocity :S?

Last edited: Feb 8, 2006
7. Feb 9, 2006

### Lisa...

I know that the velocity is the d(distance)/dt and d(distance)= d(theta)*r therefore d(distance)/dt= (d(theta)/dt) * r = w * r but this is no cross product, so how do I get to this formule WITH the cross product in it?

8. Feb 9, 2006

### vaishakh

sory. your mail came after I disconnected yesterday. So I couldn't satisfy the objective of your post as given in the title. This is a good question.
Consider r = xi + bj and if you don't know you can write angular velocity as wk. The z direction is condidered as the direction of angular velocity vector.
We know that the velocity vector is perpendicular tothe raidius vector. So find the dot product. Further we know that the magnitude of velocity is the product of the magnitude of radius vector and angula velocity. The vector you would get as velocity vector would be the same as cross product of w and r.

If you don't get it,
let v = ai + bj
V.r = 0 implies = ax + by implies b = -a(x/y)
mod v^2 = w^2(x^2 + y^2) = a^2 + b^2 = a^2 + a^2(x/y)^2
=(a^2/y^2)(x^2 + y^2)
Thus a = wy and thus b = wx.

Now from this result try yourself proving that Torque = r cross F