# Vector Mechanics

1. Oct 7, 2011

### ibysaiyan

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

A particle, mass m, moves along the path r = xi + yk where x = Rcos wt and y = Rsinwt where R and W are some constant.
a)What sort of motion is this ?
b)Find a vector expression for the velocity vector.
c) Find a vector expression for the acceleration and hence by ( appropriate substitution) give its magnitude and direction in terms of, vector r.
d) Write down the force vector, f.
e) Show by evaluating the quantity F.v are perpendicular.
f) Physically, what's the quantity F.v .

2. Relevant equations
Dot rule : A.B = mag A . B . cos theta

3. The attempt at a solution

Hi there, First of all I would like to mention that I appreciate any input which will be given on here. My attempt to the aforementioned set of problems has been as following:

a) I said it's periodic or circular motion [ since w is present]
b) For this part we know.. first derivative of vector r, is v. i.e dr/dt = v.

so vector v = -wRsin(wt)+Rwcos(wt)

c)
This is the bit which I found out to be tricky and I may have made some wrong assumptions,mistakes.

Acceleration vector = -w^2Rcos(wt)-w^2 Rsin(wt)

Now from what I have understood is that they want me to get acceleration vector in the form of x's and y's , just as r = xi +yj ...
Am I right ? If so.. by manipulation what I have got is :

a = ( -w^2 xi - w^2 yj)

I am lost beyond this...

P.S: Sorry for my lack of latex notation since I am in a sort of rush to know about this problem.

2. Oct 7, 2011

### Staff: Mentor

I think you are going okay here.

a = ( -w^2 xi - w^2 yj)

Taking out the common factor -w^2 you can write this like that equation for radial acceleration familiar from circular motion.

Tiny quibble: I noticed at the start it was given r = xi + yk .
So just checking: is that k a typo on your part, or is y really along the z axis?
Whatever, make sure you keep to that original orientation, in x-z plane, or x-y plane.

Now, using F=ma a scalar multiplication will give you F and you are as good as finished!

You are doing well.

3. Oct 8, 2011

### ibysaiyan

Continuing onto my working..

a= -w^2(xi -yj) [ sorry it was indeed a typo, i mean't (j) not component k).
So what happens to -w^2 ? am i right to ignore it since it's a constant ?

To get the magnitude of above vector : that'd simply be $\sqrt{(x^2)+ (-y^2)}$

right ? as far as finding out the direction goes.. if i have understoof the question correctly then is it asking for the scalar projrction of vector a onto the direction of vector r...
so using dot rule:
A . R = magnitude of A . mag. of R cos($\theta$
So my final answer is :
mag. of a * cos$\theta$ = $_{a}$ * r/ magnitude of r ...

4. Oct 9, 2011

### ibysaiyan

Anyone ? I suppose I am right..

edit: AH I get it thanks a lot!

5. Oct 9, 2011

### ibysaiyan

Can someone guide me on how to solve part e)
I have tried all sorts of manipulation but i can't get arc. cos to be zero! :(

What I did is : vector f = -mw^2r and vector velocity = xi+yj where x = Rcoswt and y = Rsinwt...
Now I have spend over 5 a4 sheets to proof that they are perpendicular using the dot rule to no avail. :(

6. Oct 9, 2011

### Staff: Mentor

Not right. You have to multiply this by w^2

7. Oct 9, 2011

### Staff: Mentor

Correct.
Wrong. You did not show this.

Look back to where you said:
Look closely at what you have found here. I'll write it out fully with i and j unit vectors:

v=-wRsin(wt)i + wRcos(wt)j

And you'll recognize this is saying:

v=wR(-yi +xj)

(I'll leave it up to you to put the correct squiggles and hats on the vectors and unit vectors.)

Last edited: Oct 9, 2011
8. Oct 10, 2011

### ibysaiyan

Again thanks nascent for your help. This is weird.. maybe I should stop listening to music whilst trying to solve problems. I mean't to type in v=wR(-yi +xj) which is what I have been using for the dot rule.

My problem sheet is due in the next 20 mins. I will have to submit this however once i get back I will re-attempt the question without having it rushed. I think the reason why I wasn't getting costheta = 0 could be related to the constant which I have missed.

Surely if i get the magnitude of f,v, then plug them into the formula I should be able to prove that they are perpendicular to each other.