Resolution of vectors


by logearav
Tags: resolution, vectors
logearav
logearav is offline
#1
Sep25-11, 04:30 PM
P: 335
1. The problem statement, all variables and given/known data

Revered members, i have attached two images which depicts obtaining an expression for velocity and acceleration of particle executing SHM.

2. Relevant equations

In the first attachment, velocity component vcos[itex]\Theta[/itex] is parallel to vertical diameter of the circle and vsin[itex]\Theta[/itex] is perpendicular to the vertical diameter.
In the second attachment, acceleration component vcos[itex]\Theta[/itex] is perpendicular to vertical diameter and vsin[itex]\Theta[/itex] is parallel to vertical diameter.


3. The attempt at a solution
I know perpendicular component is sine and parallel component is cosine. But i cant understand why perpendicular component to vertical diameter is taken as cosine in acceleration expression and parallel component as sine? Please help revered members, with regard to guidelines for resolving vectors in cosine and sine components.
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
Attached Thumbnails
SHM.png   SHM1.png  
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omoplata
omoplata is offline
#2
Sep26-11, 12:12 AM
P: 315
Hi dude,

Let's take the first picture first. Look at the arrow marked v. Look at the direction it's pointing and what angle it makes with OY. You'll see what they say is correct.

Feel free to ask questions if what I'm saying is not clear.
logearav
logearav is offline
#3
Sep26-11, 03:43 PM
P: 335
Omoplata,
Thanks for the reply.
I dont understand what you say.
My doubt is, the perpendicular line PN is taken as vsin[itex]\Theta[/itex] in measuring velocity
The same perpendicular line PN is taken as cosine component that is v2cos[itex]\Theta[/itex]/a in measuring acceleration.
Why the component changes from sine to cosine wrt to first and second diagram?

omoplata
omoplata is offline
#4
Sep26-11, 08:24 PM
P: 315

Resolution of vectors


That's because the velocity vector is perpendicular to the line PO, and acceleration vector lies along (is parallel to) the line PO.

i.e., the velocity vector is tangential to the circle at point P, while the acceleration vector points to the center of the circle from point P. So velocity and acceleration vectors are perpendicular to each other.
logearav
logearav is offline
#5
Sep27-11, 06:17 PM
P: 335
Thanks for the reply, omoplata. Agreed, velocity and acceleration vectors are perpendicular to each other. But PN is perpendicular to ON, so it should be sine component, Why it is taken as cosine component in the second attachment, i.e acceleration measurement?
omoplata
omoplata is offline
#6
Sep27-11, 09:11 PM
P: 315
Because the angle between the acceleration vector and PN is [itex]\theta[/itex], and the angle between the acceleration vector and ON is [itex]\pi/2 - \theta[/itex].
logearav
logearav is offline
#7
Sep27-11, 11:43 PM
P: 335
90 - [itex]\Theta[/itex] should be the sine component. Am i right?
omoplata
omoplata is offline
#8
Sep27-11, 11:59 PM
P: 315
Quote Quote by logearav View Post
90 - [itex]\Theta[/itex] should be the sine component. Am i right?
If the angle between the acceleration vector and ON is [itex]90^{\circ} - \Theta[/itex], then the component of the acceleration vector along ON is of the magnitude of the original vector times [itex]\sin{\Theta}[/itex], yes.
logearav
logearav is offline
#9
Sep29-11, 04:56 PM
P: 335
Quote Quote by omoplata View Post
That's because the velocity vector is perpendicular to the line PO, and acceleration vector lies along (is parallel to) the line PO.
I understood this point. But, why this property of velocity vector perpendicular to PO gives vsin[itex]\Theta[/itex] for PN( velocity measurement) and acceleration vector parallel to PO gives cosine component for PN?
omoplata
omoplata is offline
#10
Nov24-11, 07:32 PM
P: 315
Hi logearav. I was looking at my old posts and found that I haven't replied to this one. I must have forgotten. Sorry.

Quote Quote by logearav View Post
But, why this property of velocity vector perpendicular to PO gives vsin[itex]\Theta[/itex] for PN( velocity measurement) and acceleration vector parallel to PO gives cosine component for PN?
The angle between the velocity vector and PN is [itex]90^\circ -\theta[/itex] and the angle between the acceleration vector and PN is [itex]\theta[/itex].
logearav
logearav is offline
#11
Nov27-11, 10:37 PM
P: 335
Thanks for your kind gesture omaplata. I understood now.
logearav
logearav is offline
#12
Nov27-11, 10:40 PM
P: 335
Omaplata, one more doubt. Why vsinθ has no effect in measurement of velocity? Only vcosθ has been taken into account. It has been mentioned since vsinθ is perpendicular to the vertical diameter, it has no effect. I dont understand. Could you clarify?
omoplata
omoplata is offline
#13
Nov27-11, 11:17 PM
P: 315
They are just saying that the component of the velocity perpendicular to the vertical diameter, has in turn no component parallel to the vertical diameter. So if you change the component of the velocity perpendicular to the vertical diameter, that has no effect on the component of the velocity parallel to the vertical diameter.

As for WHY that happens, I suppose that is a property of the Euclidean vector spaces. The unit vectors of the basis making up a Euclidean vector space are orthogonal to each other. As for why THAT happens, you'll have to ask someone who knows more than me. My current understanding of Euclidean 3D space is pretty much intuitive and not that rigorous.

I can only help you understand it with an example. Suppose a boat is crossing a river which is perfectly straight and has a constant width (the edges of the river are parallel to each other). The boat crosses the river perpendicular to these edges. That is, it does not move upstream or downstream as it crosses the river. No matter how fast the boat crosses the river, it will not move upstream or downstream. In other words, the speed of the boat, which is perpendicular to the river, will have no effect on its speed upstream or downstream, which is zero (Take the velocity of the boat here to be the perpendicular component of the velocity).
logearav
logearav is offline
#14
Nov29-11, 11:16 PM
P: 335
Thanks omoplata. I raised this query in many forums but i could not get any satisfactory reply. Now with your example, i clearly understood the concept. Persons like you are an asset to this forum. Thanks a lot for your valuable help.
omoplata
omoplata is offline
#15
Nov30-11, 12:17 AM
P: 315
Well, thank YOU for the interesting question.


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