Finding Times of Particle Intersections with Equated Vectors

[jono90one]

Homework Statement

Find the times when the two particles meet
r and w are positive constants

Homework Equations

x1(t) = rcos(wt) i + rsin(wt) j
x2(t) = (2/pi) rwt j

The Attempt at a Solution

Now I can get t = 0 quite easily. I can equate :
rcos(wt) = 0 hence wt = pi/2 [1]
rsin(wt) = 2/pi rwt
hence sub in 1
and you get 1=2/pi wt
so basically you get wt = pi/2 again.
So this route is useless (?)

Alternatively, considering x2 forms a right angle with x1 you can do scalar product and you end up with:
r^2 sin(wt) wt = 0
r, w = + ve

So tsinwt = 0
t = 0, sinwt = 0
t = 0 or wt = 0, pi (or multiple) => t=0

But I am still only getting one t, is there only one t? As the question states time"s".

Thanks.

 

[fzero]

cos x is periodic so there are infinitely many points where cos x =0. Your first attempt was correct, you just didn't find all of the points.

 

[jono90one]

So in simple terms the answer is t = 0 and t = anything?
(Seems like an odd question?)

 

[fzero]

So in simple terms the answer is t = 0 and t = anything?
(Seems like an odd question?)

No, t = 0 is not even a solution to cos(wt) = 0. Sketch cos(u) vs u and see if you can write down some values of u where cos u =0. See if you can find a pattern to let you write down all values of u in a formula.

 

[jono90one]

Oh ok, then we're just looking at the x-axis intercepts, which is -3pi/2,-pi/2, pi/2, 3pi/2 (For any range).
There are no restrictions given on the question though :\
wt = ...-3pi/2,-pi/2, pi/2, 3pi/2... (I.e. alternates by pi)

But seeming i don't know w (except for that it is positive), t cannot be found?
Usually with these questions you get simultaneous equations, but not here ><

 

[fzero]

You can express t in terms of w at least. You also need verify which of these solutions satisfy the equation for the \hat{j} components.

 

[jono90one]

Ohh i see now, so basically
for the j components:
Rsinwt=(2/pi)rwt
r's cancle
sin(wt)=(2/pi) wt
Hence sub in some values for wt, pi/2 and -pi/2 produce pi/2 and -pi/2 respectively. But 3pi/2 or any higher multiple cannot exist as you get
rsinwt = nr where n is greater than one (or less than -1)

Hence we have two solutions:
t=pi/2w
t=-pi/2w

Correct?

 

[fzero]

That's correct. You could sketch sin(wt) and (2/pi) wt to see why they only intersect in two points.