How Do Rotational Transformations Affect Coordinates?

  • Thread starter Thread starter Fusilli_Jerry89
  • Start date Start date
AI Thread Summary
Rotational transformations maintain the length of objects, as demonstrated by the invariance of the equation x² + y² = x'² + y'². The transformation equations x' = xcos(θ) + ysin(θ) and y' = ycos(θ) - xsin(θ) can be derived using trigonometric principles. By visualizing a circle with rulers positioned along the axes, the relationship between the coordinates and the angle θ becomes clearer. The components x and y represent the sides of a right triangle formed by the rotation. Understanding these transformations is essential for grasping how coordinates change under rotation.
Fusilli_Jerry89
Messages
158
Reaction score
0
Just a quick question:

In my prof's lecture he drew two rotated rulers of the same length and showed that length is invariant under rotations. Obviously x^2 + y^2 = x'^2 + y'^2 = length of ruler, but how do we go from this to:

x' = xcos(theta) + ysin(theta)
y' = ycos(theta) - xsin(theta)

That's what my prof wrote down..
 
Physics news on Phys.org
Fusilli_Jerry89 said:
In my prof's lecture he drew two rotated rulers of the same length and showed that length is invariant under rotations. Obviously x² + y² = x'² + y'² = length of ruler, but how do we go from this to:

x' = xcosθ + ysinθ
y' = ycosθ - xsinθ

Hi Fusilli_Jerry89! :smile:

(have a theta: θ and a squared: ² :smile:)

Just draw a circle with two rulers in it, one along the x-axis, and the other at an angle θ.

Then use trig. :wink:
 
lol i sorry I know this is easy but for some reason I can't figure out where we get the y from. Is that a component? And is it the hypotenuse of something?
 
Fusilli_Jerry89 said:
lol i sorry I know this is easy but for some reason I can't figure out where we get the y from. Is that a component? And is it the hypotenuse of something?

No, y is the other short side of the triangle.

Yes, y is a component, just like x.

The ruler in the circle at an angle θ has its endpoint at x' = 1, y' = 0,

and simple trig shows that that is the same as x = cosθ, y = sinθ. :smile:
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top