Vectors and the Laws of Physics — Question 1

Question 1

Rotate the axes without moving the arrow

A displacement has components $(6,8)$ m in the original axes. The new axes turn counterclockwise by $30^\circ$. For perpendicular axes rotated counterclockwise by $\phi$ about the same origin, a fixed vector transforms as $A_{x\prime}=A_x\cos\phi+A_y\sin\phi$ and $A_{y\prime}=-A_x\sin\phi+A_y\cos\phi$.

1. Find the new components exactly and approximately.

2. Check the magnitude in both coordinate systems.

3. Compare the direction angle measured from each positive horizontal axis. Has the physical displacement changed?

Question 1 - Solution

Rotate the axes without moving the arrow

(a) Resolve onto the new directions.

$$\boxed{\vec A'=(3\sqrt3+4,4\sqrt3-3)\,\mathrm m}\approx(9.196,3.928)\,\mathrm m$$ $A_{x\prime}=3\sqrt3+4\approx9.196$ m and $A_{y\prime}=4\sqrt3-3\approx3.928$ m. Both remain positive because the arrow lies between the two new positive axes.

(b) Check the invariant length.

$$A'^2=(27+24\sqrt3+16)+(48-24\sqrt3+9)=100\,\mathrm{m^2}=A^2$$ The original length is 10 m. In the new system, $(3\sqrt3+4)^2+(4\sqrt3-3)^2=100\ \mathrm{m^2}$; the mixed terms cancel. The new magnitude is also 10 m.

(c) Subtract the rotation of the reference axis.

$$\theta'=\theta-\phi=\tan^{-1}(8/6)-30^\circ=23.130^\circ$$ The old angle is $\tan^{-1}(8/6)\approx53.130^\circ$. The new angle is $53.130^\circ-30^\circ=23.130^\circ$. Only the coordinate description changed. The vector still joins the same physical endpoints, with the same length and orientation in the room.