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poles of a magnet always exits in
Answer:
Example: circular motion Edit
See also: Uniform circular motion and Centripetal force
Relation between Cartesian coordinates (x,y) and polar coordinates (r,θ).
For example, consider a particle moving in a circular path. Its position is given by the displacement vector {\displaystyle r=x{\hat {\imath }}+y{\hat {\jmath }}}{\displaystyle r=x{\hat {\imath }}+y{\hat {\jmath }}}, related to the angle, θ, and radial distance, r, as defined in the figure:
{\displaystyle {\begin{aligned}x&=r\cos(\theta )\\y&=r\sin(\theta )\end{aligned}}}{\displaystyle {\begin{aligned}x&=r\cos(\theta )\\y&=r\sin(\theta )\end{aligned}}}
For this example, we assume that θ = t. Hence, the displacement (position) at any time t is given by
{\displaystyle \mathbf {r} (t)=r\cos(t){\hat {\imath }}+r\sin(t){\hat {\jmath }}}{\displaystyle \mathbf {r} (t)=r\cos(t){\hat {\imath }}+r\sin(t){\hat {\jmath }}}
This form shows the motion described by r(t) is in a circle of radius r because the magnitude of r(t) is given by
{\displaystyle |\mathbf {r} (t)|={\sqrt {\mathbf {r} (t)\cdot \mathbf {r} (t)}}={\sqrt {x(t)^{2}+y(t)^{2}}}=r\,{\sqrt {\cos ^{2}(t)+\sin ^{2}(t)}}=r}|\mathbf{r}(t)| = \sqrt{\mathbf{r}(t) \cdot \mathbf{r}(t)}=\sqrt {x(t)^2 + y(t)^2 } = r\, \sqrt{\cos^2(t) + \sin^2(t)} = r
using the trigonometric identity sin2(t) + cos2(t) = 1 and where {\displaystyle \cdot }\cdot is the usual euclidean dot product.
With this form for the displacement, the velocity now is found. The time derivative of the displacement vector is the velocity vector. In general, the derivative of a vector is a vector made up of components each of which is the derivative of the corresponding component of the original vector. Thus, in this case, the velocity vector is:
{\displaystyle {\begin{aligned}\mathbf {v} (t)={\frac {d\,\mathbf {r} (t)}{dt}}&=r\left[{\frac {d\,\cos(t)}{dt}},{\frac {d\,\sin(t)}{dt}}\right]\\&=r\ [-\sin(t),\ \cos(t)]\\&=[-y(t),x(t)].\end{aligned}}}
\begin{align}
\mathbf{v}(t) = \frac {d\,\mathbf{r}(t) }{dt} &= r \left[\frac{d\, \cos(t)}{dt}, \frac{d\, \sin(t)}{dt} \right] \\
&= r\ [ -\sin(t),\ \cos(t)] \\
&= [-y (t), x(t)].
\end{align}
Thus the velocity of the particle is nonzero even though the magnitude of the position (that is, the radius of the path) is constant. The velocity is directed perpendicular to the displacement, as can be established using the dot product:
{\displaystyle \mathbf {v} \cdot \mathbf {r} =[-y,x]\cdot [x,y]=-yx+xy=0\,.}\mathbf{v} \cdot \mathbf{r} = [-y, x] \cdot [x, y] = -yx + xy = 0\, .
Acceleration is then the time-derivative of velocity:
{\displaystyle \mathbf {a} (t)={\frac {d\,\mathbf {v} (t)}{dt}}=[-x(t),-y(t)]=-\mathbf {r} (t)\,.}\mathbf{a}(t) = \frac {d\, \mathbf{v}(t)}{dt} = [-x(t), -y(t)] = -\mathbf{r}(t)\, .
The acceleration is directed inward, toward the axis of rotation. It points opposite to the position vector and perpendicular to the velocity vector. This inward-directed acceleration is called centripetal acceleration