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Nonuniform circular motion

As a generalization of the uniform circular motion case, suppose the angular rate of rotation is not constant. The acceleration now has a tangential component, as shown the image at right. This case is used to demonstrate a derivation strategy based on a polar coordinate system.

Let r (t) be a vector that describes the position of a point mass as a function of time. Since we are assuming circular motion, let r (t) = R · u, where R is a constant (the radius of the circle) and u is the unit vector pointing from the origin to the point mass. The direction of u is described by θ, the angle between the x-axis and the unit vector, measured counterclockwise from the x-axis. The other unit vector for polar coordinates, u is perpendicular to u and points in the direction of increasing θ. These polar unit vectors can be expressed in terms of Cartesian unit vectors in the x and y directions, denoted i ^ {\displaystyle {\hat {\mathbf {i} }}} and j ^ {\displaystyle {\hat {\mathbf {j} }}} respectively: u r = cos θ i ^ + sin θ j ^ {\displaystyle \mathbf {u} _{r}=\cos \theta \ {\hat {\mathbf {i} }}+\sin \theta \ {\hat {\mathbf {j} }}} and u θ = − sin θ i ^ + cos θ j ^. {\displaystyle \mathbf {u} _{\theta }=-\sin \theta \ {\hat {\mathbf {i} }}+\cos \theta \ {\hat {\mathbf {j} }}.}

One can differentiate to find velocity: v = r d u r d t = r d d t (cos θ i ^ + sin θ j ^) = r d θ d t d d θ (cos θ i ^ + sin θ j ^) = r d θ d t (− sin θ i ^ + cos θ j ^) = r d θ d t u θ = ω r u θ {\displaystyle {\begin{aligned}\mathbf {v} &=r{\frac {d\mathbf {u} _{r}}{dt}}\\&=r{\frac {d}{dt}}\left(\cos \theta \ {\hat {\mathbf {i} }}+\sin \theta \ {\hat {\mathbf {j} }}\right)\\&=r{\frac {d\theta }{dt}}{\frac {d}{d\theta }}\left(\cos \theta \ {\hat {\mathbf {i} }}+\sin \theta \ {\hat {\mathbf {j} }}\right)\\&=r{\frac {d\theta }{dt}}\left(-\sin \theta \ {\hat {\mathbf {i} }}+\cos \theta \ {\hat {\mathbf {j} }}\right)\\&=r{\frac {d\theta }{dt}}\mathbf {u} _{\theta }\\&=\omega r\mathbf {u} _{\theta }\end{aligned}}} where ω is the angular velocity dθ / dt.

This result for the velocity matches expectations that the velocity should be directed tangentially to the circle, and that the magnitude of the velocity should be rω. Differentiating again, and noting that d u θ d t = − d θ d t u r = − ω u r , {\displaystyle {\frac {d\mathbf {u} _{\theta }}{dt}}=-{\frac {d\theta }{dt}}\mathbf {u} _{r}=-\omega \mathbf {u} _{r}\,} we find that the acceleration, a is: a = r (d ω d t u θ − ω 2 u r). {\displaystyle \mathbf {a} =r\left({\frac {d\omega }{dt}}\mathbf {u} _{\theta }-\omega ^{2}\mathbf {u} _{r}\right)\.}

Thus, the radial and tangential components of the acceleration are: a r = − ω 2 r u r = − | v | 2 r u r {\displaystyle \mathbf {a} _{r}=-\omega ^{2}r\ \mathbf {u} _{r}=-{\frac {|\mathbf {v} |^{2}}{r}}\ \mathbf {u} _{r}} and a θ = r d ω d t u θ = d | v | d t u θ , {\displaystyle \mathbf {a} _{\theta }=r\ {\frac {d\omega }{dt}}\ \mathbf {u} _{\theta }={\frac {d|\mathbf {v} |}{dt}}\ \mathbf {u} _{\theta }\,} where | v | = r ω is the magnitude of the velocity (the speed).

These equations express mathematically that, in the case of an object that moves along a circular path with a changing speed, the acceleration of the body may be decomposed into a perpendicular component that changes the direction of motion (the centripetal acceleration), and a parallel, or tangential component, that changes the speed.