for a uniform circular motion, if the position vector is given, how can one calculate the velocity and acceleration ?
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Answer:
In one-dimensional kinematics, objects with a constant speed have zero acceleration. However, in two- and three-dimensional kinematics, even if the speed is a constant, a particle can have acceleration if it moves along a curved trajectory such as a circle. In this case the velocity vector is changing, or [latex] d\overset{\to }{v}\text{/}dt\ne 0. [/latex] This is shown in (Figure). As the particle moves counterclockwise in time [latex] \text{Δ}t [/latex] on the circular path, its position vector moves from [latex] \overset{\to }{r}(t) [/latex] to [latex] \overset{\to }{r}(t+\text{Δ}t). [/latex] The velocity vector has constant magnitude and is tangent to the path as it changes from [latex] \overset{\to }{v}(t) [/latex] to [latex] \overset{\to }{v}(t+\text{Δ}t), [/latex] changing its direction only. Since the velocity vector [latex] \overset{\to }{v}(t) [/latex] is perpendicular to the position vector [latex] \overset{\to }{r}(t), [/latex] the triangles formed by the position vectors and [latex] \text{Δ}\overset{\to }{r}, [/latex] and the velocity vectors and [latex] \text{Δ}\overset{\to }{v} [/latex] are similar. Furthermore, since [latex] |\overset{\to }{r}(t)|=|\overset{\to }{r}(t+\text{Δ}t)| [/latex] and [latex] |\overset{\to }{v}(t)|=|\overset{\to }{v}(t+\text{Δ}t)|, [/latex] the two triangles are isosceles. From these facts we can make the assertion
[latex] \frac{\text{Δ}v}{v}=\frac{\text{Δ}r}{r} [/latex] or [latex] \text{Δ}v=\frac{v}{r}\text{Δ}r. [/latex]
Figure a shows a circle with center at point C. We are shown radius r of t and radius r of t, which are an angle Delta theta apart, and the chord length delta r connecting the ends of the two radii. Vectors r of t, r of t plus delta t, and delta r form a triangle. At the tip of vector r of t, the velocity is shown as v of t and points up and to the right, tangent to the circle. . At the tip of vector r of t plus delta t, the velocity is shown as v of t plus delta t and points up and to the left, tangent to the circle. Figure b shows the vectors v of t and v of t plus delta t with their tails together, and the vector delta v from the tip of v of t to the tip of v of t plus delta t. These three vectors form a triangle. The angle between the v of t and v of t plus delta t is theta.
Figure 4.18 (a) A particle is moving in a circle at a constant speed, with position and velocity vectors at times [latex] t [/latex] and [latex] t+\text{Δ}t. [/latex] (b) Velocity vectors forming a triangle. The two triangles in the figure are similar. The vector [latex] \text{Δ}\overset{\to }{v} [/latex] points toward the center of the circle in the limit [latex] \text{Δ}t\to 0. [/latex]