Question

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Discussion #3 1. Consider the motion of an object that can be treated as a point particle and is traveling counter-clockwise in a circle of radius R. This motion can (and will for the purposes of these discussion activities) be described and analyzed using a Cartesian (x-y) coordinate system with a spatial origin at the center of the particle's circular trajectory (the physical path its motion traces out in space). (a) Draw a diagram of the position vector of such an object with positive x and y components (and thus, coordinates since the components of position vectors are by definition coordinates) Determine these c that this position vector makes with the positive x-axis. mates in terms of the radius R of the circle and the angle θ (b) Write the position vector from (a) in ij. k notation. (c) Use calculus to derive this particle's velocity vector (keep in mind which quantities are constant in time and which ones are changing) in terms of R, 0, and do/dt (henceforth do/dt will be referred to as the Greek letter omega: o) in i,j, k notation. (d) What is the magnitude, v, of the velocity vector from (c) in terms of R and o and what is the unit vector tha t points in the direction of motion? (e) A particle's velocity vector is always tangent to its trajectory. In the case of circular motion (though not in general) what does this imply about the relationship between the particle's velocity vector and its position vector when the spatial origin is taken to be the center of the ticle's circular trajectory (as we're doing here) par Assume from this point on that the motion of this particle occurs at a constant speed. (f With this new restriction what can you conclude about o (the rate at which the angle 0 is changing with respect to time)? (g) Use calculus to determine this particle's acceleration vector from its velocity vector. (h) Given the fact that this particle isn't changing speed, does the acceleration come out to be zero? Why or why not? (i) what is the magnitude of the acceleration (which is a vector) in terms of R and ω? In terms of R and v G) Which way does the acceleration vector point (what's the direction of this vector)?

Answer 1

X = R cos theta y = R sin theta (b) R = R cos theta I + R sin theta j (c) V = dR/dt = -R sin theta/dt I + R cos theta d theta/dt j = -R sin theta w I + R cos w j (d) Vertical-bar v Vertical-bar = squareroot (- R sin theta w)^2 + (R cos theta w)^2 = squareroot R^2 w^2 (sin^2 theta + cos^2 theta) R w (e) this implies that the particle's velocity vector and position vector are always perpendicular to each other