Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

(A) Show that the position of a particle on a circle of radius R with center at

ID: 1707590 • Letter: #

Question

(A) Show that the position of a particle on a circle of radius R with center at the origin (x=0,y=0) is given in unit vectors by r=cosi+sinj, where is the angle of the position vector with the x axis.

B) If the particle moves with constant speed v and period T, starting on the x axis at t=0, find an expression for in terms of the time and T.

C) Differentiate the position vector twice with respect to time to find the acceleration and show that it is the centripetal acceleration, whose magnitude is given by equation (a=(v^2)/r) and whose direction is toward the center of the circle. (Not that the unit vectors i and j are constants, independent of time.)

Explanation / Answer

The position of the particle on the circle of a radius is R with center at the origin ( x = 0 , y = 0 ) The position of the particle is r = R ( cos i + sin j )........(1)     x = R cos and y = R sin r = xi + y j squaring on both sides we get               r^2 = x^2 + y^2 b ) If the particle moves with a constant speed v is v = 2r / T                         = 2 / T ( R [ cos i + sin j ] )      2 / T ( R [ cos i + sin j ] )    = R [ -sin (d / dt) i + cos ( d / dt ) j ] comparing i components in the above - sin ( d / dt) = 2 / T           d / dt = - 2 / T ( cot )         tan d = - 2 / T dt Integrating on both sides    l n cos = (2 / T ) t     = cos^-1 [ e^2 / T * t]   c ) differentiating the equation (1 ) twice with respect to x we get     a = d^3 r / dt^2          = - R ^2 [ cos i + sin j ]        a = ^2 R            = ( v/ R)^2 * R                  ( = v / R)            = v^2 / R