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Motion in two simenstions, in a plane can be studied by expressing position, velocity and acceleration as vectors in Cartestion co-ordinatesvec A = A_x hat I +A_y + A_x hat j wherehat I and hat j are unit vector alongx and y directions, respectively andA_x and A_y are corresponding compnents ofvec A (Fig. 1 (EP) .34). Motion can also also be studied by expressing vectors in circular polarco-ordinates as vec A = A_r hat r + A_(theta) hat theta wherevec r = (vec r0/r = cos hat i = sin hat j and hat theta =- in hat i + cos hat j are unti vectors along direction in which r andtheta are increasing. . (a) Expresshat i and hat j in terms ofhat r and theta (b) Show that bothhat r and hat theta are unit vectors and are perpendicular to each other. (c ) Show thatd/9dt) (hat r) = omegahat theta whereomega (dtheta0/(dt) and d/(dt)(hat theta) =- omega hat r (d) For a particle moving along a spiral given byvec r = alpha theta hat r, wherealpha =1(unit), find dimensions ofalpha. (e) Find velocity and acceleration in polar vector representation for particle moving along speral described in (d) above. |
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Answer» Solution :(a) ` hat r = cos theta + sin theta JHAT j` ` hat theta =- sin theta hat I + cos atheta hat j` ….(i) Multipluing (i) by ` sin theta ` and (ii) with ` cos theta ` and adding, we get ….(ii) ` hat e =- sin theta + hat theta cos hteta = (sin^2 theta + cos^@ theta ) hat j = hat j` or hat r sin theta + hat theta cos theta = hat j` ....(III) Mjultiplying (i) by cos theta and (ii) b y sin theta and subtracting, we get ` LTBR. ( hat r cos theta - hat theta sin theta ) = hat i` ....(iv) (b) ` hat i. hat theta = ( hat i + sin theta hat j). (- sin theta hat i + cos theta hat j)` (1) (1)` cos theta =- cos theta sin theta + sin theat cos theta =0= cos 90^@` or ` theta = 90^@` (c ) d/( dr) hat r) = d/(dr) cos theta hat i _+ sin theta hat j) =- sin theta (d theta)/(dt) hat i cos tehta (d theta )/(dt) hat i + cos theta (dtheta )/(dt) hat j = omega [- sin hat i + cos theta hat j ] = omega hat i theta` ltBrgt (d) ` vec r = alpha theta hat r ` or ` | vec r | = alpha theta ` or alpha = (|vec r|)/(theta)` Dimensions of ` alpha L/1 = L = [M^0 L^1 T^0 ]` ltBrgt (E) Given, ` alpha =1 ` unit, from above, we have ` vec r = theta hat r = theta [ cos theta hat i + sin theta hat j]` velocity , ` vec r = (d vec r)/(dt) = ( d theta )/( dt) vec r + theta [ - sin theta hat i + cos theta hat j] ( d theta)/(dt) vec r + theta hat i theta omega` [ :. (d theta )/(dt) = moega]` ` omega hat r + omega hat theta ` ACCELRATION, ` vec a = d/ (dt) [ omega theta hat theta ] = d/(dt) [ (d theta )/(dt) xx ( theta hat theta) ] = ( d^2 theta )/( dt^2)hat r + (d theta )/(dt)(d hat r)/(dt) + (d^2 theta )/( dt^2) xx (d theta )/(dt) d/(dt) ( theta hat theta)` ` = (d^2 theta )/(st^@) hat r + omega d/(dt) [ cos theta hat i + sin theta hat j ] + (d^2 theta )./( dt^2) theta hat theta + omega d/(dt) ( theta hat theta)` `= (d^2 THEA)/(dt^2) hat r omega [- sin heta hat i + cis theta hat j ] (d theta )/( dt) + (d^2 theta )/( dt^2) theta hat theta + omega [(dtehta )/(dt) hat theta + theta )d hat theta)/(dt)]` ` = (d^2 theta )/( dt^2 ) hat r + omega 2 hat theta + (d^2 theta )/(dt^2) xx theta hat theta + omega ^2 hat theta + omega^2 theta (- hat e)``[ :. (d hat theta) /(dt) =- omega hat r]` ` =(d^2 theta )/(dt^2) - omega ^2 theta ) hat r + (2 omega^2 + (d^2 theta )/(dt^2) theta )hat theta`. |
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