Question

3. (i) Find the kinetic energy of a particle of mass m with position given by the coordinates (s, u, v), related to the ordinary Cartesian coordinates by y z = 2s + 3 + u = 2u + v = 0+03 (ii) Find the kinetic energy of a particle of mass m whose position is given in cylindrical coordinates = = r cos r sine y (iii) Find the kinetic energy of a particle of mass m with position given in spherical polar coordinates C y z = = = r sin 0 cos o r sin sin o r cos

Answer 1

Page No Solution 3. (il nonst s to y : 20+V = v= du 23+ 35² s to j = 20 tv = v t3u²v now, kinatie ency. I mitt ge 42 ) - Im [(25+33°S+ ) + (2 0+ v)?+(83x2037 m 148² +95 3²+ ²+ 1255 + 6s²sut Ysů +4² + ²+ yurt et quvat 6v²1] n = droso 9 - ajiho 222 2 no 8 cos & + (-8 sino al yz jsind + rose i N. V 3) Rifiam rocoso - Sing 0 ) + (09:16 +20000) + 2) = m loi? cu o + o. Sino 62-2012/06 SMO 21+ o sino na ceste 200 Shade mot 202+

na dros o find Yo & Sirf hin0 2= oroso in & cord sind toloso cosa o orging Sino a y = 8 sinf Sind to cose sino q tosinh loro o 2 = 8 Cos o t (-8) Singö kif (2 cosy Sind toloso cony & -dsind Sinop, + (8 sind sind + 8 cosef Sinoy to sinplore o + Cocoro -o singó) no corp de trei scary 8 ² 48 sing findet + 288 cosf Sino cose 6 - 2 2 Sin Colf Sindcor og - 200 sed lose sind & & dehof hind +80s pinoy & dehid color o tlor couldipsine & + 20 Sirplay 2. dago +208 Sro sino como Ô to coso to 6?-203 Inow! 2 +2²8²8 hind q7

Answer 2

Page 1 - Solzetion : (1) Coordinate of mass m is. Coordinates (5, 40) of position of posticle Selated to the ordinary Cartesiam co-ozdinates by, x= 25+ 3+ u o y=2ut 2 - ④ z=0+ 2 - ③ - - Now, kinetic Enesgy of this pastice in Costesiam Co-ordinates. is, K.E.T= tm[ 0² + y² + ₂2] = 1 mp : V²2 - ²7 y 2 + ₃2 Hese, à = 25+ 35²5 tu ? by differentiationg with he equeftion 70, with respect to time. ż= + 302 Put å ý, i in K.E. ,so we get, 1. KETEL [2+28) 5 +ú)+(4+%)* + i]ht 302)%] T= 1 m [(2+352 3 2 + 2 (2+35) sütý ²+4 4² +4 Uſ + V2 + 22 (1+602+904] KET= 1 m/ 2+35²) 2 ² + 2 (2+35) su +592 +44 +(2+68² + golpo is given in (ii) Position Co-osdinates of particle of mass cylindrical co-ordinates, = 3050 1 Y= sino - 7=2 J -> Kinetic Energy is given by K.E.ET=_m[5? +*+227

Page 2 diffesentiating en with respect to time, we get x = scoso + sl-sino) o x = scoso-esineo V = Śsino + scoso o ₂ = Pret this values in KE equation, we will get T= 1m [åt 4²+ z 2] =&m[ ?030 - 2300so sino so + s sirloo? + 3? sino + 23sino coso só + s? cos?002 + z2 1 T = Lm [ 32 +62 +22] in Position co-ordinates of pastice of mass is given in spherical polas co-ordinates, a = sino cos 7 y = 8 simo siop y -® z = sose — NOW K.E. T= Im[52 + y2 +227 Now, differentiating ear with sespect to eto time we get a = s sinoos + 3 OSO O OS + s sino (-sinod sino sind + SCOSO O sind + ssimo (oso z = $coso - ssino o port this equation in Kinetic energy sospnika, then we get

Page - 3 =lam + +2] Im[ssimo c032° +8} ocos000326 +32? sintesino +255ô sino cose cos? - 28200 sino caso sindros -2350 sino sino cosø + ? sinto simpo +?o?ošosing +s? sinto 0320 02 +225 ósino coso siopo +28200 sino Oso sind (oso +233 o sino sindroso +3200520 + s?sio? o 02 - 2330sino caso 7 T = {m[+3?0? +s?sm?0 02]