Title Notes_Exercise_09.Center Of Mass.pdf ✅

9.1 SOLUTIONS TO CONCEPTS CHAPTER 9 1. m1 = 1kg, m2 = 2kg, m3 = 3kg, x1 = 0, x2 = 1, x3=1/2 y1 = 0, y2 = 0, y3 = 3 / 2 The position of centre of mass is C.M =                 1 2 3 1 1 2 2 3 3 1 2 3 1 1 2 2 3 3 m m m m y m y m y , m m m m x m x m x =                       1 2 3 1( )0 2( )0 3( ( 3 / 2)) , 1 2 3 1( )0 2( )1 3( /1 )2 =         12 3 3 , 12 7 from the point B. 2. Let  be the origin of the system In the above figure m1 = 1gm, x1 = – (0.96×10–10)sin 52° y1 = 0 m2 = 1gm, x2 = – (0.96×10–10)sin 52° y2 = 0 x3 = 0 y3= (0.96 × 10–10) cos 52° The position of centre of mass                 1 2 3 1 1 2 2 3 3 1 2 3 1 1 2 2 3 3 m m m m y m y m y , m m m m x m x m x =                      18 0 0 16y , 1 1 16 .0( 96 10 ) sin52 .0( 96 10 ) sin52 16 0 3 10 10 =     10 o ,0 8 / 9 .0 96 10 cos52   3. Let ‘O’ (0,0) be the origin of the system. Each brick is mass ‘M’ & length ‘L’. Each brick is displaced w.r.t. one in contact by ‘L/10’ The X coordinate of the centre of mass 7m 2 L m 10 L 2 L m 10 L 10 3L 2 L m 10 3L 2 L m 10 2L 2 L m 10 L 2 L m 2 L m Xcm                                                        = 7 2 L 10 L 2 L 5 L 2 L 10 3L 2 L 5 L 2 L 10 L 2 L 2 L            = 7 5 2L 10 5L 2 7L   = 10 7 35L 5L 4L    = 70 44L = L 35 11 4. Let the centre of the bigger disc be the origin. 2R = Radius of bigger disc R = Radius of smaller disc m1 = R2 × T ×  m2 = (2R)2 I T ×  where T = Thickness of the two discs  = Density of the two discs  The position of the centre of mass (0, 0) B C 1m (1, 1)         2 3 , 2 1 1m A 1m X Y 0.96×10–10m H m2 (0, 0) m1 Q 104° 52° 52° H m3 0.96×10–10m O X L L/10 m1 (0, 0) O m2 R (R, 0) Page 1 CENTRE OF MASS