Q.Two uniform rods A and B of lenghts 5m and 3m are placed end to end.If their linear densities are 3kg/m and 2kg/m , the position of their center of mass from their interface is
1.19/14 m on the side of heavier rod
2.8/7 m on the side of lighter rod
3.2m on side of heavier rod
4.2m on side of lighter rod
Q.A uniform disc of radius R is put over another uniform disc of radius 2R of same thickness and density.The peripheres of the two discs touch each other.The position of the center of the mass from center of bigger towards center of smaller is
Q.Particles of masses m,2m,3m...nm grams are placed on the same line at distances l,2l,3l....nl cm from a fixed point.The distance of center of mass from the fixed point is
Q.The mass of a uniform ladder 5m is 20kg.A person of mass 60kg stands on the ladder at a height of 2m from the bottom .The position of center of mass from the bottom nearly is a.1m b.2.5m c.3.5m d.2.125 m
Q.From a sphere of radius 1m , a sphere of radius 0.5 is removed from the edge.The shift in the C.M is a.13/16 b.16/13 c.14/13 d.1/14
Please show how you with the formulae too.Thanks a lot.
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2 Answers
1Q3 done in similar way as this
Particles of masses 1g,2g,3g,............100g are kept at the marks 1cm,2cm,3cm...........100cm respectively on a metre scale.Find the moment of inertia of the system of particles about a perpendicular bisector of the metre scale.
Sol:moment of inertia abt perpendicular bisector
= m1 r1^2 + m2r2^2 + m3r3^2 + .....
=1 * 49^2 + 2 * 48^2 +.....49*1^2 + 50*0^2 + 51*1^2 + .............99*49^2 + 100 * 50^2
combining the first and the second last, second and the third last , .....
we get
MI=100 * 50^2 + (1+99) * 49^2 +............(49+51)*1^2
=100(50^2 + 49^2 +.........1^2)
=100 n(n+1)(2n+1) / 6 where n= 50
=100 *50*51*101 / 6
=4292500 g cm^2
=0.42925 kg m^2
1Q4
ycm=m1y1+m2y2/m1+m2
=20*5/2 + 60*2/(5/2+2
=50+120/ 9/2
=170*2/9
=340/9
maybe calculation mistake