Question

V

Answer 1

a) center of mass - (xcm,ycm)   [initially assuming rectangle is stationary]

xcm = (2*2 + 2*4 + (-2)*2 + (-2)*3) / (2 + 4 + 2 + 3) = 2/11 m

Similarly,

ycm = (3*2 + (-3)*4 + (-3)*2 + 3*3) / (2 + 4 + 2 + 3) = -3/11 m

Therefore, initially center of mass is at (2/11,-3/11)

Now, if the system rotates about the origin, the center of mass also rotates abut the origin in a circle with radius r

r = sqrt( (2/11)^2 + (-3/11)^2 ) = sqrt(13)/11 m

b) moment of inertia

since all masses are at same distance from the origin (the point of rotation)

I = total mass x r^2 = (2+4+2+3) * (sqrt(13) )^2

I = 143 kg-m^2

c) rotational kinetic energy,

K.E.(rot) = (1/2) * I * w^2 = (1/2) * (143) * 6^2 = 2574 J

d) Force on 4 kg particle

force due to 2 kg particle ( first quadrant)   = G*2*4/6^2 (j) = 2G/9 (j)

force due to 2 kg particle ( third quadrant)   = G*2*4/4^2 (-i) = G/2 (-i)

force due to 3 kg particle ( second quadrant)   = G*3*4/52   ( (-2i + 3j) / sqrt(13) ) = 3G/13 ( (-2i + 3j) / sqrt(13) )

total force = -0.628G (i) + 0.414G (j)

magnitude = sqrt( 0.628^2 + 0.414^2)G = 0.752G = 0.752 x 6.673 x 10^-11 N = 5 x 10^-11 N