Two spheres A and B of mass mA 20 kg and mB 40 kg are conn

Two spheres A and B of mass mA = 2.0 kg and mB = 4.0 kg, are connected by a light rigid rod of length = 39.0 cm. The two spheres are resting on a horizontal frictionless surface when an impulse is applied to A giving it the velocity, Vao = 2.5i (m/s) a) determine the velocity of the center of mass of the system. b) Determine the linear momentum of the system. c) Find the angular momentum relative to the center of the mass. d) Compute the velocities of A and B after the rod has rotated 180 deg.

Solution

a) Vcm = (mAvA + mB vB) / (mA + mB)

Vcm = (2 x 2.5i + 0 ) / (2 + 4 ) = 0.83i m/s

b) linear momentum = Mtotal Vcm = (mA+ mB) Vcm

   = (2 + 4) x 0.833 = 5 kg m / s

c) distance of mA from centre of mass is r then

r = (2 x 0 + 4 x 0.39) / (2 + 4) =0.26 m

angular momentum = m v r = 2 x 2.5 x 0.26 = 1.3 kg m^2 /s

d) suppose rod is rotating with angular speed w about centre of mass then

applying angular momentum conservation,

1.3 = I w

I = (2 x 0.26^2) + (4 x (0.39 - 0.26)^2) = 0.2028 kg m^2

1.3 = 0.2028 w

w = 6.41 rad/s

speed of A = w r = 6.41 x 0.26 = 1.67 m/s

velocity = - 1.67i m/s

speed ofB = 6.41 x (0.39 - 0.26) = 0.83 m/s

velocity = 0.83i m/s