Question

4. (a) Three point masses are attached to a massless rigid rod. Mass m,-1.0 kg is located at x = 1.0 cm, mass m2-2.0 kg at x = 2.0 cm and mass m,-3.0 kg at x-3.0 m. Find the center of mass of the system. (b) Find the center of mass of the four masses as below. mi 2.0 kg at point (1,2) cm; m 3.0 kg at point (2,-3) cm; m -4.0 kg at point (3,-4) cm and m -5.0 kg at point (4,-5). Solve using a sketch. (3)

Answer 1

Solution) The center of mass is given by

Xcm = (m1x1 + m2x2 + m3x3 + ▪▪▪+mnxn)/(m1 +m2+m3+▪▪mn)

For n particles

The center of mass is given by Xcm = (m1x1 + m2x2 + m3x3 + Ellipsis + mnxn)/(m1 + m2 + m3 + mn) For n particles (a) M_1 = 1kg at X_1 = 1cm = 1 times 10^-2 M M_2 = 2kg at X_2 = 2cm = 2 times 10^-2 M M_3 = 3kg at X_3 = 3M = 3M (Note: Please check M_3 = 3 kg at X_3 = 3m, here 'M' is given but for M_1 and M_2 'X' is in cm. I'm solving this with your given values) Center of mass, X_cm = M_1 X_1 + M_2 X_2 + M_3 X_3/M_1 + M_2 + M_3 X_cm = (1) (1 times 10^-2) + (2) (2 times 10^-2) + (3) (3)/(1 times 10^-2) + (2 times 10^-2) + 3 X_cm = 2.98 M (b) M_1 = 2kg at (X_1, y_1) = (1, 2) cm M_2 = 3kg at (X_2, y_2) = (2, -3) cm M_3 = 4 kg at (X_3, y_3)