Three identical spheres each of mass 2 M are placed at the corners of a right angled triangle with mutually perpendicular sides equal to 4 m each. Taking point of intersection of these two sides as origin, the magnitude of position vector of the centre of mass of the system is (4 √2)/x, where the value of x is ___________ .

Solution

Physics sir · 4 steps

Put the corners on axes, and remember the sphere sitting at the origin still counts in the total mass.

1. Place the spheres. Take the two perpendicular sides along the x and y axes. The spheres sit at (0, 0), (4, 0) and (0, 4), each of mass 2M.
2. Equal masses. All three masses are equal, so 2M cancels. The centre of mass is just the average of the three positions.

   x_(cm) = (0 + 4 + 0)/3 = 4/3 m, y_(cm) = (0 + 0 + 4)/3 = 4/3 m

3. Find the magnitude. |r⃗_(cm)| = √((4/3)² + (4/3)²) = 4√2/3 m.
4. So. Matching with 4√2/x gives x = 3.

The common slip: 2. Dropping the sphere at the origin and dividing by 2 instead of 3 puts the centre of mass at (2, 2), so the magnitude is 4√2/2 and x comes out as 2, not 3.

The idea to keep. For equal masses, the centre of mass is the plain average of the positions. A mass at the origin adds nothing on top, but it still adds to the mass in the denominator.

Answer from the NTA final answer key. Solution by PrepWiser.