If all the particles of a system lie in X-Y plane, is it necessary that the centre of mass be in X-Y plane?

Open in App

Solution

Yes, if all the particles of a system lie in the X–Y plane, then it's necessary that its centre of mass lies in the X–Y plane.
$z_{cm} = \frac{m_{1} z_{1} + m_{2} z_{2} + . . . m_{n} z_{n}}{\sum_{} m}$

As all the particles lie in the X–Y plane, their z-coordinates are zero.
Therefore, for the whole system, z_cm = 0; i.e., its centre of mass lies in the X–Y plane.

Suggest Corrections

Similar questions

lf all the particles of a system lie in X-Y plane, is it necessary that the centre of mass be in X-Y plane?

Q. If all the particle of a system lie in a cube, is it necessary that the centre of mass be in the cube?

lf all the particles of a system lie in a cube, is it necessary that the centre of mass be in the cube?

Q. The centre of mass is defined as

\vec{R} = \frac{1}{M} \sum_{i} m_{i} \vec{r_{i}}

. Suppose we define "centre of charge" as

{\vec{R}}_{c} = \frac{1}{Q} \sum_{i} q_{i} \vec{r_{i}}

where q_i represents the ith charge placed at

{\vec{r}}_{i}

and Q is the total charge of the system.
(a) Can the centre of charge of a two-charge system be outside the line segment joining the charges?
(b) If all the charges of a system are in X-Y plane, is it necessary that the centre of charge be in X-Y plane?
(c) If all the charges of a system lie in a cube, is it necessary that the centre of charge be in the cube?

The sphere shown in figure lies on a rough plane when a particle of mass m travelling at a speed v0 collides and sticks with it. If the line of motion of the particle is at a distance h above the plane, find the linear speed of the combined system just after the collision. Assume that the mass M of the sphere is large compared to the mass of the particle so that the centre of mass of the combined system is not appreciably shifted from the centre of the sphere.

Join BYJU'S Learning Program