CBSE Class 11 Physics Chapter 7: System of Particles and Rotational Motion NCERT Solutions

For objects with uniform mass density, the centre of mass (CM) is located at their respective geometric centres.

(i) For a uniform sphere, the CM is at its geometric centre.

(ii) For a uniform cylinder, the CM is at the midpoint of its axis of symmetry.

(iii) For a uniform ring, the CM is at the centre of the ring.

(iv) For a uniform cube, the CM is at the intersection of its diagonals, which is its geometric centre.

No, the centre of mass of a body does not necessarily lie inside the body. For example, the centre of mass of a uniform ring or a hollow sphere lies at its geometric centre, which is in the empty space enclosed by the object, not within the material itself.

Let the mass of the hydrogen (H) atom be $m$. Then the mass of the chlorine (Cl) atom is $35.5m$. The distance between the nuclei of H and Cl is given as $1.27$ Å.

Let's place the Cl atom at the origin ($x=0$) and the H atom at $x = 1.27$ Å on the x-axis.

The position of the centre of mass ($x_{CM}$) for a system of two particles is given by the formula:

$$x\_\{CM\}\; =\; \backslash frac\{m\_1\; x\_1\; +\; m\_2\; x\_2\}\{m\_1\; +\; m\_2\}$$

Here, $m_1 = 35.5m$ (mass of Cl) and $x_1 = 0$ (position of Cl).

$m_2 = m$ (mass of H) and $x_2 = 1.27$ Å (position of H).

Substituting these values into the formula:

$$x\_\{CM\}\; =\; \backslash frac\{(35.5m)(0)\; +\; (m)(1.27\; \backslash text\{\; \AA \})\}\{35.5m\; +\; m\}$$

$$x\_\{CM\}\; =\; \backslash frac\{0\; +\; 1.27m\}\{36.5m\}$$

$$x\_\{CM\}\; =\; \backslash frac\{1.27\}\{36.5\}\; \backslash text\{\; \AA \}$$

$$x\_\{CM\}\; \backslash approx\; 0.0348\; \backslash text\{\; \AA \}$$

This position is measured from the Cl atom. Therefore, the approximate location of the centre of mass of the HCl molecule is about $0.035$ Å from the Chlorine atom, towards the Hydrogen atom.

The (trolley + child) system is considered. The floor is smooth, implying no external horizontal forces like friction are acting on the system. The trolley is initially moving with a uniform speed $V$. The child is initially stationary relative to the trolley.

The centre of mass (CM) of a system moves with a constant velocity if the net external force acting on the system is zero. In this scenario, the only forces acting are gravity (downwards) and the normal force from the floor (upwards), which balance each other. There are no net horizontal external forces.

When the child gets up and runs about on the trolley, the child exerts forces on the trolley, and the trolley exerts forces on the child. These are internal forces within the (trolley + child) system. Internal forces can change the distribution of mass within the system and the relative motion of its parts, but they cannot change the motion of the centre of mass of the entire system.

Since the net external force on the system is zero, the velocity of the centre of mass of the (trolley + child) system remains unchanged. Therefore, the speed of the CM of the system will continue to be $V$.

Answer: The speed of the CM of the (trolley + child) system remains $V$.