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

For objects with uniform mass density and symmetrical shapes, 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 of the body 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 atoms is given as $1.27 \text{ Å}$.

We can set up a coordinate system. Let the position of the Cl atom be at the origin ($x_{\text{Cl}} = 0$). Then, the position of the H atom is at $x_{\text{H}} = 1.27 \text{ Å}$.

The formula for the position of the center of mass ($x_{\text{CM}}$) for a system of two particles is:

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

Here, $m_1 = m_{\text{Cl}} = 35.5m$ and $x_1 = x_{\text{Cl}} = 0$. Also, $m_2 = m_{\text{H}} = m$ and $x_2 = x_{\text{H}} = 1.27 \text{ Å}$.

Substituting these values into the formula:

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

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

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

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

This distance is measured from the chlorine atom. Therefore, the approximate location of the center of mass of the HCl molecule is about $0.035 \text{ Å}$ from the chlorine atom, towards the hydrogen atom.

The speed of the centre of mass (CM) of the (trolley + child) system remains unchanged and continues to be $V$.

This is because the trolley is moving on a smooth horizontal floor, implying there are no external horizontal forces acting on the system (trolley + child). The child running about on the trolley exerts forces on the trolley, but these are internal forces within the system. According to Newton's first law and the principle of conservation of momentum, the velocity of the center of mass of an isolated system (a system on which no net external force acts) remains constant. Therefore, regardless of how the child moves within the trolley, the velocity of the CM of the combined system will not change from its initial uniform speed $V$.