System of Particles and Rotational Motion NCERT Solutions, Complete PDF Content and Important Questions

Answer: (i) Sphere: Center of the sphere. (ii) Cylinder: Mid-point of its geometrical axis. (iii) Ring: Center of the ring. (iv) Cube: Point of intersection of its diagonals. No , the center of mass does not necessarily lie inside the body. For example, the center of mass of a uniform ring or a hollow sphere lies at its geometric center, which is an empty space outside the actual material of the body.

Answer: Let the mass of the Hydrogen atom be $m$. Then, the mass of the Chlorine atom = $35.5 m$. Let us place the Hydrogen atom at the origin ($x_1 = 0$) and the Chlorine atom at $x_2 = 1.27 \text{ \AA}$. Using the center of mass formula for a two-particle system: $$X_{cm} = \frac{m_1x_1 + m_2x_2}{m_1 + m_2}$$ $$X_{cm} = \frac{m(0) + 35.5m(1.27)}{m + 35.5m}$$ $$X_{cm} = \frac{35.5m \times 1.27}{36.5m} = \frac{35.5 \times 1.27}{36.5} \approx 1.235 \text{ \AA}$$ Result: The center of mass of the HCl molecule lies approximately $1.235 \text{ \AA}$ from the Hydrogen nucleus (towards the Chlorine atom).

Answer: The child and the trolley constitute a single system. When the child gets up and runs on the trolley, the forces involved are purely internal forces . According to Newton's laws and the principles of the center of mass, internal forces cannot change the velocity of the center of mass of a system. Since there is no net external force acting on the system in the horizontal direction, the velocity of the center of mass will remain unchanged. Result: The speed of the CM of the (trolley + child) system will remain $V$ .

Answer: Let two vectors $\vec{a}$ and $\vec{b}$ represent the two adjacent sides of a triangle, with an angle $\theta$ between them. The magnitude of the cross product is given by: $$|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin\theta$$ In the triangle formed by $\vec{a}$ and $\vec{b}$, if we take the base as $|\vec{a}|$, the perpendicular height $h$ of the triangle is $|\vec{b}| \sin\theta$. Area of a triangle = $\frac{1}{2} \times \text{Base} \times \text{Height}$ $$Area = \frac{1}{2} \times |\vec{a}| \times (|\vec{b}| \sin\theta)$$ $$Area = \frac{1}{2} |\vec{a} \times \vec{b}|$$ Hence proved.

Answer: Let $\vec{a}, \vec{b}$, and $\vec{c}$ represent the three coterminous edges of a parallelepiped. The cross product $(\vec{b} \times \vec{c})$ gives a vector whose magnitude represents the area of the base of the parallelepiped, and its direction is perpendicular to the base (let's call this direction $\hat{n}$). Area of base $A = |\vec{b} \times \vec{c}|$. Now, take the dot product of $\vec{a}$ with this area vector: $$\vec{a} \cdot (\vec{b} \times \vec{c}) = \vec{a} \cdot (A \hat{n}) = |\vec{a}| A \cos\theta$$ Here, $|\vec{a}| \cos\theta$ is the projection of vector $\vec{a}$ along the normal to the base, which is exactly the vertical height ($h$) of the parallelepiped. Therefore: $$\vec{a} \cdot (\vec{b} \times \vec{c}) = A \times h = \text{Volume of the parallelepiped}$$ Hence proved.

Answer: Given: $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ and $\vec{p} = p_x\hat{i} + p_y\hat{j} + p_z\hat{k}$ Angular momentum $\vec{l} = \vec{r} \times \vec{p}$ $$\vec{l} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ x & y & z \\ p_x & p_y & p_z \end{vmatrix}$$ Expanding the determinant: $$\vec{l} = \hat{i}(yp_z - zp_y) - \hat{j}(xp_z - zp_x) + \hat{k}(xp_y - yp_x)$$ So, the components are: $l_x = yp_z - zp_y$ $l_y = zp_x - xp_z$ $l_z = xp_y - yp_x$ If the particle moves only in the x-y plane, then its z-coordinate is zero ($z = 0$) and its momentum in the z-direction is zero ($p_z = 0$). Substitute $z = 0$ and $p_z = 0$ in the components: $l_x = y(0) - (0)p_y = 0$ $l_y = (0)p_x - x(0) = 0$ $l_z = xp_y - yp_x$ Result: Since $l_x = 0$ and $l_y = 0$, the angular momentum has only a z-component. Hence proved.

Answer: Let the two parallel lines be parallel to the x-axis, separated by a distance $d$. Velocity of particle 1: $\vec{v_1} = v\hat{i}$ Velocity of particle 2: $\vec{v_2} = -v\hat{i}$ Let us take an arbitrary origin $O$ having coordinates $(0, 0)$. Let the position vector of particle 1 be $\vec{r_1} = x_1\hat{i} + y_1\hat{j}$. Let the position vector of particle 2 be $\vec{r_2} = x_2\hat{i} + y_2\hat{j}$. The separation between the lines is $d$, so $|y_1 - y_2| = d$. Total angular momentum $\vec{L} = \vec{L_1} + \vec{L_2} = (\vec{r_1} \times \vec{p_1}) + (\vec{r_2} \times \vec{p_2})$ $$\vec{L} = (x_1\hat{i} + y_1\hat{j}) \times (mv\hat{i}) + (x_2\hat{i} + y_2\hat{j}) \times (-mv\hat{i})$$ Using cross product rules ($\hat{i} \times \hat{i} = 0$, $\hat{j} \times \hat{i} = -\hat{k}$): $$\vec{L} = (0 - mvy_1\hat{k}) + (0 + mvy_2\hat{k})$$ $$\vec{L} = -mv(y_1 - y_2)\hat{k}$$ Since $|y_1 - y_2| = d$, the magnitude of total angular momentum is $|\vec{L}| = mvd$. Result: The expression $mvd$ does not depend on $x_1, x_2$, or the chosen origin. Thus, the angular momentum is independent of the point about which it is taken.

Answer: D) Angular velocity. Explanation: Moment of inertia $I = \sum m_ir_i^2$. It is an inherent property based on mass and geometry. It does not depend on how fast the body is rotating ($\omega$).

Answer: B) It decreases. Explanation: By the conservation of angular momentum ($L = I\omega = \text{constant}$), when she stretches her arms, her moment of inertia ($I$) increases. Therefore, her angular velocity ($\omega$) must decrease.

Answer: C) Null vector. Explanation: $\vec{A} \times \vec{A} = |\vec{A}||\vec{A}|\sin(0^\circ)\hat{n} = 0$.

Answer: Hollow spokes push the mass further away from the axis of rotation, increasing the moment of inertia without adding extra weight. A higher moment of inertia ensures uniform, stable motion and prevents the wheel from wobbling.

Answer: It states that if the net external torque acting on a system is zero, the total angular momentum of the system remains constant in both magnitude and direction. ($\vec{\tau}_{ext} = 0 \implies \vec{L} = \text{constant}$).

Answer: Yes, it is absolutely possible. The center of mass is a geometrical point representing average mass distribution. For a uniform circular ring or a hollow sphere, the center of mass lies exactly at the center, which is an empty space outside the physical material.

Answer: Yes. If two equal and opposite forces act on a body at different points (forming a couple), the net force is zero (translational equilibrium), but the net torque is not zero, causing the body to rotate (not in rotational equilibrium).

Answer: By definition, angular momentum is $\vec{L} = \vec{r} \times \vec{p}$. Differentiating both sides with respect to time $t$: $$\frac{d\vec{L}}{dt} = \frac{d}{dt}(\vec{r} \times \vec{p})$$ Using the product rule of differentiation: $$\frac{d\vec{L}}{dt} = \left(\frac{d\vec{r}}{dt} \times \vec{p}\right) + \left(\vec{r} \times \frac{d\vec{p}}{dt}\right)$$ We know that $\frac{d\vec{r}}{dt} = \vec{v}$ (velocity) and $\vec{p} = m\vec{v}$. Also, $\frac{d\vec{p}}{dt} = \vec{F}$ (force). $$\frac{d\vec{L}}{dt} = (\vec{v} \times m\vec{v}) + (\vec{r} \times \vec{F})$$ Since the cross product of parallel vectors ($\vec{v} \times \vec{v}$) is zero: $$\frac{d\vec{L}}{dt} = 0 + (\vec{r} \times \vec{F})$$ Since $\vec{r} \times \vec{F} = \vec{\tau}$ (Torque): $$\frac{d\vec{L}}{dt} = \vec{\tau}$$ This shows that the rate of change of angular momentum is equal to the external torque applied.

Answer: Given: $m = 20$ kg, $\omega = 100$ rad/s, $R = 0.25$ m. Moment of inertia of a solid cylinder about its axis: $$I = \frac{1}{2}mR^2 = \frac{1}{2} \times 20 \times (0.25)^2 = 10 \times 0.0625 = 0.625 \text{ kg m}^2$$ Rotational Kinetic Energy: $$K = \frac{1}{2}I\omega^2 = \frac{1}{2} \times 0.625 \times (100)^2 = \frac{1}{2} \times 0.625 \times 10000 = 3125 \text{ J}$$ Angular Momentum: $$L = I\omega = 0.625 \times 100 = 62.5 \text{ J s}$$

Answer: Condition for rolling without slipping: The point of contact of the rolling body with the surface must be instantaneously at rest. This implies $v = R\omega$, where $v$ is linear velocity of the COM, $R$ is radius, and $\omega$ is angular velocity. Kinetic Energy: A rolling body possesses both translational and rotational kinetic energy. Total Kinetic Energy ($K$) = Translational KE + Rotational KE $$K = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$$ Let $I = mk^2$ (where $k$ is the radius of gyration). Also substitute $\omega = \frac{v}{R}$: $$K = \frac{1}{2}mv^2 + \frac{1}{2}(mk^2)\left(\frac{v}{R}\right)^2$$ $$K = \frac{1}{2}mv^2 \left( 1 + \frac{k^2}{R^2} \right)$$

Answer: (a) $\tau = r \times F = (20 \times 10^{-2} \text{ m}) \times 50 \text{ N} = 10 \text{ N m}$. (b) We need $\tau = 10 \text{ N m}$, with $F' = 25 \text{ N}$. Let the new length be $r'$. $10 = r' \times 25 \implies r' = \frac{10}{25} = 0.4 \text{ m} = 40 \text{ cm}$.

Answer: Torque ($\tau = r \times F$) depends directly on the perpendicular distance ($r$) from the axis of rotation (hinges). By pushing at the edge, $r$ is maximized, requiring less force to produce the necessary rotational torque to open the door.

Answer: (A) Both A and R are true, and R is the correct explanation for A.

Answer: (D) A is false, but R is true. Even if torque is zero (Angular momentum $L = I\omega$ is constant), the angular velocity ($\omega$) can change if the moment of inertia ($I$) changes (e.g., a diver tucking their body).