System of Particles and Rotational Motion Previous Year Questions — JEE Main Physics
58 past-year System of Particles and Rotational Motion questions from JEE Main (Physics) — page 6.
Q17 (2025)
A uniform rod of mass $250\mathrm{g}$ having length $100\mathrm{cm}$ is balanced on a sharp edge at $40\mathrm{cm}$ mark[cite: 150, 151]. A mass of $400\mathrm{g}$ is suspended at $10\mathrm{cm}$ mark. To maintain the balance of the rod, the mass to be suspended at $90\mathrm{cm}$ mark, is [cite: 154, 156]
- $300\mathrm{g}$
- $190\mathrm{g}$
- $200\mathrm{g}$
- $290\mathrm{g}$
### Related Formula
For rotational equilibrium, the \sum of all counter-clockwise torques about the pivot point must exactly balance the \sum of all clockwise torques:
$$\sum \tau_{\text{pivot}} = 0 \implies \sum (m_i \cdot g \cdot x_i) = 0$$
### Core Logic
The rod is uniform, meaning its mass ($250\text{ g}$) acts exactly at its geometric center of mass, the $50\text{ cm}$ mark[cite: 150, 151]. Let the pivot point be the sharp edge at the $40\text{ cm}$ mark .
Calculate the relative lever arms from the pivot [cite: 775, 776, 777]:
* $400\text{ g}$ mass at $10\text{ cm}$ mark: lever arm $= 40 - 10 = 30\text{ cm}$ (counter-clockwise)
* $250\text{ g}$ rod mass at $50\text{ cm}$ mark: lever arm $= 50 - 40 = 10\text{ cm}$ (clockwise)
* Unknown mass $M$ at $90\text{ cm}$ mark: lever arm $= 90 - 40 = 50\text{ cm}$ (clockwise)
Setting up the torque balance equation:
$$400 \times 30 = (250 \times 10) + (M \times 50)$$
$$12000 = 2500 + 50M$$
$$50M = 9500 \implies M = \frac{9500}{50} = 190\text{ g}$$
### Step 1: Visual Context
The structural layout of forces acting on the balanced rod system is shown below:
{{SOL_IMG1}}
### Pattern Recognition
Never forget to include the weight of a uniform rod itself in equilibrium equations. It is a common oversight to omit the rod's mass, which always acts at its geometric center.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: System of Particles and Rotational Motion
Q1296 (2025)
The coordinates of a particle with respect to origin in a given reference frame is (1, 1, 1) meters. If a force of $\vec{\mathrm{F}} = \hat{\mathrm{i}} -\hat{\mathrm{j}} +\hat{\mathrm{k}}$ acts on the particle, then the magnitude of torque (with respect to origin) in $z$ -direction is
### Related Formula
$$\vec{\tau} = \vec{r} \times \vec{F}$$
### Core Logic
Given position vector $\vec{r} = \hat{i} + \hat{j} + \hat{k}$ and force $\vec{F} = \hat{i} - \hat{j} + \hat{k}$:
$$\vec{\tau} = \left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & -1 & 1 \end{array} \right|$$
### Step 1: Isolate z-component
$$\tau_z = \hat{k}(1(-1) - 1(1)) = -2\hat{k}$$
The absolute magnitude of the torque component in the $z$-direction equals $2 \mathrm{~N \cdot m}$.
### Chapter Mix
Class 11 Physics: System of Particles and Rotational Motion
Q1376 (2024)
The identical spheres each of mass 2M are placed at the corners of a right angled triangle with mutually perpendicular sides equal to $4\mathrm{~m}$ 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 $\frac{4\sqrt{2}}{x}$, where the value of $x$ is ______.
### Related Formula
Position vector of the Centre of Mass (COM):
$$\vec{r}_{\text{COM}} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2 + m_3\vec{r}_3}{m_1 + m_2 + m_3}$$
### Core Logic
Assign coordinates to the three masses ($m_1=m_2=m_3=2M$):
- Origin mass: $\vec{r}_1 = 0\hat{i} + 0\hat{j}$
- X-axis mass: $\vec{r}_2 = 4\hat{i} + 0\hat{j}$
- Y-axis mass: $\vec{r}_3 = 0\hat{i} + 4\hat{j}$
Substitute these into the COM formula:
$$\vec{r}_{\text{COM}} = \frac{2M(0) + 2M(4\hat{i}) + 2M(4\hat{j})}{2M + 2M + 2M} = \frac{8M\hat{i} + 8M\hat{j}}{6M} = \frac{4}{3}\hat{i} + \frac{4}{3}\hat{j}$$
### Step 1: Calculate Position Vector Magnitude
$$|\vec{r}_{\text{COM}}| = \sqrt{\left(\frac{4}{3}\right)^2 + \left(\frac{4}{3}\right)^2} = \frac{4\sqrt{2}}{3}$$
Matching this directly with the given template $\frac{4\sqrt{2}}{x}$ shows that $x = 3$.
### Pattern Recognition
Since the mass layout is completely symmetric along both right-angle legs, the COM coordinates are identical ($x_{\text{COM}} = y_{\text{COM}}$).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: System of Particles and Rotational Motion
Q60 (2024)
A body of mass $5\text{ kg}$ moving with a uniform speed $3\sqrt{2}\text{ ms}^{-1}$ in X–Y plane along the line $y = x + 4$. The angular momentum of the particle about the origin will be ______ $\text{kg m}^2\text{s}^{-1}$.
### Related Formula
The magnitude of the angular momentum $L$ of a particle of mass $m$ moving with velocity $v$ is:
$L = m v d$
where:
* $d$ is the perpendicular distance from the axis of rotation (origin) to the line of motion of the particle.
### Core Logic
Given parameters:
* Mass, $m = 5\text{ kg}$
* Velocity, $v = 3\sqrt{2}\text{ ms}^{-1}$
* Line of motion: $y = x + 4 \implies x - y + 4 = 0$
### Step 1: Calculate Perpendicular Distance
The perpendicular distance $d$ from the origin $(0,0)$ to the line $Ax + By + C = 0$ is:
$$d = \frac{|A(0) + B(0) + C|}{\sqrt{A^2 + B^2}}$$
For the line $x - y + 4 = 0$:
$$d = \frac{|4|}{\sqrt{1^2 + (-1)^2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}\text{ m}$$
### Step 2: Calculate Angular Momentum
Substitute the values into the angular momentum formula:
$L = m v d$
$$L = 5\text{ kg} \times (3\sqrt{2}\text{ ms}^{-1}) \times (2\sqrt{2}\text{ m})$$
$$L = 5 \times 3 \times 4 = 60\text{ kg m}^2\text{s}^{-1}$$
Thus, the angular momentum of the particle about the origin is $60\text{ kg m}^2\text{s}^{-1}$.
### Pattern Recognition
Instead of complicated vector cross products, find the perpendicular distance of the straight line from the origin using standard coordinate geometry. $L = mvd$ is extremely fast and reliable.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: System of Particles and Rotational Motion
Q1556 (2024)
Four particles each of mass $1\text{ kg}$ are placed at four corners of a square of side $2\text{ m}$. The moment of inertia of the system about an axis perpendicular to its plane and passing through one of its vertices is ______ $\text{kg}\cdot\text{m}^{2}$.
{{IMG}}
### Related Formula
$$I = \sum m_i r_i^2$$
### Core Logic
Let the axis pass through vertex 1. Evaluate distances ($r$) for each corner particle:
- Particle at vertex 1: $r_1 = 0$
- Particle at adjacent vertex 2: $r_2 = a$
- Particle at adjacent vertex 4: $r_4 = a$
- Particle at diagonally opposite vertex 3: $r_3 = \sqrt{2}a$
### Step 1: Set up substitution formula
$$I = m(0)^2 + m(a)^2 + m(a)^2 + m(\sqrt{2}a)^2$$
$$I = ma^2 + ma^2 + 2ma^2 = 4ma^2$$
### Step 2: Numeric Evaluation
Substitute $m = 1\text{ kg}$ and side length $a = 2\text{ m}$:
$$I = 4 \times 1 \times (2)^2 = 4 \times 4 = 16\text{ kg}\cdot\text{m}^2$$
### Pattern Recognition
For a standard planar configuration system, total orthogonal moment components map predictably via basic summation configurations matching $4ma^2$ exactly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: System of Particles and Rotational Motion
