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 3.
Q15 (2025)
Three equal masses $m$ are kept at vertices (A, B, C) of an equilateral triangle of side $a$ in free space. At $t = 0$ , they are given an initial velocity $\vec{V}_{\mathrm{A}} = V_{0}\overrightarrow{\mathrm{AC}}$ , $\vec{V}_{\mathrm{B}} = V_{0}\overrightarrow{\mathrm{BA}}$ and $\vec{V}_{\mathrm{C}} = V_{0}\overrightarrow{\mathrm{CB}}$ . Here, $\overrightarrow{\mathrm{AC}}, \overrightarrow{\mathrm{CB}}$ and $\overrightarrow{\mathrm{BA}}$ are unit vectors along the edges of the triangle. If the three masses interact gravitationally, then the magnitude of the net angular momentum of the system at the point of collision is:
{{IMG1}}
- $\frac{1}{2} \mathrm{~a} \mathrm{~mV}_{0}$
- $3\mathrm{amV}_0$
- $\frac{\sqrt{3}}{2} \mathrm{~a} \mathrm{~mV}_{0}$
- $\frac{3}{2} \mathrm{~a} \mathrm{~mV}_{0}$
### Related Formula
$$\vec{L} = \sum \left(\vec{r}_i \times m \vec{v}_i\right)$$
$$\vec{\tau}_{\text{ext}} = \frac{d\vec{L}}{dt}$$
### Core Logic
Since the three masses interact purely through mutual internal gravitational forces, the net external torque acting on the system about any central reference point is zero:
$$\vec{\tau}_{\text{ext}} = 0 \implies \vec{L}_{\text{initial}} = \vec{L}_{\text{final}}$$
Let us compute the total angular momentum about the centroid of the equilateral triangle:
{{SOL_IMG1}}
From trigonometry, the perpendicular distance from the centroid to the velocity vector along any edge is:
$$r_{\perp} = \frac{a}{2\sqrt{3}}$$
The initial angular momentum for one mass about the centroid is $L_1 = m V_0 r_{\perp}$. Since all three particles move cyclically in the same direction, their angular momenta reinforce cleanly:
$$L_{\text{net}} = 3 \cdot \left(m V_0 \frac{a}{2\sqrt{3}}\right) = \frac{3}{2\sqrt{3}} m V_0 a = \frac{\sqrt{3}}{2} a m V_0$$
By conservation of angular momentum, this configuration value remains unchanged up to the point of collision.
### Pattern Recognition
Mutual internal central forces can never alter the angular momentum of a system. Hence, the solution completely reduces to measuring the static configuration values at $t=0$ about the center of mass symmetry.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: System of Particles and Rotational Motion
Class 11 Physics: Gravitation
Q19 (2025)
The centre of mass of a thin rectangular plate (fig - x) with sides of length a and b, whose mass per unit area $(\sigma)$ varies as $\sigma = \frac{\sigma_0x}{ab}$ (where $\sigma_0$ is a constant), would be
{{IMG1}}
- $\left(\frac{2}{3} a, \frac{b}{2}\right)$
- $\left(\frac{2}{3} a, \frac{2}{3} b\right)$
- $\left(\frac{a}{2},\frac{b}{2}\right)$
- $\left(\frac{1}{3} a, \frac{b}{2}\right)$
### Core Logic
Since density $\sigma$ is independent of the y-coordinate, the vertical center of mass resolves directly by symmetry:
$$\mathrm{y}_{\mathrm{cm}} = \frac{\mathrm{b}}{2}$$
{{SOL_IMG1}}
To find the horizontal center of mass, evaluate the continuous mass integral along the x-axis:
$$\mathrm{x}_{\mathrm{cm}} = \frac{\int_0^{\mathrm{a}} \mathrm{x \, dm}}{\int_0^{\mathrm{a}} \mathrm{dm}} = \frac{\int_0^{\mathrm{a}} \mathrm{x} \left(\frac{\sigma_0 \mathrm{x}}{\mathrm{ab}}\right) \mathrm{b \, dx}}{\int_0^{\mathrm{a}} \left(\frac{\sigma_0 \mathrm{x}}{\mathrm{ab}}\right) \mathrm{b \, dx}}$$
$$\mathrm{x}_{\mathrm{cm}} = \frac{\int_0^{\mathrm{a}} \mathrm{x}^2 \, \mathrm{dx}}{\int_0^{\mathrm{a}} \mathrm{x} \, \mathrm{dx}} = \frac{\left[ \frac{\mathrm{x}^3}{3} \right]_0^{\mathrm{a}}}{\left[ \frac{\mathrm{x}^2}{2} \right]_0^{\mathrm{a}}} = \frac{\mathrm{a}^3 / 3}{\mathrm{a}^2 / 2} = \frac{2\mathrm{a}}{3}$$
### Step 1: Final Position Coordinates
The center of mass coordinates are $\left(\frac{2}{3} \mathrm{a}, \frac{\mathrm{b}}{2}\right)$, which matches option (1).
### Pattern Recognition
When density varies linearly with position ($\sigma \propto \mathrm{x}$), the mass distribution shifts outward, moving the center of mass from the geometric midpoint $\frac{\mathrm{a}}{2}$ to the $\frac{2}{3}\mathrm{a}$ mark.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: System of Particles and Rotational Motion
Q22 (2025)
The moment of inertia of a solid disc rotating along its diameter is 2.5 times higher than the moment of inertia of a ring rotating in similar way. The moment of inertia of a solid sphere which has same radius as the disc and rotating in similar way, is n times higher than the moment of inertia of the given ring. Here, $\mathrm{n} =$ _____. Consider all the bodies have equal masses.
### Related Formula
$$\mathrm{I}_{\text{disc}} = \frac{\mathrm{M}\mathrm{R}_1^2}{4}, \quad \mathrm{I}_{\text{ring}} = \frac{\mathrm{M}\mathrm{R}_2^2}{2}, \quad \mathrm{I}_{\text{sphere}} = \frac{2\mathrm{M}\mathrm{R}_1^2}{5}$$
### Core Logic
Let's list the relevant moment of inertia formulas based on their rotation axes:
{{SOL_IMG1}}
{{SOL_IMG2}}
{{SOL_IMG3}}
From the given problem statements:
$$\frac{\mathrm{I}_{\text{disc}}}{\mathrm{I}_{\text{ring}}} = 2.5 \implies \frac{\frac{\mathrm{M}\mathrm{R}_1^2}{4}}{\frac{\mathrm{M}\mathrm{R}_2^2}{2}} = \frac{5}{2} \implies \frac{\mathrm{R}_1^2}{\mathrm{R}_2^2} = 5$$
Now, evaluating the second geometric layout ratio:
$$\frac{\mathrm{I}_{\text{sphere}}}{\mathrm{I}_{\text{ring}}} = \mathrm{n} \implies \frac{\frac{2\mathrm{M}\mathrm{R}_1^2}{5}}{\frac{\mathrm{M}\mathrm{R}_2^2}{2}} = \mathrm{n} \implies \frac{4\mathrm{R}_1^2}{5\mathrm{R}_2^2} = \mathrm{n}$$
Substituting our radius parameter ($\frac{\mathrm{R}_1^2}{\mathrm{R}_2^2} = 5$):
$$\mathrm{n} = \frac{4}{5} \cdot 5 = 4$$
### Step 1: Final Value Conclusion
The scale value parameter is found to be:
$\mathrm{n} = 4$
### Pattern Recognition
Be careful with rotation axis descriptions. Disc and ring components rotating along their structural diameter axes use values that are half of their standard perpendicular planar formulas.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: System of Particles and Rotational Motion
Q24 (2025)
Two iron solid discs of negligible thickness have radii $\mathbf{R}_1$ and $\mathbf{R}_2$ and moment of inertia $\mathrm{I}_1$ and $\mathrm{I}_2$ , respectively. For $\mathrm{R}_2 = 2\mathrm{R}_1$ , the ratio of $\mathrm{I}_1$ and $\mathrm{I}_2$ would be $1 / \mathrm{x}$ , where $\mathrm{x} =$
### Core Logic
Since mass scales with the face surface area for discs of identical thickness and material composition:
$$\mathrm{M} = \sigma \cdot \pi \mathrm{R}^2 \implies \mathrm{M} \propto \mathrm{R}^2$$
{{SOL_IMG1}}
{{SOL_IMG2}}
$$\mathrm{M}_1 = \mathrm{M}_0, \quad \mathrm{M}_2 = \sigma \pi (2\mathrm{R}_1)^2 = 4\mathrm{M}_0$$
The moment of inertia formula for a disc is:
$$mathrm{I} = \frac{1}{2}\mathrm{M}\mathrm{R}^2 \implies \mathrm{I} \propto \mathrm{M}\mathrm{R}^2 \propto \mathrm{R}^4$$
Calculating the ratio for the given radii configuration:
$$\frac{\mathrm{I}_1}{\mathrm{I}_2} = \frac{\mathrm{M}_1 \mathrm{R}_1^2}{\mathrm{M}_2 \mathrm{R}_2^2} = \frac{\mathrm{M}_0 \cdot \mathrm{R}_1^2}{4\mathrm{M}_0 \cdot (2\mathrm{R}_1)^2} = \frac{1}{16}$$
### Step 1: Value Convergence
Comparing this fraction to $1/\mathrm{x}$ yields:
$\mathrm{x} = 16$
### Pattern Recognition
For 2D uniform laminar objects, scaling the radius changes both the mass factor (by $\mathrm{R}^2$) and the distribution distance (by $\mathrm{R}^2$), resulting in an overall $\mathrm{R}^4$ dependency rule.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: System of Particles and Rotational Motion
Q10 (2025)
A force of $49\mathrm{~N}$ acts tangentially at the highest point of a sphere (solid) of mass $20\mathrm{~kg}$, kept on a rough horizontal plane. If the sphere rolls without slipping, then the acceleration of the center of the sphere is:
{{IMG1}}
- $3.5\mathrm{~m/s}^2$
- $0.35\mathrm{~m/s}^2$
- $2.5\mathrm{~m/s}^2$
- $0.25\mathrm{~m/s}^2$
### Related Formula
Torque equation about the instantaneous center of zero velocity (bottom contact point $P$):
$$\tau_P = I_P \alpha$$
For a solid sphere, the moment of inertia about the center is $I_c = \frac{2}{5}MR^2$. By the parallel axis theorem:
$$I_P = I_c + MR^2 = \frac{7}{5}MR^2$$
### Core Logic
Since the sphere rolls without slipping, we can conveniently write the torque equation about the lowest point of contact $P$ because static friction passes through this point and exerts zero torque.
- Distance from point $P$ to the top highest point is $2R$.
- Tangential force $F = 49\mathrm{~N}$.
- Mass of solid sphere, $M = 20\mathrm{~kg}$.
$$\tau_P = F \times 2R$$
Substitute $\tau_P$ and $I_P$ into the torque equation:
$$F \times 2R = \left(\frac{7}{5}MR^2\right) \alpha$$
### Step 1: Solving for Linear Acceleration
For pure rolling, the acceleration of the center of mass $a$ is related to angular acceleration $\alpha$ by $a = R\alpha$:
$$2F R = \frac{7}{5}MR^2 \left(\frac{a}{R}\right)$$
$$2F = \frac{7}{5} M a \implies a = \frac{10F}{7M}$$
Substitute the numerical values ($F = 49\mathrm{~N}$ and $M = 20\mathrm{~kg}$):
$$a = \frac{10 \times 49}{7 \times 20} = \frac{490}{140} = 3.5\mathrm{~m/s}^2$$
### Step 2: Analysis of Friction Force Direction
Let's write force equations to verify consistency:
$F + f = M a$
$$49 + f = 20 \times 3.5 = 70 \implies f = 21\mathrm{~N}$$
Since $f$ is positive, static friction acts in the forward direction. Rolling without slipping is fully maintained since the required static friction coefficient is well within realistic limits.
### Pattern Recognition
Calculating torque about the bottom contact point is a powerful shortcut for rolling-without-slipping questions! It completely bypasses having to guess or set up equations for the friction direction.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: System of Particles and Rotational Motion
