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 11.
Q49 (2026)
Suppose there is a uniform circular disc of mass M kg and radius r m shown in figure. The shaded regions are cut out from the disc. The moment of inertia of the remainder about the axis A of the disc is given by $\frac{x}{256}$ Mr $^{2}$ . The value of x is ____.
{{IMG1}}
### Related Formula
$$I_{\text{remaining}} = I_{\text{total}} - \sum I_{\text{cut}}$$
Parallel axis theorem: $I = I_{\text{cm}} + md^2$
### Core Logic
Total mass of original disc is $M = \sigma \pi r^2$.
Two identical smaller discs are removed. From the diagram, each small cut-out disc has radius $r' = \frac{r}{4}$.
The center of each cut-out disc is at distance $d = \frac{3}{4}r$ from the main axis A.
Mass of each cut-out disc:
$$m = \sigma \pi \left(\frac{r}{4}\right)^2 = \frac{\sigma \pi r^2}{16} = \frac{M}{16}$$
### Step 1: Moment of Inertia of Total Disc
For the main solid disc about its central axis A:
$$I_{\text{total}} = \frac{1}{2} M r^2$$
### Step 2: Moment of Inertia of Cut-out Discs
Using parallel axis theorem for one cut-out disc about axis A:
$$I_{\text{cut}} = \frac{1}{2} m (r')^2 + m d^2$$
$$I_{\text{cut}} = \frac{1}{2} m \left(\frac{r}{4}\right)^2 + m \left(\frac{3}{4}r\right)^2$$
$$I_{\text{cut}} = \frac{mr^2}{32} + \frac{9mr^2}{16} = m r^2 \left( \frac{1 + 18}{32} \right) = \frac{19}{32} m r^2$$
Since there are 2 cut-out discs, the total subtracted inertia is:
$$I_{\text{removed}} = 2 \times \frac{19}{32} m r^2 = \frac{19}{16} m r^2$$
### Step 3: Calculate Final Inertia in Terms of M
Substitute $m = \frac{M}{16}$:
$$I_{\text{removed}} = \frac{19}{16} \left( \frac{M}{16} \right) r^2 = \frac{19}{256} M r^2$$
Now, subtract from total:
$$I_{\text{remaining}} = \frac{1}{2} M r^2 - \frac{19}{256} M r^2$$
$$I_{\text{remaining}} = \frac{128 - 19}{256} M r^2 = \frac{109}{256} M r^2$$
Thus, $x = 109$.
### Pattern Recognition
In cavity problems, mass is strictly proportional to area ($R^2$). Use parallel axis theorem perfectly on the 'negative mass' segments and subtract.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: System of Particles and Rotational Motion
Q41 (2026)
Two masses 400 g and 350 g are suspended from the ends of a light string passing over a heavy pulley of radius 2 cm. When released from rest the heavier mass is observed to fall 81 cm in 9 s. The rotational inertia of the pulley is ____ $\text{kg}\cdot\text{m}^{2}$.
(g = $9.8 \text{ m/s}^{2}$)
- $9.5 \times 10^{-3}$
- $4.75 \times 10^{-3}$
- $1.86 \times 10^{-2}$
- $8.3 \times 10^{-3}$
### Related Formula
$$s = ut + \frac{1}{2}at^2$$
$$a = \frac{(m_1 - m_2)g}{m_1 + m_2 + \frac{I}{R^2}}$$
### Core Logic
{{SOL_IMG1}}
First, calculate the acceleration of the system using kinematics:
$$s = ut + \frac{1}{2}at^2$$
$$0.81 = 0 + \frac{1}{2} a (9)^2$$
$$a = \frac{2 \times 0.81}{81} = 0.02 \text{ m/s}^2$$
Applying Newton's second law for masses and rotation:
$$m_1 g - T_1 = m_1 a$$
$$T_2 - m_2 g = m_2 a$$
$$(T_1 - T_2)R = I \cdot \alpha = I \left(\frac{a}{R}\right)$$
This leads to the standard Atwood machine acceleration with massive pulley:
$$a = \frac{(m_1 - m_2)g}{m_1 + m_2 + \frac{I}{R^2}}$$
### Step 1: Calculate Moment of Inertia
Substitute known values ($m_1=0.4\text{ kg}$, $m_2=0.35\text{ kg}$, $R=0.02\text{ m}$):
$$0.02 = \frac{(0.4 - 0.35) \times 9.8}{(0.4 + 0.35) + \frac{I}{R^2}}$$
$$0.02 = \frac{0.05 \times 9.8}{0.75 + \frac{I}{R^2}}$$
$$0.75 + \frac{I}{R^2} = \frac{0.49}{0.02} = 24.5$$
$$\frac{I}{R^2} = 24.5 - 0.75 = 23.75$$
$$I = 23.75 \times (0.02)^2 = 23.75 \times 4 \times 10^{-4}$$
$$I = 95 \times 10^{-4} = 9.5 \times 10^{-3} \text{ kg}\cdot\text{m}^2$$
### Pattern Recognition
For massive pulley problems, the "effective mass" of the system increases by the pulley's equivalent translating mass $I/R^2$. Simply use $a = F_{\text{net}} / M_{\text{effective}}$.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: System of Particles and Rotational Motion
Class 11 Physics: Laws of Motion
Q31 (2026)
A thin uniform rod (X) of mass M and length L is pivoted at a height $\left(\frac{L}{3}\right)$ as shown in the figure. The rod is allowed to fall from a vertical position and lie horizontally on the table. The angular velocity of this rod when it hits the table top, is ____. (g = gravitational acceleration)
{{IMG1}}
- $\sqrt{\frac{3}{2}\frac{\mathrm{g}}{\mathrm{L}}}$
- $\frac{3}{\sqrt{2}}\sqrt{\frac{\mathrm{g}}{\mathrm{L}}}$
- $\frac{1}{\sqrt{2}}\sqrt{\frac{\mathrm{g}}{\mathrm{L}}}$
- $\sqrt{\frac{3\mathrm{g}}{\mathrm{L}}}$
### Related Formula
$$\Delta K.E. = \Delta P.E.$$
$$mg \Delta h_{\text{com}} = \frac{1}{2} I \omega^2$$
### Core Logic
The rod falls such that its center of mass lowers by a distance.
The rod is pivoted at $\frac{L}{3}$ from the bottom, meaning the distance from the pivot to the center of mass (which is at $\frac{L}{2}$ from either end) is:
$$h_{\text{com}} = \frac{L}{2} - \frac{L}{3} = \frac{L}{6}$$
### Step 1: Moment of Inertia
Using the parallel axis theorem, the moment of inertia about the pivot is:
$$I = I_{\text{com}} + m d^2$$
$$I = \frac{mL^2}{12} + m\left(\frac{L}{6}\right)^2 = \frac{mL^2}{12} + \frac{mL^2}{36} = \frac{mL^2}{9}$$
### Step 2: Energy Conservation
Equating the loss in potential energy to the gain in rotational kinetic energy:
$$mg \frac{L}{6} = \frac{1}{2} \left(\frac{mL^2}{9}\right) \omega^2$$
$$mg \frac{L}{6} = \frac{mL^2}{18} \omega^2$$
$$\omega^2 = \frac{3g}{L}$$
$$\omega = \sqrt{\frac{3g}{L}}$$
### Pattern Recognition
For a hinged rod falling from a vertical orientation to horizontal, always track the displacement of the center of mass and calculate rotational inertia strictly about the hinge using $I_{\text{pivot}} = I_{\text{cm}} + md^2$.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: System of Particles and Rotational Motion
Q49 (2026)
A uniform solid cylinder of length L and radius R has moment of inertia about its axis equal to $I_{1}$ . A small co-centric cylinder of length L/2 and radius R/3 carved from this cylinder has moment of inertia about its axis equals to $I_{2}$ . The ratio $I_{1}/I_{2}$ is
### Related Formula
$$I = \frac{1}{2} M R^2$$
$$M = \rho \cdot V = \rho \cdot \pi R^2 L$$
### Core Logic
{{SOL_IMG1}}
For the original cylinder (mass $M$):
$$I_1 = \frac{1}{2} M R^2$$
For the carved cylinder, its mass $m$ is:
$$m = \rho \times \pi \left(\frac{R}{3}\right)^2 \times \frac{L}{2}$$
### Step 1: Calculate Mass of Carved Cylinder
$$m = \frac{\rho \pi R^2 L}{18} = \frac{M}{18}$$
### Step 2: Calculate Inertia of Carved Cylinder
$$I_2 = \frac{1}{2} m \left(\frac{R}{3}\right)^2$$
$$I_2 = \frac{1}{2} \left( \frac{M}{18} \right) \left( \frac{R^2}{9} \right)$$
$$I_2 = \frac{1}{324} M R^2$$
### Step 3: Finding the Ratio
$$\frac{I_1}{I_2} = \frac{\frac{1}{2} M R^2}{\frac{1}{324} M R^2} = \frac{324}{2} = 162$$
### Pattern Recognition
For similar geometries, mass scales as $R^2 L$. Inertia scales as $M R^2$ which ultimately means $I \propto R^4 L$. Here $R \rightarrow R/3$ (factor of $1/81$) and $L \rightarrow L/2$ (factor of $1/2$), so $I_2$ is $1/162$ of $I_1$.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: System of Particles and Rotational Motion
Q29 (2026)
Two circular discs of radius each 10 cm are joined at their centres by a rod of length 30 cm and mass 600 gm as shown in figure.
{{IMG1}}
If the mass of each disc is 600 gm and applied torque between two discs is $43 \times 10^{5} \mathrm{~dyne\cdot cm}$, the angular acceleration of the discs about the given axis AB is ____ $\mathrm{rad/s}^{2}$ .
- $22$
- $11$
- $100$
- $27$
### Related Formula
$\tau = I \alpha$
where $\tau$ is the torque, $I$ is the moment of inertia about the axis of rotation, and $\alpha$ is the angular acceleration.
### Core Logic
First, calculate the total moment of inertia of the system (two discs + one rod) about axis AB using the parallel axis theorem. Let $m = 600 \mathrm{~gm}$ and $R = 10 \mathrm{~cm}$.
### Step 1: Moment of Inertia of Discs
For the left disc (distance $R$ from AB):
$$I_{\text{left}} = \frac{1}{4}mR^2 + mR^2 = \frac{5}{4}mR^2$$
For the right disc (distance $2R$ from AB):
$$I_{\text{right}} = \frac{1}{4}mR^2 + m(2R)^2 = \frac{17}{4}mR^2$$
Total for discs:
$$I_{\text{discs}} = \frac{5}{4}mR^2 + \frac{17}{4}mR^2 = \frac{22}{4}mR^2 = \frac{11}{2}mR^2$$
### Step 2: Moment of Inertia of Rod
Rod has mass $m$ and length $3R = 30 \mathrm{~cm}$. Its center of mass is at a distance $R/2$ from axis AB.
$$I_{\text{rod}} = \frac{m(3R)^2}{12} + m\left(\frac{R}{2}\right)^2$$
$$= \frac{9mR^2}{12} + \frac{mR^2}{4} = \frac{3}{4}mR^2 + \frac{1}{4}mR^2 = mR^2$$
### Step 3: Total Moment of Inertia
$$I_{\text{total}} = \left(\frac{11}{2} + 1\right) mR^2 = \frac{13}{2} mR^2$$
Substitute $m = 600 \mathrm{~g}$ and $R = 10 \mathrm{~cm}$:
$$I_{\text{total}} = \frac{13}{2} \times 600 \times (10)^2 = 39 \times 10^4 \mathrm{~g \cdot cm^2}$$
### Step 4: Calculating Angular Acceleration
$$\alpha = \frac{\tau}{I} = \frac{43 \times 10^5}{39 \times 10^4} \mathrm{~rad/s^2}$$
$$= \frac{430}{39} \approx 11.02 \mathrm{~rad/s^2}$$
Rounding off, $\alpha \approx 11 \mathrm{~rad/s^2}$.
### Pattern Recognition
In compound systems, decompose into basic shapes (rod, disc). Determine the parallel distance to the required axis for each center of mass. Keep everything in CGS units since torque is given in dyne-cm.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: System of Particles and Rotational Motion
