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 5.
Q25 (2025)
M and R be the mass and radius of a disc. A small disc of radius R/3 is removed from the bigger disc as shown in figure. The moment of inertia of remaining part of bigger disc about an axis AB passing through the centre O and perpendicular to the plane of disc is $\frac{4}{\mathrm{x}}\mathrm{MR}^2$ The value of x is
{{IMG1}} [cite: 197, 198, 199, 200, 208]
### Related Formula
$$I_{\text{disc}} = \frac{1}{2} M R^2$$ [cite: 833]
$$I = I_{\text{cm}} + M d^2 \quad \text{(Parallel Axis Theorem)}$$ [cite: 848]
### Core Logic
Let the original mass density per unit area be $\sigma$. Without any cavity, the moment of inertia is: [cite: 198, 833]
$$I_1 = \frac{MR^2}{2}$$ [cite: 833]
The mass of the removed small section scales directly with its cut-out area profile: [cite: 198, 834]
$$m = \frac{M}{\pi R^2} \times \pi \left(\frac{R}{3}\right)^2 = \frac{M}{9}$$ [cite: 198, 834, 844]
The center of mass of the removed disk sits at a distance $d = R - \frac{R}{3} = \frac{2R}{3}$ away from the primary center $O$[cite: 198, 205, 848]. Calculating its partial moment of inertia about $O$ via the parallel axis theorem: [cite: 199, 848]
$$I_2 = \frac{m r^2}{2} + m d^2 = \frac{\frac{M}{9}\left(\frac{R}{3}\right)^2}{2} + \frac{M}{9}\left(\frac{2R}{3}
ight)^2$$ [cite: 848]
$$I_2 = \frac{MR^2}{162} + \frac{4MR^2}{81} = \frac{MR^2 + 8MR^2}{162} = \frac{9MR^2}{162} = \frac{MR^2}{18}$$ [cite: 848, 850]
Subtracting the removed component from the original configuration: [cite: 851]
$$I = I_1 - I_2 = \frac{MR^2}{2} - \frac{MR^2}{18} = \frac{9MR^2 - MR^2}{18} = \frac{8MR^2}{18} = \frac{4}{9}MR^2$$ [cite: 851]
Matching this with $\frac{4}{x}MR^2$, we find $x = 9$[cite: 200, 208, 851, 853].
### Pattern Recognition
For uniform planar surfaces, mass always scales squarely with linear dimension changes ($r \rightarrow \frac{r}{3} \implies m \rightarrow \frac{m}{9}$)[cite: 198, 834, 844]. Always apply the parallel axis theorem to bring the component values to a unified reference point before executing addition or subtraction[cite: 848, 851].
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: System of Particles and Rotational Motion
Q4 (2025)
A solid sphere is rolling without slipping on a horizontal plane. The ratio of the linear kinetic energy of the centre of mass of the sphere and rotational kinetic energy is:
- $\frac{2}{5}$
- $\frac{5}{2}$
- $\frac{3}{4}$
- $\frac{4}{3}$
### Related Formula
$$K_{\text{linear}} = \frac{1}{2} m v_{\text{cm}}^2$$
$$K_{\text{rotational}} = \frac{1}{2} I \omega^2$$
### Core Logic
For a solid sphere, the moment of inertia about the center of mass is $I = \frac{2}{5}mR^2$.
Since it rolls without slipping, the condition $v_{\text{cm}} = \omega R$ holds.
Substituting $I$ and $\omega$ into the ratio:
$$\frac{K_{\text{linear}}}{K_{\text{rotational}}} = \frac{\frac{1}{2} m v_{\text{cm}}^2}{\frac{1}{2} \left(\frac{2}{5}mR^2\right) \left(\frac{v_{\text{cm}}}{R}\right)^2}$$
$$\frac{K_{\text{linear}}}{K_{\text{rotational}}} = \frac{1}{\frac{2}{5}} = \frac{5}{2}$$
### Pattern Recognition
The ratio of translational to rotational kinetic energy for any rolling body is given by $\frac{mR^2}{I_{\text{cm}}}$. For a solid sphere, this becomes $\frac{1}{2/5} = \frac{5}{2}$.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: System of Particles and Rotational Motion
Q9 (2025)
A solid sphere and a hollow sphere of the same mass and of same radius are rolled on an inclined plane. Let the time taken to reach the bottom by the solid sphere and the hollow sphere be $t_1$ and $t_2$, respectively, then
- $t_1 < t_2$
- $t_1 = t_2$
- $t_1 = 2t_2$
- $t_1 > t_2$
### Related Formula
$$t = \sqrt{\frac{2\ell}{a_{\text{cm}}}}$$
$$a_{\text{cm}} = \frac{g \sin \theta}{1 + \frac{I_{\text{cm}}}{MR^2}}$$
### Core Logic
For a solid sphere: $I_{\text{solid}} = \frac{2}{5}MR^2 \implies a_1 = \frac{g\sin\theta}{1 + 2/5} = \frac{5}{7}g\sin\theta$.
For a hollow sphere: $I_{\text{hollow}} = \frac{2}{3}MR^2 \implies a_2 = \frac{g\sin\theta}{1 + 2/3} = \frac{3}{5}g\sin\theta$.
Comparing accelerations:
$a_1 > a_2$
Since acceleration of the solid sphere is greater, it takes less time to descend the incline:
$t_1 < t_2${{SOL_IMG1}}
### Pattern Recognition
Smaller moment of inertia mass distribution (more concentrated at the center) yields larger acceleration down an incline, meaning a quicker descent.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: System of Particles and Rotational Motion
Q1133 (2025)
An object of mass 'm' is projected from origin in a vertical xy plane at an angle $45^{\circ}$ with the x-axis with an initial velocity $v_{0}$ The magnitude and direction of the angular momentum of the object with respect to origin, when it reaches at the maximum height, will be
[g is acceleration due to gravity]
- $\frac{mv_{0}^{3}}{2\sqrt{2}g}$ along negative z-axis
- $\frac{mv_{0}^{3}}{2\sqrt{2}g}$ along positive z-axis
- $\frac{mv_{0}^{3}}{4\sqrt{2}g}$ along positive z-axis
- $\frac{mv_{0}^{3}}{4\sqrt{2}g}$ along negative z-axis
### Related Formula
The definition of angular momentum vector $\vec{L}$ relative to the origin is:
$$\vec{L} = \vec{r} \times \vec{p} = m(\vec{r} \times \vec{v})$$
In scalar form for a horizontal speed component at a maximum altitude $H$:
$L = m v_{x} H$
### Core Logic
At maximum height, the vertical speed component drops to zero, and the projectile travels entirely horizontally along the curve profile as shown in {{SOL_IMG1}} :
$$v_{x} = v_{0} \cos 45^{\circ} = \frac{v_{0}}{\sqrt{2}}$$
$$H = \frac{v_{0}^{2} \sin^{2} 45^{\circ}}{2g} = \frac{v_{0}^{2}}{4g}$$
### Step 1: Calculating Magnitude and Vector Direction
Evaluate the horizontal vector cross components:
$$L = m \left(\frac{v_{0}}{\sqrt{2}}\right) \left(\frac{v_{0}^{2}}{4g}\right) = \frac{mv_{0}^{3}}{4\sqrt{2}g}$$
Using the right-hand rule, $\vec{r}$ points into quadrant-1 while velocity points towards $+\hat{i}$. Therefore, $\vec{r} \times \vec{v}$ tracks clockwise, yielding a negative $\hat{k}$ orientation (along the negative z-axis).
### Pattern Recognition
At peak height, always map $\vec{L}$ using $m \cdot v_{\text{peak}} \cdot y_{\text{max}}$. This scalar form simplifies the calculation significantly.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: System of Particles and Rotational Motion
Q11 (2025)
A uniform solid cylinder of mass 'm' and radius 'r' rolls along an inclined rough plane of inclination $45^{\circ}$ If it starts to roll from rest from the top of the plane then the linear acceleration of the cylinder axis will be :-
- $\frac{1}{\sqrt{2}} g$
- $\frac{1}{3\sqrt{2}} g$
- $\frac{\sqrt{2} g}{3}$
- $\sqrt{2} g$
### Related Formula
The linear acceleration $a$ for pure rolling motion down an incline profile is:
$$a = \frac{g\sin\theta}{1 + \frac{I}{mr^{2}}}$$
### Core Logic
For a uniform solid cylinder, the moment of inertia around its central axis is:
$$I = \frac{1}{2}mr^{2} \implies \frac{I}{mr^{2}} = \frac{1}{2}$$
### Step 1: Calculating Acceleration
Substitute $\theta = 45^{\circ}$ and the cylinder inertial factor into the formula :
$$a = \frac{g\sin 45^{\circ}}{1 + \frac{1}{2}} = \frac{\frac{g}{\sqrt{2}}}{\frac{3}{2}}$$
$$a = \frac{2g}{3\sqrt{2}} = \frac{\sqrt{2}g}{3}$$
### Pattern Recognition
Solid cylinder rolls with acceleration matching $\frac{2}{3} g\sin\theta$. Since $\sin 45^{\circ} = \frac{1}{\sqrt{2}}$, this simplifies directly to $\frac{\sqrt{2}g}{3}$.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: System of Particles and Rotational Motion
