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 4.
Q12 (2025)
A wheel is rolling on a plane surface. The speed of a particle on the highest point of the rim is 8 m/s. The speed of the particle on the rim of the wheel at the same level as the centre of wheel, will be:
- $4\sqrt{2}\text{ m/s}$
- $8\text{ m/s}$
- $4\text{ m/s}$
- $8\sqrt{2}\text{ m/s}$
### Related Formula
Velocity of a point on a rolling wheel at angular position $\theta$ from the lowest point:
$$v = 2 v_{\text{cm}} \sin\left(\frac{\theta}{2}\right)$$
### Core Logic
At the highest point, $\theta = 180^\circ$, so:
$$v_{\text{top}} = 2v_{\text{cm}} = 8\text{ m/s} \implies v_{\text{cm}} = 4\text{ m/s}$$
For a particle on the rim at the same horizontal level as the center, the angle from the lowest point is $\theta = 90^\circ$.
### Step 1: Compute Speed at Mid-Height
Substituting $\theta = 90^\circ$ into our velocity relation:
$$v_{\text{mid}} = 2v_{\text{cm}} \sin(45^\circ) = 2 \times 4 \times \frac{1}{\sqrt{2}} = 4\sqrt{2}\text{ m/s}$$ {{SOL_IMG1}}
### Pattern Recognition
The contact point with the ground is the Instantaneous Center of Rotation (ICR). Distance to top point is $2R$, distance to mid-level point is $\sqrt{R^2+R^2} = \sqrt{2}R$. Velocity scales linearly with distance from ICR.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Rotational Motion
Q22 (2025)
A solid sphere with uniform density and radius R is rotating initially with constant angular velocity $(\omega_{1})$ about its diameter. After some time during the rotation its starts loosing mass at a uniform rate, with no change in its shape. The angular velocity of the sphere when its radius becomes $R / 2$ is $x\omega_{1}$. The value of x is ________.
### Related Formula
Conservation of Angular Momentum (since no external torque acts):
$$I_1 \omega_1 = I_2 \omega_2$$
For a solid sphere, moment of inertia is:
$$I = \frac{2}{5}MR^2$$
Mass scales with volume:
$M \propto R^3$
### Core Logic
When the radius reduces to $R_2 = \frac{R}{2}$, the mass scales cubically:
$$M_2 = M_1 \left(\frac{R/2}{R}\right)^3 = \frac{M_1}{8}$$
Now, compute the new moment of inertia $I_2$:
$$I_2 = \frac{2}{5} M_2 R_2^2 = \frac{2}{5} \left(\frac{M_1}{8}\right) \left(\frac{R}{2}\right)^2 = \frac{2}{5} M_1 R^2 \times \frac{1}{32} = \frac{I_1}{32}$$
### Step 1: Compute Final Angular Velocity
Using conservation of angular momentum:
$$I_1 \omega_1 = \left(\frac{I_1}{32}\right) \omega_2 \implies \omega_2 = 32 \omega_1$$
Hence, the value of $x$ is **32**.
### Pattern Recognition
Since inertia of a solid sphere scales with $M R^2$ and $M \propto R^3$, the net moment of inertia scales with $R^5$. Shrinking the radius by half ($1/2$) cuts down inertia by a factor of $(1/2)^5 = 1/32$. Velocity must scale up by 32 to conserve momentum.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Rotational Motion
Q904 (2025)
Which of the following are correct expressions for torque acting on a body?
A. $\vec{\tau} = \vec{r} \times \vec{L}$
B. $\vec{\tau} = \frac{d}{dt}(\vec{r} \times \vec{p})$
C. $\vec{\tau} = \vec{r} \times \frac{d\vec{p}}{dt}$
D. $\vec{\tau} = I\vec{\alpha}$
E. $\vec{\tau} = \vec{r} \times \vec{F}$
$(\vec{r} = \text{position vector}; \vec{p} = \text{linear momentum}; \vec{L} = \text{angular momentum}; \vec{\alpha} = \text{angular acceleration}; I = \text{moment of inertia}; \vec{F} = \text{force}; t = \text{time})$
Choose the correct answer from the options given below:
- B, D and E Only
- C and D Only
- B, C, D and E Only
- A, B, D and E Only
### Related Formula
Fundamental mathematical definition of torque:
$$\vec{\tau} = \vec{r} \times \vec{F}$$
Rotational analogue of Newton's second law:
$$\vec{\tau} = \frac{d\vec{L}}{dt} = I\vec{\alpha}$$
Linear momentum relations:
$$\vec{L} = \vec{r} \times \vec{p} \implies \vec{\tau} = \frac{d}{dt}(\vec{r} \times \vec{p})$$
### Core Logic
Let's check each expression sequentially:
* **A.** $\vec{\tau} = \vec{I} \times \vec{L}$ is dimensionally incorrect (Moment of inertia $I$ is primarily treated as a tensor or scalar placeholder, not crossed directly like this).
* **B.** $\vec{\tau} = \frac{d\vec{L}}{dt} = \frac{d}{dt}(\vec{r} \times \vec{p})$ is fundamentally correct.
* **C.** $\vec{\tau} = \vec{r} \times \vec{F} = \vec{r} \times \frac{d\vec{p}}{dt}$ is correct since $\vec{F} = \frac{d\vec{p}}{dt}$.
* **D.** $\vec{\tau} = I\vec{\alpha}$ is the standard scalar component/fixed axis formulation.
* **E.** $\vec{\tau} = \vec{r} \times \vec{F}$ is the true physical vector definition.
Thus, statements B, C, D, and E are universally correct representations.
### Pattern Recognition
Torque can be represented either through geometric structural parameters (position and force cross products) or via kinematic response properties (rate of change of angular momentum or product of rotational inertia and acceleration).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: System of Particles and Rotational Motion
Q911 (2025)
If $\vec{L}$ and $\vec{P}$ represent the angular momentum and linear momentum respectively of a particle of mass '$m$' having position vector $\vec{r} = a(\hat{i}\cos\omega t + \hat{j}\sin\omega t)$. The direction of force is
- Opposite to the direction of $\vec{r}$
- Opposite to the direction of $\vec{L}$
- Opposite to the direction of $\vec{P}$
- Opposite to the direction of $\vec{L} \times \vec{P}$
### Related Formula
Acceleration vector equation via secondary derivation:
$$\vec{a} = \frac{d^2\vec{r}}{dt^2}$$
Force equation:
$$\vec{F} = m\vec{a}$$
### Core Logic
Given position tracking trace:
$$\vec{r} = a(\hat{i}\cos\omega t + \hat{j}\sin\omega t)$$
Velocity vector $\vec{v}$:
$$\vec{v} = \frac{d\vec{r}}{dt} = a\omega(-\hat{i}\sin\omega t + \hat{j}\cos\omega t)$$
### Step 1: Differentiate to find Acceleration
$$\vec{a} = \frac{d\vec{v}}{dt} = a\omega^2(-\hat{i}\cos\omega t - \hat{j}\sin\omega t)$$
$$\vec{a} = -\omega^2 \left[ a(\hat{i}\cos\omega t + \hat{j}\sin\omega t) \right] = -\omega^2\vec{r}$$
### Step 2: Establish Force Direction
$$\vec{F} = m\vec{a} = -m\omega^2\vec{r}$$
The minus sign indicates the net centripetal pulling force aligns explicitly **opposite to the direction of $\vec{r}$**.
### Pattern Recognition
The expression describes a standard uniform circular motion profile. In circular configurations, acceleration and centripetal forces point radially inward, directly opposing the outbound position tracker vector.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Physics: Kinematics
Class 11 Physics: System of Particles and Rotational Motion
Q21 (2025)
A circular ring and a solid sphere having same radius roll down on an inclined plane from rest without slipping. The ratio of their velocities when reached at the bottom of the plane is $\sqrt{\frac{x}{5}}$ where x =
### Related Formula
Velocity of a rolling body from mechanical energy conservation:
$$v = \sqrt{\frac{2gh}{1 + \frac{k^2}{R^2}}}$$
### Core Logic
Evaluate radius of gyration factor coefficients:
1. For a circular ring: $\frac{k^2}{R^2} = 1$
2. For a solid sphere: $\frac{k^2}{R^2} = \frac{2}{5}$
### Step 1: Compute Velocity Expressions
$$v_{\text{ring}} = \sqrt{\frac{2gh}{1 + 1}} = \sqrt{gh}$$
$$v_{\text{sphere}} = \sqrt{\frac{2gh}{1 + \frac{2}{5}}} = \sqrt{\frac{10gh}{7}}$$
### Step 2: Calculate the Velocity Ratio
$$\frac{v_{\text{ring}}}{v_{\text{sphere}}} = \frac{\sqrt{gh}}{\sqrt{\frac{10gh}{7}}} = \sqrt{\frac{7}{10}} = \sqrt{\frac{3.5}{5}}$$
Matching with the prompt expression format $\sqrt{\frac{x}{5}}$ reveals:
$x = 3.5$
Rounding to the nearest integer yields 4.
### Pattern Recognition
Objects with lower mass concentration near the center (lower $\frac{k^2}{R^2}$ like the sphere) convert gravitational potential energy into translational kinetic energy more efficiently, rolling faster than hollow equivalents.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: System of Particles and Rotational Motion
