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 7.
Q1644 (2024)
A cylinder is rolling down on an inclined plane of inclination $60^{\circ}$. It's acceleration during rolling down will be $ rac{x}{\sqrt{3}} \mathrm{~m / s^{2}}$, where $x =$ ________ (use $g = 10 \mathrm{~m/s}^{2}$).
### Related Formula
The linear acceleration ($a$) of a symmetric body performing pure rolling down an inclined plane of angle $\theta$ is given by:
$$a = \frac{g \sin \theta}{1 + \frac{I_{\text{cm}}}{M R^2}}$$
### Core Logic
For a solid cylinder, the moment of inertia about its central longitudinal axis is:
$$I_{\text{cm}} = \frac{1}{2} M R^2 \implies \frac{I_{\text{cm}}}{M R^2} = \frac{1}{2}$$
Given inclination angle, $\theta = 60^\circ$, and $g = 10 \mathrm{~m/s^2}$.
{{SOL_IMG1}}
### Step 1: Calculate Linear Acceleration
Substituting the values into the acceleration template:
$$a = \frac{10 \times \sin 60^\circ}{1 + \frac{1}{2}} = \frac{10 \times \frac{\sqrt{3}}{2}}{\frac{3}{2}}$$
$$a = \frac{10 \sqrt{3}}{3} = \frac{10}{\sqrt{3}} \mathrm{~m/s^2}$$$
### Step 2: Solve for x
Comparing this evaluated value with the expression $ rac{x}{\sqrt{3}}$:
$$\frac{10}{\sqrt{3}} = \frac{x}{\sqrt{3}} \implies x = 10$$
Therefore, the value of $x$ is $10$.
### Pattern Recognition
Pure rolling problems reduce down to tracking the shape factor fraction $\beta = 1 + \frac{I}{MR^2}$. For solid cylinders it is $1.5$, for solid spheres it is $1.4$, and for hoops it is $2.0$. This value acts as an effective inertial scaling factor for gravity.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: System of Particles and Rotational Motion
Q53 (2024)
Two discs of moment of inertia $\mathrm{I}_1 = 4 \mathrm{~kg} \mathrm{~m}^2$ and $\mathrm{I}_2 = 2 \mathrm{~kg} \mathrm{~m}^2$ about their central axes & normal to their planes, rotating with angular speeds $10 \mathrm{~rad/s}$ & $4 \mathrm{~rad/s}$ respectively are brought into contact face to face with their axe of rotation coincident. The loss in kinetic energy of the system in the process is ________ $\mathrm{J}$.
### Related Formula
$$\text{C.O.A.M: } I_1 \omega_1 + I_2 \omega_2 = (I_1 + I_2) \omega_{\text{final}}$$
$$\Delta K.E. = \frac{1}{2} \frac{I_1 I_2}{I_1 + I_2} (\omega_1 - \omega_2)^2$$
### Core Logic
When two rotating discs are brought into contact, they exert friction on each other until they reach a common angular velocity. Angular momentum is conserved about the central axis. We can use the conservation of angular momentum to find the final angular velocity, or use the direct formula for loss in kinetic energy.
### Step 1: Calculate Final Angular Velocity (Alternative Method)
$$I_1 \omega_1 + I_2 \omega_2 = (I_1 + I_2)\omega_0$$
$$4(10) + 2(4) = (4 + 2)\omega_0$$
$$40 + 8 = 6\omega_0 \implies \omega_0 = 8 \mathrm{~rad/s}$$
### Step 2: Calculate Kinetic Energy Loss
Initial Energy $\mathrm{E}_1 = \frac{1}{2} I_1 \omega_1^2 + \frac{1}{2} I_2 \omega_2^2$
$$\mathrm{E}_1 = \frac{1}{2}(4)(100) + \frac{1}{2}(2)(16) = 200 + 16 = 216 \mathrm{~J}$$
Final Energy $\mathrm{E}_2 = \frac{1}{2} (I_1 + I_2) \omega_0^2$
$$\mathrm{E}_2 = \frac{1}{2} (6) (8^2) = 3 \times 64 = 192 \mathrm{~J}$$
Loss $\Delta E = \mathrm{E}_1 - \mathrm{E}_2 = 216 - 192 = 24 \mathrm{~J}$$
### Pattern Recognition
The loss formula $\Delta K = \frac{1}{2} \frac{I_1 I_2}{I_1 + I_2} (\omega_1 - \omega_2)^2$ is incredibly fast:
$\frac{1}{2} \times \frac{4 \times 2}{6} \times (10 - 4)^2 = \frac{1}{2} \times \frac{8}{6} \times 36 = \frac{4}{6} \times 36 = 24 \mathrm{~J}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: System of Particles and Rotational Motion
Q47 (2024)
A spherical body of mass $100 \mathrm{~g}$ is dropped from a height of $10 \mathrm{~m}$ from the ground. After hitting the ground, the body rebounds to a height of $5 \mathrm{~m}$. The impulse of force imparted by the ground to the body is given by: (given $g = 9.8 \mathrm{~m / s^2}$)
- $4.32 \mathrm{~kg\,ms^{-1}}$
- $43.2 \mathrm{~kg\,ms^{-1}}$
- $23.9 \mathrm{~kg\,ms^{-1}}$
- $2.39 \mathrm{~kg\,ms^{-1}}$
### Related Formula
$v = \sqrt{2gh}$
$$\vec{I} = \Delta \vec{P} = m(\vec{v}_f - \vec{v}_i)$$
### Core Logic
Impulse delivered by the ground equals the total change in momentum of the body during the collision. We must compute the velocity immediately before impact and immediately after rebound, respecting their opposite vector directions.
### Step 1: Calculate Velocities
Velocity just before hitting the ground ($v_i$):
$$v_i = \sqrt{2 \times 9.8 \times 10} = \sqrt{196} = -14 \mathrm{~m/s} \quad (\text{downwards})$$
Velocity just after rebounding ($v_f$):
$$v_f = \sqrt{2 \times 9.8 \times 5} = \sqrt{98} = +7\sqrt{2} \mathrm{~m/s} \quad (\text{upwards})$$
### Step 2: Calculate Impulse
Mass $M = 100 \mathrm{~g} = 0.1 \mathrm{~kg}$
$$\vec{I} = \Delta P = m(v_f - v_i)$$
$$\vec{I} = 0.1 \times [7\sqrt{2} - (-14)]$$
$$\vec{I} = 0.1(14 + 7\sqrt{2})$$
Since $\sqrt{2} \approx 1.414$:
$$\vec{I} = 0.1(14 + 7(1.414)) = 0.1(14 + 9.898) = 0.1(23.898)$$
$$\vec{I} \approx 2.39 \mathrm{~kg\,ms^{-1}}$$
### Pattern Recognition
When dealing with rebounds, $\Delta v$ is the sum of magnitudes $|v_1| + |v_2|$ because the direction reverses. A common trap is to subtract the magnitudes instead of adding.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: System of Particles and Rotational Motion
Class 11 Physics: Kinematics
Q48 (2024)
A particle of mass m projected with a velocity 'u' making an angle of $30^{\circ}$ with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height h is :
- $\frac{\sqrt{3}}{16} \frac{mu^3}{g}$
- $\frac{\sqrt{3}}{2} \frac{mu^2}{g}$
- $\frac{mu^3}{\sqrt{2}g}$
- $\text{zero}$
### Related Formula
$$L = m v_\perp r_{\perp} = m (u \cos \theta) H$$
$$H_{\max} = \frac{u^2 \sin^2 \theta}{2g}$$
### Core Logic
At maximum height, the vertical component of velocity is zero, so the only velocity is horizontal ($u \cos \theta$). The perpendicular distance from the line of action of this velocity to the origin is exactly the maximum height $H$.
### Step 1: Calculate Angular Momentum
$$L = m \cdot (u \cos \theta) \cdot H$$
Substitute $H = \frac{u^2 \sin^2 \theta}{2g}$:
$$L = m u \cos \theta \left( \frac{u^2 \sin^2 \theta}{2g} \right)$$
### Step 2: Plug in Angles
For $\theta = 30^\circ$:
$$L = \frac{m u^3}{2g} \cos(30^\circ) \sin^2(30^\circ)$$
$$L = \frac{m u^3}{2g} \times \frac{\sqrt{3}}{2} \times \left(\frac{1}{2}\right)^2$$
$$L = \frac{\sqrt{3} m u^3}{16g}$$
### Pattern Recognition
Angular momentum of a projectile at apex about the launch point strictly uses the horizontal velocity component coupled with the maximum height as the lever arm: $L = mv_x H$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: System of Particles and Rotational Motion
Class 11 Physics: Kinematics
Q58 (2024)
Consider a Disc of mass $5 \mathrm{~kg}$, radius $2 \mathrm{~m}$, rotating with angular velocity of $10 \mathrm{~rad/s}$ about an axis perpendicular to the plane of rotation. An identical disc is kept gently over the rotating disc along the same axis. The energy dissipated so that both the discs continue to rotate together without slipping is ________ J.
{{IMG1}}
### Related Formula
$$I = \frac{MR^2}{2} \quad (\text{for a solid disc})$$
$L = I \omega$
$$E = \frac{1}{2} I \omega^2$$
### Core Logic
Since no external torque acts on the system along the axis of rotation, the angular momentum of the system is conserved. After coupling, the total moment of inertia doubles, slowing the common angular velocity. The kinetic energy lost goes into frictional heat between the discs.
### Step 1: Calculate Initial State
Moment of inertia of one disc:
$$I = \frac{MR^2}{2} = \frac{5 \times 2^2}{2} = 10 \mathrm{~kg\,m^2}$$
Initial angular momentum:
$$L_i = I \omega_i = 10 \times 10 = 100 \mathrm{~kg\,m^2/s}$$
Initial kinetic energy:
$$E_i = \frac{1}{2} I \omega_i^2 = \frac{1}{2}(10)(10)^2 = 500 \mathrm{~J}$$
### Step 2: Conservation of Angular Momentum
$$\vec{L}_i = \vec{L}_f$$
$$100 = 2I \omega_f = 2(10) \omega_f$$
$$100 = 20 \omega_f \Rightarrow \omega_f = 5 \mathrm{~rad/s}$$
### Step 3: Calculate Final Energy and Loss
Final kinetic energy (for both discs):
$$E_f = \frac{1}{2} (2I) \omega_f^2 = \frac{1}{2} (20) (5)^2 = 10 \times 25 = 250 \mathrm{~J}$$
Energy dissipated:
$$\Delta E = E_i - E_f = 500 - 250 = 250 \mathrm{~J}$$
### Pattern Recognition
When an identical object drops onto a spinning object ($I_f = 2I_i$), conservation of $L$ dictates $\omega_f = \omega_i / 2$. Rotational KE ($L^2 / 2I$) is inversely proportional to $I$. Thus, doubling $I$ halves the KE. The loss is exactly half the initial energy.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: System of Particles and Rotational Motion
