Exams › NEET › Physics › System of Particles and Rotational Motion
137 questions with worked solutions.
Answer: the momentum of the particle is shared between the particle and the rod and remains conserved
The particle sticks to the rod, so the collision is perfectly inelastic, but the plane is smooth and cannot provide horizontal impulse. Therefore the total linear momentum of the particle-rod system is conserved, and that momentum is shared by both bodies after impact. Angular momentum depends on the chosen point, so it is not automatically conserved about arbitrary points like the rod’s midpoint.
Q2. The dimensional formula of torque is
Answer: [ML²T⁻²]
Torque is the product of force and perpendicular distance. The dimensional formula of force is [MLT⁻²], and distance is [L]. Multiplying these gives [ML²T⁻²].
Answer: \(-7\mathbf{i} - 4\mathbf{j} - 8\mathbf{k}\)
The moment of a force is given by the cross product of the position vector (from the point about which the moment is calculated to the point of application of the force) and the force vector. The position vector is \\(\mathbf{r} = (2 - 2)\mathbf{i} + (0 - (-2))\mathbf{j} + (-3 - (-2))\mathbf{k} = 0\mathbf{i} + 2\mathbf{j} - 1\mathbf{k}\\. The moment is \\(\mathbf{r} \times \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 2 & -1 \\ 4 & 5 & -6 \end{vmatrix} = -7\mathbf{i} - 4\mathbf{j} - 8\mathbf{k}\\.
Answer: 4i + 23j − 16k
Using conservation of momentum, the total momentum before and after the break must be equal. The initial momentum is the mass times velocity of the object. After the break, the momentum of the smaller and larger pieces is calculated using their respective masses and velocities. Solving for the velocity of the larger piece gives 4i + 23j − 16k.
Answer: 1/2 mv² + 1/2 Iω² = 1/2 kx²
At maximum compression, the total mechanical energy is conserved. The initial kinetic energy of the cylinder, which includes both translational (1/2 mv²) and rotational (1/2 Iω²) components, is converted into the potential energy of the spring (1/2 kx²).
Answer: zero
The center of mass of the system (boat + two persons) does not shift because there are no external forces acting on the system in the horizontal direction. The internal movement of the 55 kg man is balanced by the boat's movement in the opposite direction to conserve the center of mass.
Answer: (B) and (D)
Statement (B) is correct because the center of mass is the point where the total gravitational torque is zero. Statement (D) is also correct because a mechanical advantage greater than one implies that a smaller effort can lift a larger load. Statements (A) and (C) are incorrect as the center of mass does not always coincide with the center of gravity, and a couple only produces rotational motion, not translational motion.
Answer: 67 cm
The center of mass is calculated using the formula: \( x_{cm} = \frac{m_1x_1 + m_2x_2}{m_1 + m_2} \). Taking the 5 kg particle as the origin (\(x_1 = 0\)) and the 10 kg particle at \(x_2 = 1\) m, \(x_{cm} = \frac{(5 \times 0) + (10 \times 1)}{5 + 10} = \frac{10}{15} = 0.67\) m or 67 cm.
Answer: MLω² / 2
The force at the other end is due to the centrifugal force acting on the liquid. The mass element at a distance x from the axis experiences a force proportional to x. Integrating this force over the length of the tube gives the total force as MLω² / 2.
Answer: 1 : √2
The rotational kinetic energy is given by (1/2)Iω². Since the kinetic energies are equal, (1/2)I₁ω₁² = (1/2)I₂ω₂², which simplifies to I₁ω₁² = I₂ω₂². Angular momentum L = Iω, so L₁/L₂ = √(I₁/I₂). If I₁/I₂ = 1/2, then L₁/L₂ = 1/√2.
Answer: π / 15 N-m
The angular velocity is converted from 60 rpm to rad/s as ω = 2π × (60/60) = 2π rad/s. Using the equation τ = Iα, where α = (ω - 0)/t = 2π/60 rad/s², and I = 2 kg-m², we get τ = 2 × (2π/60) = π/15 N-m.
Answer: m₁ / m₂ d
To keep the center of mass stationary, the displacement of m₂ must balance the displacement of m₁. Using the center of mass condition, m₁ * d = m₂ * x, solving for x gives x = (m₁ / m₂) * d.
Q13. A disc is rotating with angular velocity ω. If a child sits on it, what is conserved?
Answer: (b) Angular momentum
When no external torque acts on the system, angular momentum is conserved. The child sitting on the disc does not introduce an external torque, so angular momentum remains constant.
Q14. A boy suddenly comes and sits on a circular rotating table. What will remain conserved?
Answer: (b) Angular momentum
When the boy sits on the rotating table, there is no external torque acting on the system. Hence, angular momentum is conserved. However, angular velocity, linear momentum, and kinetic energy may change.
Answer: (c) 10 rad/s
The angular acceleration is given by α = τ/I, where τ is the torque and I is the moment of inertia. Substituting the values, α = 1000/200 = 5 rad/s². Using ω = αt, where t = 3 s, we get ω = 5 × 3 = 15 rad/s.
Q16. A couple produces
Answer: (c) purely rotational motion
A couple consists of two equal and opposite forces whose lines of action do not coincide, creating a pure rotational effect without any net linear motion.
Q17. Angular momentum is
Answer: (a) vector (axial)
Angular momentum is an axial vector because it is the cross product of position and linear momentum, and its direction is determined by the right-hand rule.
Answer: (a) Iω
Angular momentum (L) is given by the formula L = Iω, where I is the moment of inertia and ω is the angular velocity. This is a direct application of the definition of angular momentum.
Answer: (a) 60 units
The angular momentum is given by L = mvr sinθ, where r is the perpendicular distance from the origin to the line of motion. The line y = √3x + 4 has a perpendicular distance of 2 units from the origin. Substituting m = 5, v = √3√2, and r = 2, we get L = 5 × √3√2 × 2 = 60 units.
Answer: (d) angular momentum about the centre of mass is conserved
During rolling and sliding motion, external forces like friction act on the sphere, which can change the angular momentum about the point of contact. However, angular momentum about the center of mass is conserved because no external torque acts about the center of mass.