System Of Particles And Rotational Motion – NEET UG PhysicsPractice Questions & PYQs

The stability of the coin on a rotating record depends on the static friction force providing the necessary centripetal force required for circular motion. For the coin to avoid slipping, the maximum available frictional force $\mu mg$ must at least equal the required centripetal force $mr\omega^2$. Solving the inequality $\mu mg \ge mr\omega^2$ for the distance $r$ confirms that the coin remains on the record as long as $r \le \frac{\mu g}{\omega^2}$.

The fundamental definition of torque acting on a particle is given by the cross product of the position vector and the force vector, expressed as $\vec{\tau} = \vec{r} \times \vec{F}$. A key mathematical property of the cross product is that the resulting vector is always mutually perpendicular to both of its constituent vectors. Because the torque vector $\vec{\tau}$ lies in a direction orthogonal to both the position vector $\vec{r}$ and the force vector $\vec{F}$, the dot product of the torque with either of these vectors must equate to zero, yielding the relations $\vec{r} \cdot \vec{\tau} = 0$ and $\vec{F} \cdot \vec{\tau} = 0$.

The parallel axis theorem, which states $I = I_{CM} + Md^2$, allows for the determination of the moment of inertia of a rigid body about any axis parallel to one passing through its center of mass. For a single thin rod of mass $M$ and length $l$, the moment of inertia about an axis through its midpoint and perpendicular to its length is $\frac{Ml^2}{12}$. Each of the four rods in the square frame is positioned such that its center is at a distance $d = \frac{l}{2}$ from the axis passing through the center of the square and perpendicular to its plane.

Applying the parallel axis theorem to each rod, the moment of inertia contribution for an individual rod is $\frac{Ml^2}{12} + M(\frac{l}{2})^2$, which simplifies to $\frac{Ml^2}{3}$. Summing the contributions of all four identical rods, the total moment of inertia for the entire square frame is:

$I = \frac{4}{3} Ml^2$

The position of the centre of mass for a system of discrete particles is determined by calculating the weighted average of the position vectors of each mass. This principle is mathematically expressed by the formula $\vec{r}_{CM} = \frac{\sum m_i \vec{r}_i}{\sum m_i}$, where the numerator represents the vector sum of each individual mass multiplied by its corresponding position vector, and the denominator represents the total mass of the system.

Given the masses $1\,\text{kg}$ and $3\,\text{kg}$ with position vectors $\hat{i} + 2\hat{j} + \hat{k}$ and $-3\hat{i} - 2\hat{j} + \hat{k}$, we first compute the weighted sum of these vectors. Multiplying the first mass by its position gives $1(\hat{i} + 2\hat{j} + \hat{k}) = \hat{i} + 2\hat{j} + \hat{k}$, while multiplying the second mass by its position yields $3(-3\hat{i} - 2\hat{j} + \hat{k}) = -9\hat{i} - 6\hat{j} + 3\hat{k}$. Adding these two expressions results in $-8\hat{i} - 4\hat{j} + 4\hat{k}$. Dividing this resultant vector by the total system mass of $4\,\text{kg}$ gives the final position vector of the centre of mass, which is $-2\hat{i} - \hat{j} + \hat{k}$.

Because the masses are attached gently, no external torque acts on the rotating ring system, which means the total angular momentum remains constant. Initially, the ring has a moment of inertia of $I = MR^2$ rotating with angular velocity $\omega$, so the initial angular momentum is $L = MR^2 \omega$. When the two masses are placed at opposite ends of a diameter, they effectively become point masses at a distance $R$ from the center, creating a new total moment of inertia of $I' = MR^2 + 2mR^2 = (M + 2m)R^2$. By equating the initial angular momentum to the final angular momentum, $MR^2 \omega = (M + 2m)R^2 \omega'$, and solving for the new angular velocity, we find $\omega' = \frac{M\omega}{M + 2m}$.

The radius of gyration $k$ of a rotating body relates its moment of inertia $I$ and mass $M$ through the expression
$I = Mk^2$
For a circular disc of radius $R$, the moment of inertia about its central axis is
$I_{disc} = \frac{1}{2}MR^2$
which leads to a radius of gyration of $k_{disc} = \frac{R}{\sqrt{2}}$. In contrast, the moment of inertia for a circular ring of the same radius about its central axis is
$I_{ring} = MR^2$
which results in a radius of gyration of $k_{ring} = R$. Comparing these two values by taking the ratio of the disc's radius of gyration to the ring's radius of gyration yields
$\frac{k_{disc}}{k_{ring}} = \frac{R/\sqrt{2}}{R} = \frac{1}{\sqrt{2}}$

The moment of inertia for a uniform rod of mass $m$ and length $l$ rotating about an axis passing through one end perpendicular to its length is defined by $I = \frac{ml^2}{3}$. When the rod of total mass $M$ and total length $L$ is bent at its midpoint, it divides into two identical segments, each possessing a mass of $M/2$ and a length of $L/2$. Because both segments rotate about the common bending point perpendicular to the plane formed by the halves, their individual moments of inertia are additive. Combining the contributions of both segments leads to the calculation $2 \times \frac{(M/2)(L/2)^2}{3}$, which simplifies to $\frac{ML^2}{12}$.

The angular acceleration of a rotating object is determined by the relationship between the applied torque and its moment of inertia, as defined by the rotational version of Newton's second law, $\tau = I \alpha$. For a uniform rod released from a horizontal position, the gravitational torque arises from the weight acting at the center of mass, which is located at the midpoint $l/2$, resulting in a torque magnitude of $\tau = mgl/2$. By substituting this torque value and the provided moment of inertia $I = ml^2/3$ into the rotational equation, we get the relationship:

$\frac{mgl}{2} = \left( \frac{ml^2}{3} \right) \alpha$

Solving this equation for $\alpha$ by multiplying both sides by the reciprocal of the moment of inertia yields the initial angular acceleration of $3g/2l$.

The total angular displacement of a rotating object is determined by the kinematic equation $\theta = \omega_i t + \frac{1}{2} \alpha t^2$, which relates the final position to the initial angular velocity, angular acceleration, and elapsed time. Given an initial angular velocity of 2 rad/s, an angular acceleration of 3 rad/s^2, and a time duration of 2 s, we substitute these values into the formula to find the total angle:
$\theta = 2(2) + \frac{1}{2}(3)(2^2)$
Calculating this expression results in a total angular displacement of 10 rad.

The angular momentum of a particle with respect to an origin is defined by the product of its mass, velocity, and the perpendicular distance from the origin to the line of motion. Since the particle is moving along a fixed straight line, the perpendicular distance from the origin to the line of travel remains constant at every point on the path. Because the mass and velocity also remain unchanged throughout this motion, the angular momentum at any position is identical, meaning $L_A = L_B$.

The parallel axis theorem is the fundamental principle used to determine the moment of inertia about an axis that is shifted from the center of mass. For a uniform circular disc, the moment of inertia about an axis passing through its center and perpendicular to its plane is given as $I_{C} = \frac{1}{2} MR^{2}$. When the axis of rotation is shifted to the edge of the disc, the distance between the parallel axes becomes $h = R$. Applying the parallel axis theorem, which states that the total moment of inertia is the sum of the central moment of inertia and the term $Mh^{2}$, yields the equation $I = I_{C} + Mh^{2}$. Substituting the known values into this formula gives $I = \frac{1}{2} MR^{2} + MR^{2}$, which simplifies to $\frac{3}{2} MR^{2}$.

The distribution of mass within a uniform, incompressible liquid column rotating about one end can be simplified by treating the total mass as if it were concentrated entirely at the geometric center of the tube. This center of mass is situated at a distance $r = \frac{L}{2}$ from the pivot point. By applying the standard definition of centrifugal force $F = M r \omega^{2}$, the force exerted at the outer end is calculated by substituting this effective radius into the expression, which yields a magnitude of $\frac{M L \omega^{2}}{2}$.

Rotational kinetic energy is determined by the expression $K = \frac{1}{2} I \omega^2$, where $I$ is the moment of inertia and $\omega$ is the angular velocity. For two bodies with equal kinetic energies, we equate their respective expressions:

$\frac{1}{2} I_1 \omega_1^2 = \frac{1}{2} I_2 \omega_2^2$

Substituting the given values $I_1 = I$ and $I_2 = 2I$ results in the relationship $I \omega_1^2 = 2I \omega_2^2$. Dividing both sides by $I$ reveals $\omega_1^2 = 2 \omega_2^2$, indicating that the ratio of the squares of the angular velocities is 2 : 1, which simplifies to $\sqrt{2} : 1$ upon taking the square root.

The Parallel Axis Theorem provides the most direct approach for calculating the moment of inertia when the axis of rotation is shifted away from the object's center of mass. For a circular disc rotating about an axis passing through its center and perpendicular to its plane, the standard moment of inertia is $I_{cm} = \frac{1}{2}MR^2$. Because the new axis of rotation is located at the edge of the disc, it is parallel to the central axis at a perpendicular distance equal to the radius $R$. By applying the theorem, we add the product of the mass and the square of this distance to the central moment of inertia, yielding $I = \frac{1}{2}MR^2 + MR^2$, which simplifies to $\frac{3}{2}MR^2$.

When a drum rolls down an incline without slipping, gravity acts to accelerate its center of mass, driving translational motion down the slope. Static friction at the contact point exerts a torque that opposes this sliding tendency, forcing the object to spin. This interaction effectively converts a portion of the drum's translational kinetic energy into rotational kinetic energy, ensuring the drum rolls smoothly rather than sliding.

Newton's second law for rotation dictates that applying a torque to an object results in angular acceleration, or in this case, angular deceleration. To bring the wheel to rest, the applied torque must counteract its initial angular momentum, where the magnitude of this torque relates to the moment of inertia and the angular acceleration by the relation $\tau = I\alpha$.

Converting the rotational rate of $60\,\text{rpm}$ into radians per second is necessary to work with standard units. Since one revolution covers $2\pi$ radians and a minute consists of $60\,\text{s}$, the initial angular velocity is $\omega = \frac{60 \times 2\pi}{60} = 2\pi\,\text{rad/s}$. Because the wheel must stop, the required angular deceleration is the change in angular velocity over time, which gives $\alpha = \frac{2\pi\,\text{rad/s}}{60\,\text{s}} = \frac{\pi}{30}\,\text{rad/s}^2$. Using the given moment of inertia of $2\,\text{kg}\,\text{m}^2$, the magnitude of the torque required to stop the wheel is:

$\tau = I \alpha = 2\,\text{kg}\,\text{m}^2 \times \frac{\pi}{30}\,\text{rad/s}^2 = \frac{\pi}{15}\,\text{Nm}$

The radius of gyration provides a measure of how an object's mass is distributed relative to its axis of rotation, governed by the relationship $I = MK^2$. For a circular disc of radius $R$ rotating about an axis tangential to its plane, the radius of gyration is $K_1 = \frac{\sqrt{5}}{2} R$, while for a circular ring of the same radius $R$ about a similar tangential axis in its plane, the radius of gyration is $K_2 = \sqrt{\frac{3}{2}} R$. Comparing these two values by dividing $K_1$ by $K_2$ causes the radius $R$ to cancel out, simplifying the expression to $\frac{\sqrt{5}}{2} \div \sqrt{\frac{3}{2}}$, which yields the ratio $\sqrt{5} : \sqrt{6}$.

Because no external torque acts on the rotating system, the total angular momentum remains constant throughout the interaction between the two discs. Initially, the system's angular momentum is simply the product of the first disc's moment of inertia and its angular velocity, denoted as $I_{1} \omega$. Once the second disc is placed on top, the two discs rotate together as a single body with a combined moment of inertia of $I_{1} + I_{2}$. Setting the initial angular momentum equal to the final angular momentum provides the relation $I_{1} \omega = (I_{1} + I_{2}) \omega_{f}$, which confirms that the final angular velocity is $\omega_{f} = \frac{I_{1} \omega}{I_{1} + I_{2}}$.

The position of the centre of mass of a two-particle system remains unchanged if the sum of the weighted displacements of the individual masses is zero, which is represented by the equation $m_1 \Delta x_1 + m_2 \Delta x_2 = 0$. When the particle of mass $m_1$ moves a distance $d$ towards the centre of mass, its displacement is effectively $-d$. To maintain the system's centre of mass at its original position, we substitute these values into the balance equation to get $m_1(-d) + m_2 \Delta x_2 = 0$. Solving this for the displacement of the second particle, $\Delta x_2$, results in $\Delta x_2 = \frac{m_1}{m_2} d$.

The moment of inertia for a system of point particles about a specific axis is determined by the sum of each mass multiplied by the square of its perpendicular distance from that axis, expressed as $I = \sum m_i r_i^2$. In this configuration, three particles of mass $m$ occupy the vertices of an equilateral triangle with side length $l$. The axis of rotation, AX, passes through vertex A and is perpendicular to side AB. Since particle A lies directly on this axis, its perpendicular distance $r_A$ is $0$, contributing nothing to the rotational inertia.

Particle B is positioned on side AB, which is perpendicular to the axis AX, placing it at a distance of $r_B = l$. For particle C, we find the perpendicular distance by considering the geometry of the triangle; the angle between side AC and the axis AX is $30^\circ$, making the perpendicular distance $r_C = l \sin 30^\circ = \frac{l}{2}$. Combining these values into the formula provides the total moment of inertia:

$I = m(0)^2 + m(l)^2 + m\left(\frac{l}{2}\right)^2$
$I = ml^2 + \frac{1}{4}ml^2 = \frac{5}{4} m l^2$