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

When a solid sphere rolls without slipping, its motion comprises both translational and rotational kinetic energy. We determine the translational component using $K_t = \frac{1}{2}mv^2$ and the rotational component using $K_r = \frac{1}{2}I\omega^2$, where the moment of inertia for a solid sphere is $I = \frac{2}{5}mr^2$. Substituting the rolling condition $\omega = \frac{v}{r}$ into the rotational energy expression simplifies it to $K_r = \frac{1}{5}mv^2$.

Combining these two components yields a total kinetic energy of $K = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$. By taking the ratio of the rotational kinetic energy to the total kinetic energy, we find:
$\frac{K_r}{K} = \frac{\frac{1}{5}mv^2}{\frac{7}{10}mv^2} = \frac{2}{7}$

A sphere rolling without slipping possesses two types of kinetic energy: translational kinetic energy due to the movement of its center of mass, and rotational kinetic energy due to its spin about that center. Translational kinetic energy is defined as $K_{\text{trans}} = \frac{1}{2}mv^2$, while rotational kinetic energy is defined as $K_{\text{rot}} = \frac{1}{2}I\omega^2$.

For a solid sphere, the moment of inertia about the center is $I = \frac{2}{5}mr^2$. Using the rolling condition $\omega = \frac{v}{r}$, the rotational energy becomes:
$K_{\text{rot}} = \frac{1}{2} \left( \frac{2}{5}mr^2 \right) \left( \frac{v}{r} \right)^2 = \frac{1}{5}mv^2$
The total kinetic energy is the sum of these two contributions:
$K_{\text{total}} = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$
Dividing the translational kinetic energy by the total kinetic energy yields the ratio:
$\frac{K_{\text{trans}}}{K_{\text{total}}} = \frac{\frac{1}{2}mv^2}{\frac{7}{10}mv^2} = \frac{5}{7}$

The moment of inertia for a solid sphere is given by $\frac{2}{5} \times \text{mass} \times \text{radius}^2$ about an axis passing through its center. To determine the moment of inertia about an axis shifted from the center, we apply the parallel axis theorem, which adds the product of the sphere's mass and the square of the distance between the two parallel axes to the original moment of inertia. Since the spheres are attached, the distance between their centers is $R+r$.

When calculating the total moment of inertia about the axis passing through the center of sphere $A$, we combine the moment of inertia of sphere $A$ with the parallel-shifted moment of inertia of sphere $B$, resulting in $I_A = \frac{2}{5}MR^2 + \frac{2}{5}mr^2 + m(R+r)^2$. Similarly, for the axis passing through the center of sphere $B$, we add the parallel-shifted moment of inertia of sphere $A$ to the moment of inertia of sphere $B$, resulting in $I_B = \frac{2}{5}mr^2 + \frac{2}{5}MR^2 + M(R+r)^2$. Subtracting these two expressions causes the terms representing the individual moments of inertia about their own centers to cancel out, leaving the final difference as $(m-M)(R+r)^2$.

The Parallel Axis Theorem is the fundamental principle used to determine the moment of inertia for an object rotating about an axis offset from its center of mass. For a thin circular ring, the moment of inertia about a diameter passing through its center is $I_{\text{cm}} = \frac{1}{2} M r^2$. Because the axis $yy^{\prime}$ is tangent to the ring and parallel to a diameter, the perpendicular distance $d$ between the center axis and $yy^{\prime}$ is exactly the radius $r$ of the ring. Applying the theorem $I = I_{\text{cm}} + Md^2$, the total moment of inertia becomes $\frac{1}{2}Mr^2 + Mr^2$, resulting in $\frac{3}{2}Mr^2$.

Defining the physical parameters based on the wire's length $L$ and linear mass density $m$ leads to a total mass of $M = mL$. Geometrically, the circumference relation $2\pi r = L$ establishes the radius as $r = \frac{L}{2\pi}$. Substituting these expressions into the formula $\frac{3}{2}Mr^2$ results in the calculation $\frac{3}{2}(mL)\left(\frac{L}{2\pi}\right)^2$. Expanding the square term produces $\frac{3}{2}mL \cdot \frac{L^2}{4\pi^2}$, which simplifies to $\frac{3 mL^3}{8 \pi^2}$.

The centre of mass of a system is defined geometrically based solely on the spatial coordinates and the individual masses of the particles, as represented by the vector formula $\vec{R}_{cm} = \frac{\sum m_i \vec{r}_i}{\sum m_i}$. Because this calculation relies entirely on the distribution of mass rather than the dynamics of motion, the instantaneous location of the centre of mass remains unaffected by any external or internal forces acting on the system. Therefore, regardless of how forces influence the movement or acceleration of these particles, they do not play a role in determining the balance point of the total mass distribution.

The center of mass for a system of objects represents the weighted average position of the mass distribution, which for each sphere of radius $r$ is situated at its geometric center. Since the three spheres are identical and their centers form an equilateral triangle, the system exhibits perfect rotational symmetry around the centroid of this triangle. This balance ensures that the center of mass of the entire assembly coincides with the point where the medians of the triangle intersect.

Understanding the time evolution of the total linear momentum in this system requires decomposing the velocity vectors of the two particles at any arbitrary angular position $\theta$. Since both particles possess mass $m=1$ and move with uniform speed $v$ along the circular path, their positions remain symmetric about the vertical axis throughout the motion. At any angle $\theta$, the horizontal components of their velocities are equal in magnitude but opposite in direction, effectively cancelling each other out, while the vertical components add together, resulting in a net momentum magnitude of $P = 2v|\cos\theta|$. As the particles traverse the path from $\theta=\frac{\pi}{2}$ to $\theta=-\frac{\pi}{2}$, this magnitude starts at zero, rises to a peak value of $2v$ when the particles align horizontally at $\theta=0$, and decreases symmetrically back to zero.

The moment of inertia of an object is determined by the integral $I = \int r^2 \, dm$, which quantifies the distribution of mass relative to the axis of rotation. For a rectangle with side lengths satisfying $BC = 2AB$, the axis $EG$ is uniquely positioned so that the mass elements are kept at the minimum average squared distance from the center. This specific geometric alignment ensures that the resistance to rotational motion is at its lowest possible value compared to other potential axes of rotation.

When objects roll down an inclined plane without slipping, their linear acceleration depends on their mass distribution relative to the axis of rotation, which is captured by the ratio $K^2/R^2$. A smaller value for this ratio means the object is more efficient at converting potential energy into translational motion, resulting in higher acceleration and a shorter descent time. Since the solid sphere has a ratio $K^2/R^2 = 2/5$, which is lower than the value of $1/2$ shared by both the disc and solid cylinder, the sphere gains the highest acceleration and arrives at the bottom first.

The kinetic energy of any object undergoing rotational motion is calculated using the expression $K = \frac{1}{2} I \omega^2$, where $I$ represents the moment of inertia relative to the rotation axis. For a ring rotating about an axis passing through its center and perpendicular to its plane, the moment of inertia is known to be $I = m r^2$. Substituting this relationship into the kinetic energy formula leads to $K = \frac{1}{2} (m r^2) \omega^2$, which simplifies to $\frac{1}{2} m r^2 \omega^2$.

Whenever a rigid body spins around a fixed axis that passes through its center of mass, the angular momentum remains invariant regardless of which point along that axis is chosen for calculation. This occurs because the center of mass is at rest, meaning the linear momentum term in the general expression for angular momentum, $\vec{L} = \vec{r} \times \vec{P} + \vec{L}_{CM}$, becomes zero. Consequently, the angular momentum about both point $A$ and point $B$ simplifies to the intrinsic spin angular momentum, $I_{CM} \omega$. Since this quantity is identical for any point on the axis of rotation, the ratio $\frac{L_A}{L_B}$ is equal to 1.

When a rotating object experiences constant angular acceleration, its average speed over a given interval is equal to the arithmetic mean of its initial and final values. Converting the flywheel's operating speeds from revolutions per minute to revolutions per second simplifies the arithmetic: the initial speed is 10 revolutions per second, while the final speed is 20 revolutions per second. Because the acceleration is uniform, the average speed is $\frac{10 + 20}{2} = 15$ revolutions per second. Multiplying this average speed by the duration of 10 seconds determines that the flywheel completes 150 revolutions.

The parallel axis theorem allows us to calculate the moment of inertia about an axis by adding the product of the mass and the square of the distance between the axes to the moment of inertia about a parallel axis passing through the center of mass. For a circular disc, the moment of inertia about its diameter is given by $I = \frac{1}{4} M R^{2}$. Since the tangential axis is parallel to the diameter at a distance of $R$, we apply the theorem by adding $M R^{2}$ to the initial value to obtain the final result, $I = \frac{1}{4} M R^{2} + M R^{2} = \frac{5}{4} M R^{2}$.

A couple consists of two equal, parallel, and oppositely directed forces acting at different points on a rigid body, resulting in a net force of zero and a non-zero net torque. Since the resultant force is zero, the body remains in translational equilibrium and does not undergo linear motion. Therefore, the net torque, given by the formula $\tau = F \cdot d$ where $F$ is the magnitude of one of the forces and $d$ is the perpendicular distance between their lines of action, causes the body to experience purely rotational motion.

A rigid body spinning around a fixed axis stores energy due to its rotation, which is defined by the relationship $K.E. = \frac{1}{2} I \omega^2$, where $I$ denotes the moment of inertia and $\omega$ corresponds to the angular speed. We rearrange this formula to solve for the moment of inertia as $I = \frac{2 K.E.}{\omega^2}$. By substituting the known energy of 360 J and the angular velocity of 30 rad/sec, we obtain the calculation $I = \frac{2 \times 360}{30^2}$. This expression simplifies to $I = \frac{720}{900}$, which results in $I = 0.8$ kg m².

The force of kinetic friction on the sphere acts entirely at the point of contact with the horizontal surface. Because this force vector passes directly through the point of contact, it generates zero torque about that specific location, implying that the net external torque about the point of contact is zero. Consequently, the total angular momentum, expressed as $\vec{L} = I_{cm}\vec{\omega} + \vec{r} \times M\vec{v}$, remains constant throughout the motion. This conservation holds even though linear momentum is not preserved, as the frictional force continuously acts on the system.

The perpendicular axis theorem dictates that the moment of inertia of the disc about an axis perpendicular to its plane and passing through its center is the sum of its moments of inertia about two perpendicular diameters, resulting in $I + I = 2I$. To calculate the moment of inertia about a parallel axis passing through the rim, apply the parallel axis theorem, which states that the total moment of inertia is the sum of the moment of inertia about the center of mass and the product $MR^2$. Given that the moment of inertia about a diameter is $I = \frac{1}{4}MR^2$, the term $MR^2$ equals $4I$, which means the final moment of inertia becomes $2I + 4I = 6I$.

When two objects of identical mass undergo a perfectly elastic collision, they exchange their velocities completely. Consequently, when mass $A$ strikes the stationary bob $B$ at $10\ \mathrm{ms^{-1}}$, bob $B$ instantaneously acquires that same speed of $10\ \mathrm{ms^{-1}}$ while mass $A$ comes to a rest. As bob $B$ swings upward, it converts its initial kinetic energy into gravitational potential energy at the highest point of its trajectory. This energy transformation is governed by the relation:

$mgh = \frac{1}{2}mv^2$

Canceling the mass $m$ from both sides, we isolate the height variable as $h = \frac{v^2}{2g}$. Inserting the given values of $v = 10\ \mathrm{ms^{-1}}$ and $g = 10\ \mathrm{ms^{-2}}$ yields $h = \frac{100}{20}$, which determines that bob $B$ rises to a height of $5\ \mathrm{m}$.

The moment of inertia of a uniform triangular plate about any of its sides is directly proportional to the square of the altitude dropped from the opposite vertex to that side, represented by the relationship $I = \frac{1}{6}mh^2$. In this specific triangular geometry, the altitude from vertex $A$ to side $BC$ is greater than the altitude from vertex $C$ to side $AB$. Because the mass is effectively distributed further from the axis $BC$ compared to the axis $AB$, the resulting moment of inertia about side $BC$ is larger, leading to the relation $I_{BC} > I_{AB}$.

Static equilibrium requires that the total force and total torque acting on the rod must be zero. The rod is subject to its weight acting downwards at the center of mass, a normal force from the floor, a friction force from the floor, and a normal force from the smooth wall. Vertical equilibrium implies the floor's normal force balances the rod's weight, while horizontal equilibrium dictates that the friction force at the floor equals the normal force from the wall.

Since the rod makes an angle of $60^{\circ}$ with the vertical wall, it makes an angle of $30^{\circ}$ with the horizontal floor. Setting the torque about the point of contact on the floor, the torque from the weight must be balanced by the torque from the wall's normal force:
$Mg \cdot \frac{L}{2} \cos 30^{\circ} = N_{wall} \cdot L \sin 30^{\circ}$
Solving this equation for the wall force gives $N_{wall} = \frac{Mg}{2} \cot 30^{\circ}$. Substituting the values $M = 20$ kg, $g = 10 \text{ m/s}^2$, and $\cot 30^{\circ} = \sqrt{3}$, the friction force is calculated to be $100 \sqrt{3}$ N.