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

To determine the speed of the cylinder at the base of the incline, we employ the law of conservation of energy. Because the object rolls without slipping, its initial gravitational potential energy transforms into both translational and rotational kinetic energy at the bottom.

The energy balance is expressed as the potential energy at height $h$ equal to the total kinetic energy, where the moment of inertia for a solid cylinder is $I = \frac{1}{2}MR^2$ and the angular velocity is $\omega = \frac{v}{R}$:

$Mgh = \frac{1}{2}Mv^{2} + \frac{1}{2}I\omega^{2} = \frac{1}{2}Mv^{2} + \frac{1}{2} \left( \frac{1}{2}MR^{2} \right) \left( \frac{v}{R} \right)^{2}$

By simplifying the terms on the right side, we obtain:

$Mgh = \frac{1}{2}Mv^{2} + \frac{1}{4}Mv^{2} = \frac{3}{4}Mv^{2}$

Solving this equation for $v$ yields the final velocity at the bottom of the incline:

$v = \sqrt{\frac{4}{3}gh}$

Since no external torque acts on the rotating ring when the additional masses are gently placed, the total angular momentum of the system remains conserved throughout the process. The initial angular momentum of the ring spinning about its central axis is the product of its moment of inertia, $I = Mr^2$, and its initial angular velocity, $\omega$.

After placing the four objects of mass $m$ at the opposite ends of two perpendicular diameters, each mass is situated at a distance $r$ from the center. The new moment of inertia for the system is the sum of the ring's moment of inertia and the four point masses, which is $I' = Mr^2 + 4mr^2 = (M + 4m)r^2$. By applying the principle of conservation of angular momentum, the initial angular momentum must equal the final angular momentum, expressed as $Mr^2 \omega = (M + 4m)r^2 \omega'$. Dividing both sides by the radius term $r^2$ and solving for the new angular velocity yields $\frac{M \omega}{M + 4m}$.

Since gravity acts as a conservative force, the total mechanical energy of the stone remains constant as it moves from the lowest point to the horizontal position, meaning the kinetic energy lost corresponds to the potential energy gained. Equating the energies at the bottom and horizontal positions gives
$\frac{1}{2} m u^2 = \frac{1}{2} m v^2 + mgl$
which reveals the velocity at the horizontal point to be $v = \sqrt{u^2 - 2gl}$. Because the stone's velocity is horizontal at the bottom and vertical at the horizontal position, these vectors are perpendicular. Consequently, the magnitude of the change in velocity is calculated as $\sqrt{u^2 + v^2}$, which yields $\sqrt{u^2 + (u^2 - 2gl)} = \sqrt{2(u^2 - gl)}$ upon substitution.

When an object rolls without slipping, its total kinetic energy is the sum of its translational kinetic energy, $\frac{1}{2}mv^2$, and its rotational kinetic energy, $\frac{1}{2}I\omega^2$. By substituting the definition of the radius of gyration $I = mK^2$ and the no-slip condition $\omega = \frac{v}{R}$, the rotational energy term becomes $\frac{1}{2}mv^2(\frac{K^2}{R^2})$. Consequently, the total energy of the system is the sum of these two components, resulting in $\frac{1}{2}mv^2(1 + \frac{K^2}{R^2})$. Dividing the rotational kinetic energy by this total energy value yields $\frac{\frac{K^2}{R^2}}{1 + \frac{K^2}{R^2}}$, which simplifies to $\frac{K^2}{R^2+K^2}$.

To locate the centre of gravity of a rod with a non-uniform mass distribution, we treat the object as a continuous system and calculate the weighted average position of all its mass elements. Since the mass per unit length varies linearly with distance $x$ as $\lambda = kx$, the mass of an infinitesimal element $dx$ at distance $x$ is given by $dm = kx \, dx$. The centre of gravity $x_c$ is determined by the ratio of the integral of the moment of mass to the total mass:

$x_c = \frac{\int_{0}^{3} x(kx \, dx)}{\int_{0}^{3} kx \, dx}$

Simplifying this expression by canceling the proportionality constant $k$ leaves the ratio of the integrals of $x^2$ and $x$ over the rod's length of 3 m. Evaluating the integral in the numerator gives $\left[ \frac{x^3}{3} \right]_0^3 = 9$, while evaluating the integral in the denominator gives $\left[ \frac{x^2}{2} \right]_0^3 = 4.5$. Dividing these results locates the centre of gravity at 2 m from the reference end.

Displacement represents the straight-line distance between the starting position of point $P$ at the bottom and its final position at the top after the wheel rotates halfway. Since the wheel rolls without slipping, the horizontal distance covered is equal to half the circumference, $\pi R$, which results in a horizontal shift of $\pi$ meters given the radius of 1 meter. Simultaneously, the point moves vertically from the ground to the top of the wheel, covering a distance equal to the diameter, $2R$, which is 2 meters. By applying the Pythagorean theorem to these perpendicular components, the total displacement is calculated as $\sqrt{\pi^2 + 2^2}$, which simplifies to $\sqrt{\pi^2 + 4}$ meters.

According to Newton's second law of motion, the linear acceleration of an object's center of mass is determined solely by the net external force acting on it and its total mass, which is expressed by the equation $a = F/m$. Since the sphere is placed on a perfectly smooth surface, there is no frictional force to influence the translation of the center of mass, regardless of where the horizontal force is applied. While varying the height $h$ changes the torque and thus the angular acceleration of the sphere, it exerts no influence on the net force causing linear motion. Therefore, the acceleration of the center of mass is independent of $h$, meaning there is no specific relation between the height of application and the radius $R$.

The moment of inertia of a circular disc about its geometrical axis is the sum of the moments of inertia of its constituent concentric rings, where the contribution of each ring is defined by $I = M R^2$. Since iron has a higher density than aluminium, the mass $M$ of a ring of a specific thickness is greater if it is made of iron. To maximize the total moment of inertia, the larger mass contribution provided by the iron must be placed at the largest possible radius, effectively positioning the denser material at the exterior and the lighter aluminium at the interior.

The principle of conservation of angular momentum dictates that a system's angular momentum remains constant as long as there is no net external torque acting upon it. When a child sits on a rotating disc, they become part of the system, and since no external torques are introduced, the total angular momentum is conserved. Consequently, because the system's moment of inertia increases as the child moves toward the center, the angular speed must decrease to maintain the balance defined by $L = I\omega$.

Rolling motion without slipping is the superposition of translational motion of the center of mass with velocity $v_{\text{cm}}$ and rotational motion about the center of mass with tangential velocity $v_{\text{rot}} = \omega R = v_{\text{cm}}$. At the topmost point, the translational and rotational velocity vectors point in the same direction such that $v_{\text{top}} = v_{\text{cm}} + v_{\text{cm}} = 2v_{\text{cm}}$, while at the point of contact with the ground, these vectors act in opposite directions resulting in $v_{\text{contact}} = v_{\text{cm}} - v_{\text{cm}} = 0$.

Torque is the rotational equivalent of force and is computed using the cross product $\vec{\tau} = \vec{r} \times \vec{F}$. Placing the position vector $\vec{r} = 7\hat{i} + 3\hat{j} + \hat{k}$ in the second row and the force vector $\vec{F} = -3\hat{i} + \hat{j} + 5\hat{k}$ in the third row, we construct the following determinant:

$\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 7 & 3 & 1 \\ -3 & 1 & 5 \end{vmatrix}$

The $\hat{i}$ component results from $(3)(5) - (1)(1) = 14\hat{i}$, while the $\hat{j}$ component is the negative of $(7)(5) - (1)(-3)$, yielding $-38\hat{j}$. Finally, the $\hat{k}$ component comes from $(7)(1) - (3)(-3) = 16\hat{k}$, which completes the calculation as $14\hat{i} - 38\hat{j} + 16\hat{k}$.

Torque is defined as the product of the force magnitude and the perpendicular distance from the axis of rotation, and for a system to be in rotational equilibrium, the net torque must equal zero. In an equilateral triangle, the perpendicular distance from the centre O to each of the three sides is identical; let this constant distance be denoted by $x$. Since the forces $\vec{F}_1, \vec{F}_2,$ and $\vec{F}_3$ act along the sides, each produces a torque about the centre equal to the product of the force and distance $x$. Setting the total torque to zero implies that the combined rotational effect of two forces must be balanced by the third, expressed as $F_1 x + F_2 x = F_3 x$, which simplifies to $F_3 = F_1 + F_2$.

The moment of inertia of a rigid body is governed by the distribution of its mass relative to the axis of rotation, defined by the relationship $I = \sum mr^2$. Because this value relies on the square of the distance between each mass element and the axis, any configuration that positions the majority of the object's mass further away will result in a higher moment of inertia. Comparing the geometry of the two axes, the mass is distributed at a greater effective distance from the axis represented by $I_2$ than from the axis represented by $I_1$, establishing that $I_2 > I_1$.

When a solid sphere rolls down an inclined plane from rest without sliding, the mechanical energy is conserved, meaning the initial potential energy is entirely converted into a combination of translational and rotational kinetic energy at the bottom. The potential energy is given by $mgh$, and the total kinetic energy comprises translational kinetic energy $\frac{1}{2}mv^2$ and rotational kinetic energy $\frac{1}{2}I\omega^2$, where the moment of inertia $I$ for a solid sphere is $\frac{2}{5}mr^2$ and the angular velocity $\omega$ equals $v/r$ due to the no-slip condition.

Substituting the moment of inertia and angular velocity into the energy equation provides the total kinetic energy:
$\frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{2}{5}mr^2\right)\left(\frac{v}{r}\right)^2 = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$
Equating the potential energy to the total kinetic energy, $mgh = \frac{7}{10}mv^2$, allows us to solve for $v$, resulting in $v = \sqrt{\frac{10}{7}gh}$.

In vertical circular motion, the tension in the string is determined by the combined effects of the gravitational force and the centripetal force required to maintain the path. At the lowest point, which is point B, the string must support the full weight of the mass in addition to providing the necessary centripetal force, resulting in a maximum tension given by $T = mg + \frac{mv^2}{R}$. Because the tension is inherently greatest at this location, any increase in velocity that pushes the force beyond the string's limit will cause it to snap precisely at this point.

Torque, which measures the rotational effect of a force, is defined by the cross product of the position vector relative to the origin and the force vector applied at that position. Using the relation $\vec{\tau} = \vec{r} \times \vec{F}$, we organize the components of $\vec{r} = 3\hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{F} = 2\hat{i} - 3\hat{j} + 4\hat{k}$ into a determinant to perform the multiplication. Evaluating the determinant

involves expanding the terms to get $\hat{i}(2 \times 4 - 3 \times -3) - \hat{j}(3 \times 4 - 3 \times 2) + \hat{k}(3 \times -3 - 2 \times 2)$. Simplifying this expression leads to $17\hat{i} - 6\hat{j} - 13\hat{k}\,\text{N}\cdot\text{m}$, which represents the final torque vector.

When objects roll down an inclined plane without slipping, their acceleration is governed by the rotational inertia factor $\frac{K^2}{R^2}$, where objects with a smaller ratio experience higher acceleration. For a solid cylinder, this ratio is $\frac{1}{2}$, whereas for a hollow cylinder, it is $1$. Since the solid cylinder possesses a lower value for this factor, it accelerates more rapidly and reaches the bottom of the incline before the hollow cylinder.

To determine the duration for which the angular acceleration acts, we must connect the rotational kinetic energy to the principles of rotational kinematics. Since the body starts from rest and reaches a specific amount of rotational kinetic energy, we can calculate the final angular velocity using the relationship $E_r = \frac{1}{2} I \omega^2$. Rearranging this formula to solve for angular velocity gives $\omega = \sqrt{\frac{2 E_r}{I}}$. By substituting the values of 1500 J for energy and 1.2 kg·m² for the moment of inertia, we find the final angular velocity to be 50 rad/s.

With the final angular velocity known, we apply the kinematic equation for constant angular acceleration, $\omega_f = \omega_i + \alpha t$. Given that the initial angular velocity is 0 rad/s, this simplifies to $\omega_f = \alpha t$. By dividing the final velocity of 50 rad/s by the angular acceleration of 25 rad/s², we find the time required to reach this state is 2 s.

The angular momentum of a particle relative to the origin is defined as the product of its linear momentum and the perpendicular distance from the origin to the particle's line of motion. For a particle with mass 5 moving with a speed of $3\sqrt{2}$, the linear momentum is $5 \times 3\sqrt{2} = 15\sqrt{2}$. The particle follows the path $Y = X + 4$, which intersects the Y-axis at a distance of 4 units from the origin. Because the slope of this line is 1, it forms an angle of 45 degrees with the X-axis. Using trigonometry, the perpendicular distance from the origin to this path is $4 \sin(45^{\circ}) = 4 \times \frac{1}{\sqrt{2}} = 2\sqrt{2}$. The total angular momentum is the product of the linear momentum $15\sqrt{2}$ and this perpendicular distance $2\sqrt{2}$, which results in 60 units.