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

The work-energy theorem states that the work done by the torque to stop the rotation equals the change in rotational kinetic energy, expressed as $\tau \theta = \frac{1}{2} I \omega^2$. The moment of inertia for a solid cylinder rotating about its axis is $I = \frac{1}{2} m r^2 = \frac{1}{2} (2)(0.04)^2 = 1.6 \times 10^{-3} \text{ kg m}^2$. The initial angular velocity is $\omega = 3 \times \frac{2\pi}{60} = \frac{\pi}{10} \text{ rad/s}$ and the total angular displacement required is $\theta = 2\pi \times 2\pi = 4\pi^2 \text{ rad}$. Substituting these values into the expression, $\tau (4\pi^2) = \frac{1}{2} (1.6 \times 10^{-3}) (\frac{\pi}{10})^2$, which simplifies to $4\pi^2 \tau = 0.8 \times 10^{-3} \times \frac{\pi^2}{100}$. Solving for the torque gives $\tau = \frac{0.8 \times 10^{-5}}{4} = 2 \times 10^{-6} \text{ N m}$.

The work-energy theorem establishes that the work required to stop a moving object is equal to its total kinetic energy. For a rolling disc, this energy comprises both translational and rotational motion, which results in the formula $KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = \frac{3}{4}mv^2$ After converting the speed from 20 cm/s to 0.2 m/s, we can substitute the mass of 100 kg into the energy expression. This calculation yields $KE = \frac{3}{4} \times 100 \times (0.2)^2 = 3 \, \text{J}$ Because the initial kinetic energy is 3 J, an equal amount of work must be performed to stop the disc.

The cylinder's kinetic energy consists of both translational and rotational components, which are converted into gravitational potential energy as it rises to its maximum height. Because a solid cylinder has a moment of inertia $I = \frac{1}{2}mr^2$, its total energy can be expressed as $E = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$, which simplifies to $\frac{3}{4}mv^2$. Setting this equal to the potential energy $mgh$, we determine the vertical height reached using the speed of 4 m/s and gravity of 10 m/s^2:

$h = \frac{3v^2}{4g} = \frac{3(4)^2}{4(10)} = 1.2$

With the vertical height determined to be 1.2 m, the distance $L$ traveled along the incline is found using the geometry of the slope, where $L = \frac{h}{\sin 30^{\circ}}$. Given that the sine of 30 degrees is 0.5, the total distance traveled is 2.4 m.

The work-energy theorem establishes that the work needed to bring a rotating object to rest is equal to its rotational kinetic energy, expressed as:
$W = \frac{1}{2}I\omega^2$
Since all three objects spin with the same angular velocity $\omega$, the work required to stop each object is directly proportional to its moment of inertia $I$.

The moments of inertia for the objects are $I_A = \frac{2}{5}MR^2$ for the solid sphere, $I_B = \frac{1}{2}MR^2$ for the thin circular disk, and $I_C = MR^2$ for the circular ring. Comparing these coefficients numerically, $1 > 0.5 > 0.4$ clearly indicates that the circular ring has the largest moment of inertia, followed by the disk, and then the sphere. Because the work scales directly with these inertia values, the relation follows the order $W_C > W_B > W_A$.

The moment of a force about a specific pivot point is determined by the cross product of the position vector, which signifies the location of the force's application relative to the pivot, and the force vector itself. This calculation characterizes the rotational impact that a force exerts on an object about a given axis.

Subtracting the pivot coordinates of (2, -2, -2) from the point of application coordinates of (2, 0, -3) yields the relative position vector $\vec{r} = 0\hat{i} + 2\hat{j} - 1\hat{k}$. The moment vector $\vec{\tau}$ is then obtained from the cross product $\vec{\tau} = \vec{r} \times \vec{F}$ using the force vector $\vec{F} = 4\hat{i} + 5\hat{j} - 6\hat{k}$, expressed as the following determinant:
$\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 2 & -1 \\ 4 & 5 & -6 \end{vmatrix}$
Expanding this determinant along the top row produces $\hat{i}(2 \times -6 - (-1) \times 5) - \hat{j}(0 \times -6 - (-1) \times 4) + \hat{k}(0 \times 5 - 2 \times 4)$, which evaluates to the final result of $-7\hat{i} - 4\hat{j} - 8\hat{k}$.

In pure rolling motion, the total kinetic energy of a body is the sum of its translational kinetic energy, given by $K_t = \frac{1}{2}mv^2$, and its rotational kinetic energy, $K_r = \frac{1}{2}I\omega^2$. For a solid sphere with moment of inertia $I = \frac{2}{5}mr^2$ rolling without slipping, where the angular velocity is $\omega = \frac{v}{r}$, the rotational energy term simplifies to $K_r = \frac{1}{5}mv^2$. Consequently, the total kinetic energy is:

$K_t + K_r = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$

The ratio of translational kinetic energy to this total kinetic energy is given by $\frac{\frac{1}{2}mv^2}{\frac{7}{10}mv^2}$, which simplifies to $\frac{5}{7}$.

Since the sphere is rotating in free space without the influence of external forces, the net external torque acting on it is $\tau_{\text{ex}} = 0$. Because the rate of change of angular momentum is directly proportional to the applied torque, as shown by the relation $\frac{dL}{dt} = 0$, the angular momentum $L$ must remain constant regardless of any internal changes to the sphere's radius. Therefore, while parameters like the moment of inertia and rotational kinetic energy adjust as the radius increases, the angular momentum stays unchanged.

Rotational dynamics is defined by the relationship where applied torque equals the moment of inertia multiplied by angular acceleration, $\tau = I \alpha$. Because the rope pulls on the rim of the hollow cylinder, the applied torque is $RF$, and since the moment of inertia for this geometry is $I = mR^2$, the relationship simplifies to $RF = mR^2 \alpha$. Solving for the unknown variable results in the expression $\alpha = \frac{F}{mR}$, and substituting the given values of 30 N for force, 3 kg for mass, and 0.4 m for radius leads to $\alpha = \frac{30}{3 \times 0.4}$, which results in an angular acceleration of 25 rad/s$^2$.

The centre of mass is the point representing the mean location of all matter in a system, which is physically distinct from the centre of gravity, the point where the net gravitational force effectively acts. While these points often coincide in uniform gravitational fields, the defining property of the centre of mass is that the net gravitational torque acting on the body about this point is zero. Consequently, statement (b) is a correct physical definition, whereas statement (a) is inaccurate because their alignment is not a universal requirement.

In the context of rigid body dynamics, a couple consists of two equal and opposite forces applied at distinct points, resulting in a net force of zero and thus causing only rotational motion without any change in translational velocity. Because a couple cannot produce translational motion, statement (c) is incorrect. Mechanical advantage is determined by the ratio $\frac{\text{Load}}{\text{Effort}}$, and when this ratio is greater than $1$, it confirms that the $\text{Load}$ is greater than the $\text{Effort}$. This demonstrates that a smaller input force can effectively overcome a much larger resistance, making statement (d) correct.

When two rotating discs with the same moment of inertia are brought into contact, the system adheres to the principle of conservation of angular momentum, as no external torque is applied to the discs. By equating the initial total angular momentum, $I\omega_1 + I\omega_2$, to the final angular momentum of the combined discs, $2I\omega$, we find the common final angular velocity of the system to be $\omega = \frac{\omega_1 + \omega_2}{2}$.

The initial kinetic energy of the system is the sum of the rotational energies of the two separate discs, given by $\frac{1}{2}I\omega_1^2 + \frac{1}{2}I\omega_2^2$. Once they are coupled, the discs rotate together as a single body with a new kinetic energy of $\frac{1}{2}(2I)\omega^2$, which simplifies to $I\left(\frac{\omega_1 + \omega_2}{2}\right)^2$. Calculating the energy lost during this process requires subtracting the final kinetic energy from the initial kinetic energy. Performing this subtraction, $\frac{1}{2}I(\omega_1^2 + \omega_2^2) - \frac{I(\omega_1 + \omega_2)^2}{4}$, and expanding the squared term leads to the expression $\frac{1}{4}I(\omega_1^2 - 2\omega_1\omega_2 + \omega_2^2)$, which simplifies to $\frac{1}{4}I\left(\omega_1-\omega_2\right)^2$.

The rotational kinetic energy of a body depends on both its moment of inertia and its angular velocity, defined by the formula $E = \frac{1}{2}I\omega^2$. For a solid sphere rotating about its diameter, the moment of inertia is $\frac{2}{5}mR^2$. Applying the kinetic energy formula with an angular speed of $\omega$, the energy of the sphere becomes $\frac{1}{2} (\frac{2}{5}mR^2) \omega^2$, which simplifies to $\frac{1}{5}mR^2\omega^2$.

A solid cylinder rotating about its geometrical axis possesses a moment of inertia of $\frac{1}{2}mR^2$. Since the cylinder rotates with an angular speed of $2\omega$, its kinetic energy is calculated as $\frac{1}{2} (\frac{1}{2}mR^2) (2\omega)^2$. Expanding the term for angular velocity results in $4\omega^2$, which makes the cylinder's energy $\frac{1}{4}mR^2 (4\omega^2)$, simplifying to $mR^2\omega^2$. Comparing the kinetic energy of the sphere to that of the cylinder yields the ratio $\frac{1}{5}$.

The moment of inertia of the remaining disc is found by subtracting the moment of inertia of the removed hole from that of the original full disc. For the full disc, the initial moment of inertia is $\frac{1}{2}MR^2$, while the removed hole-having a radius of $R/2$ and mass of $M/4$-has a moment of inertia about the original centre of $\frac{3MR^2}{32}$ after accounting for the parallel axis theorem. Subtracting the latter from the former, $\frac{1}{2}MR^2 - \frac{3MR^2}{32}$, provides a resultant moment of inertia of $\frac{13MR^2}{32}$.

The rotational kinetic energy $K$ of a body can be expressed in terms of its angular momentum $L$ and moment of inertia $I$ using the relationship $K = \frac{L^2}{2I}$. Given that both bodies have identical kinetic energies, we can equate their expressions to get $\frac{L_A^2}{2I_A} = \frac{L_B^2}{2I_B}$, which simplifies to the ratio $\frac{L_A^2}{L_B^2} = \frac{I_A}{I_B}$. Because the problem states that $I_B > I_A$, the ratio on the right side is less than one, which necessitates that $L_A^2 < L_B^2$. Consequently, the angular momentum of body B must be greater than that of body A.

The moment of inertia for a system of point masses about an axis is the sum of the products of each mass and the square of its perpendicular distance from that axis. With two masses $m_1$ and $m_2$ separated by length $l$, their distances from the center of mass are $r_1 = \frac{m_2 l}{m_1 + m_2}$ and $r_2 = \frac{m_1 l}{m_1 + m_2}$. Substituting these values into the formula $I = m_1 r_1^2 + m_2 r_2^2$ results in $I = m_1\left(\frac{m_2l}{m_1 + m_2}\right)^2 + m_2\left(\frac{m_1l}{m_1 + m_2}\right)^2$. Factoring common terms leads to $I = \frac{m_1 m_2 l^2}{(m_1 + m_2)^2} (m_2 + m_1)$, which simplifies to the final expression $I = \frac{m_1 m_2}{m_1 + m_2}l^2$.

The time an object takes to roll down an inclined plane is independent of its mass, relying instead on its geometric distribution of mass represented by the ratio $\frac{k^2}{R^2}$. Given that the time required to complete the descent is proportional to $\sqrt{1+\frac{k^2}{R^2}}$, the object with the smallest value for this ratio will arrive at the bottom first. A sphere has a smaller $\frac{k^2}{R^2}$ ratio than a disk, which allows it to accelerate faster and complete the journey in the least amount of time.

A point on a rotating circular disc with angular acceleration experiences two perpendicular components of linear acceleration: the tangential acceleration, which changes the speed, and the centripetal acceleration, which changes the direction of motion. First, express the radius as $r = 0.5$ meters. The tangential acceleration follows the formula $a_t = r\alpha$, which calculates to 1.0 m s$^{-2}$.

The centripetal acceleration depends on the instantaneous angular velocity $\omega = \alpha t$. After 2.0 seconds, this velocity reaches 4.0 rad s$^{-1}$. Using the expression $a_c = \omega^2 r$, the centripetal acceleration is 8.0 m s$^{-2}$. Since the tangential and centripetal components are orthogonal, the magnitude of the net acceleration is found using $a_{\text{net}} = \sqrt{a_t^2 + a_c^2}$. Substituting the calculated values gives $\sqrt{1.0^2 + 8.0^2}$, which results in approximately 8.0 m s$^{-2}$.

When a rigid body like this rod rests in horizontal equilibrium, the net torque acting on it about any pivot point must be zero. Calculating the rotational balance around knife edge B effectively eliminates the reaction force at that point from the equation, simplifying the determination of the normal reaction at A. The upward reaction force $N_A$ acts at a distance $d$ from B, producing a counter-clockwise torque of $N_A d$, while the rod's weight $W$ acts downwards at the centre of mass, located a distance $(d - x)$ from B, producing a clockwise torque of $W(d - x)$. Equating these torques results in the expression $N_A d = W(d - x)$, which reveals that $N_A = \frac{W(d - x)}{d}$.

Newton's second law for rotation relates the torque applied to an object to its moment of inertia and angular acceleration through the equation $\tau = I \alpha$. To solve this, first convert the linear speed of the automobile from 54 km/h to 15 m/s, then use the relation $\omega = \frac{v}{R}$ to find the initial angular velocity of the wheel, which results in 100/3 rad/s when divided by the radius of 0.45 m. Because the vehicle stops in 15 s, the magnitude of the angular retardation is determined by $\alpha = \frac{\omega}{t}$, yielding 100/45 rad/s². Multiplying this retardation by the wheel's moment of inertia of 3 kg m² provides the magnitude of the braking torque, expressed as $\tau = 3 \times \frac{100}{45}$, which simplifies to approximately 6.66 kg m² s⁻².

The total moment of inertia of the system is the sum of the individual moments of inertia of the three identical spherical shells about the given axis $XX'$. For the central shell, the axis passes directly through its diameter, so its contribution is simply $\frac{2}{3}mr^2$. Regarding the two shells that are tangent to the axis, the parallel axis theorem allows us to find their moment of inertia by shifting the rotation axis from the center to the tangent line at a distance $r$ from the center. This adds an extra term $mr^2$ to each of their moments, resulting in $\frac{2}{3}mr^2 + mr^2 = \frac{5}{3}mr^2$ for each. Combining all three contributions, the total moment of inertia is:

$I = \frac{2}{3}mr^2 + \left(\frac{2}{3}mr^2 + mr^2\right) + \left(\frac{2}{3}mr^2 + mr^2\right) = 4mr^2$

The work required to set a rigid rod in rotation with a specific angular velocity $\omega_0$ is equal to its rotational kinetic energy, which is defined by the expression $K = \frac{1}{2} I \omega_0^2$. Since the angular velocity is fixed, minimizing the work is equivalent to finding the axis position $x$ that minimizes the moment of inertia, $I = m_1 x^2 + m_2 (L-x)^2$. Differentiating this expression for $I$ with respect to $x$ and setting the derivative to zero yields the condition:
$2 m_1 x - 2 m_2 (L-x) = 0$
This equation simplifies to $m_1 x = m_2 L - m_2 x$, which, upon rearranging terms to isolate $x$, leads to $x(m_1 + m_2) = m_2 L$. Solving for the final position gives:
$x = \frac{m_2 L}{m_1+m_2}$