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

Because the tension force is directed toward the center of the rotation, it produces zero torque on the mass, meaning angular momentum must remain constant throughout the motion. Equating the initial angular momentum $m v_0 R_0$ to the final angular momentum $m v_f (\frac{R_0}{2})$ reveals that the final velocity becomes $v_f = 2 v_0$. Calculating the final kinetic energy with this updated velocity using the expression $\frac{1}{2} m v_f^2$ results in a final value of $2 m v_0^2$.

Angular momentum remains conserved when the net torque acting on a body is zero, as torque represents the time rate of change of angular momentum. Since torque is defined by the cross product of the position vector and the force vector, we require $\vec{r} \times \vec{F} = 0$ to satisfy the conservation requirement. Substituting the given vectors $\vec{r} = 2 \hat{i}-6 \hat{j}-12 \hat{k}$ and $\vec{F} = \alpha \hat{i}+3 \hat{j}+6 \hat{k}$, the cross product determinant yields:

$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -6 & -12 \\ \alpha & 3 & 6 \end{vmatrix} = \hat{i}(-36 - (-36)) - \hat{j}(12 - (-12\alpha)) + \hat{k}(6 - (-6\alpha)) = 0$

This expression simplifies to $0\hat{i} - (12 + 12\alpha)\hat{j} + (6 + 6\alpha)\hat{k} = 0$. Because the coefficients for the unit vectors must vanish, we require $12 + 12\alpha = 0$ and $6 + 6\alpha = 0$, both of which consistently determine that $\alpha = -1$.

When an object slides down a frictionless incline, gravity acts as the sole force driving the motion, resulting in an acceleration of $a_{\text{slipping}} = g \sin \theta$. In contrast, a solid sphere rolling without slipping must convert a portion of its potential energy into rotational kinetic energy, which restricts its linear acceleration to $a_{\text{rolling}} = \frac{5}{7} g \sin \theta$. By dividing the rolling acceleration by the slipping acceleration, we can determine how rotational inertia affects the rate of descent:

$\frac{a_{\text{rolling}}}{a_{\text{slipping}}} = \frac{\frac{5}{7} g \sin \theta}{g \sin \theta} = \frac{5}{7}$

This comparison confirms the acceleration ratio between the rolling and slipping sphere is $5:7$.

To determine the tension required to accelerate the rotating cylinder, apply the rotational version of Newton's second law, which relates the net torque to the product of the moment of inertia and angular acceleration. The torque exerted by the tension $T$ at radius $r$ is $\tau = Tr$, while the moment of inertia for a solid cylinder is $I = \frac{1}{2}mr^2$. Because the angular acceleration is provided in revolutions per second squared, convert this value to radians per second squared by multiplying by $2\pi$, which results in an acceleration of $4\pi$ radians per second squared. Substituting these expressions into the dynamic equation $Tr = I\alpha$ allows the tension to be expressed as $T = \frac{1}{2}mr\alpha$. Calculating this with the mass of 50 kg, radius of 0.5 m, and angular acceleration of $4\pi$ radians per second squared leads to the expression $T = \frac{50 \times 0.5 \times 4\pi}{2}$, which evaluates to 157 N.

The rotational motion of the rod is determined by the fundamental relationship $\tau = I\alpha$, which defines torque as the product of the moment of inertia and angular acceleration. Because the rod rotates about the hinge at its endpoint, gravity provides the torque acting at the rod's center of mass, which is located at a distance of $\frac{L}{2}$ from the pivot point.

The torque magnitude is given by $\tau = Mg \frac{L}{2}$, and the moment of inertia for a rod rotating about one end is $I = \frac{ML^2}{3}$. Equating the two using the dynamic relation $Mg \frac{L}{2} = \left( \frac{ML^2}{3} \right) \alpha$ allows us to isolate the angular acceleration. Canceling common terms on both sides results in $\alpha = \frac{3g}{2L}$.

Since no external torques act on the system when the two rotating discs are brought together, the principle of conservation of angular momentum applies. This means the total angular momentum before the discs are coupled must equal the total angular momentum of the combined system after they are joined, allowing the final angular velocity to be determined by equating these values.

The first step involves calculating the moment of inertia for each disc using the formula $I = \frac{1}{2}MR^2$. For the first disc, $I_1 = \frac{1}{2} \times 2 \times (0.2)^2 = 0.04\,\text{kg}\,\text{m}^2$. For the second disc, $I_2 = \frac{1}{2} \times 4 \times (0.1)^2 = 0.02\,\text{kg}\,\text{m}^2$.

Using the conservation law, the total initial angular momentum is the sum of the individual angular momenta, while the final angular momentum is the product of the total moment of inertia and the final angular velocity $\omega$. The relationship is defined as:
$I_1 \omega_1 + I_2 \omega_2 = (I_1 + I_2) \omega$
Substituting the known values, the equation becomes $(0.04)(50) + (0.02)(200) = (0.04 + 0.02) \omega$. Simplifying the arithmetic gives $2 + 4 = 0.06 \omega$, which results in $6 = 0.06 \omega$. Solving this equation yields a final angular velocity of $100\,\text{rad}\,\text{s}^{-1}$.

The principle of conservation of energy governs this motion, where the total initial kinetic energy, comprising both translational and rotational components, is entirely converted into gravitational potential energy at the peak height $h$. Expressed mathematically, the initial energy state is $\frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2$, which must equate to $Mgh$ at the highest point. Incorporating the rolling condition $\omega = \frac{v}{R}$ and the specified height $h = \frac{3v^2}{4g}$ allows us to determine the object's moment of inertia.

Inserting these expressions leads to the equation $\frac{1}{2}Mv^2 + \frac{1}{2}I(\frac{v}{R})^2 = Mg(\frac{3v^2}{4g})$, which simplifies to $\frac{1}{2}Mv^2 + \frac{1}{2}I\frac{v^2}{R^2} = \frac{3}{4}Mv^2$. Subtracting the translational kinetic energy from both sides isolates the rotational term as $\frac{1}{2}I\frac{v^2}{R^2} = \frac{1}{4}Mv^2$. Solving for the moment of inertia yields $I = \frac{1}{2}MR^2$, a value characteristic of a disc.

Finding the center of mass for a system of point particles requires calculating the weighted average of their positions based on their individual masses, expressed as:
$X_{CM} = \frac{m_1x_1 + m_2x_2 + m_3x_3}{m_1 + m_2 + m_3}$
This equation identifies the single point that effectively represents the distribution of mass for the entire system along the axis of placement.

Placing the three masses-300 g at 0 cm, 500 g at 40 cm, and 400 g at 70 cm-results in a total system mass of 1200 g. Multiplying each mass by its position gives a total moment of 0 + 20000 + 28000, which sums to 48000 g·cm. Dividing this total moment by the sum of the masses yields the center of mass position at 40 cm.

The principle of conservation of angular momentum is central here because the frictionless axle ensures that no external torque influences the rotation of the platform-man system. Since the system is initially at rest, the total angular momentum is zero, implying that the angular momentum generated by the man must be exactly offset by the angular momentum acquired by the platform in the opposite direction. Calculating the man's angular momentum relative to the axle as $m v R = 100$, we equate this to the platform's angular momentum $I \omega_p = 200 \omega_p$. Solving $200 \omega_p = 100$ gives an angular velocity for the platform of $\omega_p = 0.5$ rad/s. The man moves with an angular velocity relative to the ground of $\omega_g = v / R = 0.5$ rad/s. Because the platform rotates in the direction opposite to the man's motion, his angular velocity relative to the platform is the sum of these two magnitudes, resulting in $\omega_{\text{rel}} = 0.5 + 0.5 = 1$ rad/s. Finally, the time taken for one full revolution is determined by dividing the full angular rotation of $2\pi$ radians by this relative angular velocity, yielding $T = \frac{2\pi}{1} = 2\pi$ seconds.

Angular momentum ($\vec{L}$) is fundamentally defined by the cross product of the position vector ($\vec{r}$) and the linear momentum ($\vec{p}$), mathematically represented as $\vec{L} = \vec{r} \times \vec{p}$. Because both the position and linear momentum vectors are confined to the plane of rotation, their vector product-determined by the right-hand rule-must inherently point along a line perpendicular to that plane. This configuration ensures the angular momentum vector remains oriented normally to the plane of motion throughout the entire rotation.

The motion of the center of mass of any system is governed by the net external force acting on it, represented by the equation $\vec{F}_{ext} = M_{total} \vec{a}_{cm}$. In this scenario, the boat, which weighs 100 kg, and the two individuals of 55 kg and 65 kg, function as an isolated system within still water, meaning there is no net external horizontal force present. Since the 55 kg person moves across the 3.0 m length of the boat by exerting only internal forces, the acceleration of the center of mass remains zero. Consequently, the position of the system's center of mass does not change, resulting in a total shift of zero.

The moment of inertia for any axis parallel to the center of mass is governed by the parallel axis theorem, expressed as $I = I_{CM} + Ma^2$. In this formula, $I_{CM}$ represents the moment of inertia about the center of mass, $M$ is the total mass of the disc, and $a$ is the perpendicular distance separating the two parallel axes. Since both $I_{CM}$ and $M$ remain constant for the disc, the total moment of inertia increases directly with the distance $a$, which reaches its maximum value at point B.

Torque relates to angular motion through the expression $\tau = I\alpha$, meaning the torque becomes zero at the exact moment when the angular acceleration vanishes. By differentiating the given angular position $\theta(t) = 2t^3 - 6t^2$ with respect to time, the angular velocity is calculated as $\frac{d\theta}{dt} = 6t^2 - 12t$. Differentiating once more yields the angular acceleration $\alpha = \frac{d^2\theta}{dt^2} = 12t - 12$. Setting this expression for angular acceleration to zero leads to $12t - 12 = 0$, which solves to $t = 1\,\text{s}$.

The Parallel Axis Theorem provides a straightforward way to calculate the moment of inertia about any axis once you know the moment of inertia about a parallel axis passing through the center of mass. This theorem states that the total moment of inertia equals the sum of the moment of inertia about the center of mass and the product of the mass and the square of the distance between the two axes, expressed as $I = I_{CM} + Md^2$. For a uniform rod, the perpendicular distance between the axis at the midpoint and the axis at the end is exactly half the total length, or $\frac{L}{2}$. By substituting the known value $I_0$ for the center of mass and the distance $\frac{L}{2}$ into the formula, we arrive at the following result:

$I = I_0 + M\left(\frac{L}{2}\right)^2 = I_0 + \frac{ML^2}{4}$

Because the pulling force is directed toward the center of rotation, it exerts no torque on the mass, which implies that angular momentum is conserved. We express this conservation as $mvr = mv'r'$, where $r$ and $v$ are the initial radius and velocity, and $r'$ and $v'$ are the final values. Given that the radius is reduced to half its original value, $r' = \frac{r}{2}$, substituting this into the equation $v'r' = vr$ gives $v'(\frac{r}{2}) = vr$, which simplifies to $v' = 2v$. Since kinetic energy is defined as $K = \frac{1}{2}mv^2$, doubling the velocity leads to a new kinetic energy $K' = \frac{1}{2}m(v')^2 = \frac{1}{2}m(2v)^2$. This calculation results in $K' = 4(\frac{1}{2}mv^2)$, demonstrating that the kinetic energy increases by a factor of 4.

In physics, the motion of a system depends entirely on the net external force acting upon it. When two particles interact only through their own internal mutual attraction, there is no external force present to change the system's collective state of motion, meaning the center of mass cannot accelerate. Because the system is initially at rest, the initial total momentum is zero, and due to the conservation of momentum, the velocity of the center of mass must remain zero throughout the entire process, regardless of the individual speeds $v$ and $2v$ that the particles may acquire.

Since no external torque acts on the rotating system, the principle of conservation of angular momentum applies, meaning $I_1\omega = I_2\omega'$. The initial moment of inertia of the ring is $I_1 = Mr^2$, creating an initial angular momentum of $Mr^2\omega$. When the two masses are attached, the total moment of inertia becomes $I_2 = (M+2m)r^2$, leading to the following equality for the conserved angular momentum:
$Mr^2\omega = (M+2m)r^2\omega'$
Dividing both sides by $(M+2m)r^2$ provides the final angular velocity as $\omega' = \frac{M\omega}{M+2m}$.

Conservation of angular momentum dictates that since no external torque acts on the system of two disks, the total angular momentum before and after the collision remains constant. Equating the initial angular momentum, $I_t \omega_i$, to the final angular momentum, $(I_t + I_b) \omega_f$, allows for the determination of the final angular speed as $\omega_f = \frac{I_t \omega_i}{I_t + I_b}$. The energy lost is equal to the decrease in rotational kinetic energy, calculated as $\Delta K = K_i - K_f = \frac{1}{2} I_t \omega_i^2 - \frac{1}{2} (I_t + I_b) \omega_f^2$. Substituting the expression for $\omega_f$ into the final kinetic energy term yields $K_f = \frac{1}{2} (I_t + I_b) \left( \frac{I_t \omega_i}{I_t + I_b} \right)^2 = \frac{1}{2} \frac{I_t^2 \omega_i^2}{I_t + I_b}$. Subtracting this from the initial kinetic energy results in $\Delta K = \frac{1}{2} I_t \omega_i^2 - \frac{1}{2} \frac{I_t^2 \omega_i^2}{I_t + I_b} = \frac{1}{2} \omega_i^2 \left( \frac{I_t(I_t + I_b) - I_t^2}{I_t + I_b} \right) = \frac{1}{2} \frac{I_t I_b}{I_t + I_b} \omega_i^2$.

Calculating the moment of inertia of a disc with a removed section is best approached by applying the principle of superposition, where the moment of inertia of the remaining body equals the inertia of the original solid disc minus the inertia of the removed portion, provided they share the same central axis. The moment of inertia for a uniform disc about its central axis is given by the formula $I = \frac{1}{2}MR^2$.

The complete disc with mass $9M$ and radius $R$ has an initial moment of inertia of $I_{total} = \frac{1}{2}(9M)R^2 = \frac{9}{2}MR^2$. The removed portion has a mass $M$ and radius $\frac{R}{3}$, so its moment of inertia about the same central axis is $I_{removed} = \frac{1}{2}M\left(\frac{R}{3}\right)^2 = \frac{1}{18}MR^2$. Subtracting the removed portion from the total yields the final inertia of the remaining disc, calculated as $\frac{9}{2}MR^2 - \frac{1}{18}MR^2 = \frac{81}{18}MR^2 - \frac{1}{18}MR^2 = \frac{80}{18}MR^2 = \frac{40}{9}MR^2$.

When objects roll down an inclined plane without slipping, the duration of the descent is governed by the relation $t = \sqrt{\frac{2l(1+K^2/R^2)}{g\sin\theta}}$. A smaller value of the $K^2/R^2$ ratio results in a shorter time to reach the bottom. Since the solid cylinder has a $K^2/R^2 = 1/2$ and the hollow cylinder has a $K^2/R^2 = 1$, the solid cylinder experiences less rotational resistance and reaches the bottom first.