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

Since the density of the sphere is uniform, the mass is directly proportional to its volume. The larger sphere has a radius of $2R$, so its volume is eight times that of the smaller sphere with radius $R$, meaning the mass of the smaller sphere is $M' = \frac{M}{8}$. To determine the moment of inertia of the smaller sphere about the Y-axis, which is an axis passing through the center of the large sphere, we apply the parallel axis theorem with a shift distance $d = R$. This yields $I_{\text{small}} = \frac{2}{5} M' R^2 + M' R^2 = \frac{7}{5} M' R^2$, and substituting $M' = \frac{M}{8}$ gives $I_{\text{small}} = \frac{7}{40} MR^2$. The moment of inertia of the entire large sphere about its diameter is $I_{\text{whole}} = \frac{2}{5} M (2R)^2 = \frac{8}{5} MR^2$. The moment of inertia of the remaining portion of the sphere is found by subtracting the smaller sphere's contribution from the whole, resulting in $I_{\text{remaining}} = \frac{8}{5} MR^2 - \frac{7}{40} MR^2 = \frac{57}{40} MR^2$. Comparing these values by taking the ratio of $I_{\text{small}}$ to $I_{\text{remaining}}$ gives $\frac{7/40}{57/40}$, which simplifies to $\frac{7}{57}$.

Since the Sun expands without any external torque, its total angular momentum must be conserved. Angular momentum is defined as the product of the moment of inertia and angular velocity, expressed as $L = I\omega$. For a solid sphere, the moment of inertia is given by $I = \frac{2}{5}mR^2$. If the radius increases to twice its original size while the mass remains constant, the new moment of inertia becomes $I' = \frac{2}{5}m(2R)^2$, which simplifies to $I' = 4I$.

Because the angular momentum $L$ must remain constant, an increase in the moment of inertia by a factor of 4 requires the angular velocity to decrease to one-fourth of its initial value. Given the relationship $\omega = \frac{2\pi}{T}$, a fourfold reduction in angular velocity leads to a fourfold increase in the time period. By multiplying the initial rotation period by this factor, the new time period becomes:

$T' = 4 \times 27 = 108 \text{ days}$

Impulse is equivalent to the change in momentum, where the direction of motion must be carefully accounted for in the calculation. The speed just before the ball hits the ground is determined by the relationship $v = \sqrt{2gh_1}$, yielding $\sqrt{2 \times 9.8 \times 40} = 28$ m/s downward. After rebounding to a height of 10 m, the ball attains a speed of $\sqrt{2 \times 9.8 \times 10} = 14$ m/s upward. By adopting the upward direction as positive, the initial velocity is $-28$ m/s and the final velocity is $14$ m/s. The impulse imparted to the ball is calculated using the equation $I = m(v_f - v_i)$, which results in $0.5 \times (14 - (-28)) = 21$ Ns.

When objects collide, the principle of conservation of linear momentum dictates that the system's total momentum remains unchanged. Before the impact, only body $A$ contributes to the momentum while body $B$ is at rest, meaning the total initial momentum is $m v_1$. Since the collision is completely inelastic, both bodies move together as a single entity with a combined mass of $2m$ at a final velocity $v_2$. Equating the initial and final states gives the equation $m v_1 = 2 m v_2$, which leads directly to the ratio $v_1 : v_2 = 2 : 1$.

When a wheel undergoes pure rolling, its motion is a combination of forward translation at speed $\nu$ and rotation about its center. At the lowest point Q, the rotational velocity is directed opposite to the translational velocity, causing the two components to cancel out and resulting in a net velocity of $0$. At the highest point P, the rotational velocity is directed in the same sense as the forward translation, so the two components add together to equal $2\nu$. Since $2\nu$ is strictly greater than $0$, the topmost point moves significantly faster than the point of contact with the road.

The moment of inertia for a uniform thin rod rotating about an axis through its center, perpendicular to its length, is determined by the formula $I = \frac{M L^2}{12}$, where $M$ is the mass and $L$ is the length of the rod. Given an inertia of 2400 g cm^2 and a mass of 400 g, we can substitute these values into the equation to get $2400 = \frac{400 \times L^2}{12}$ to solve for the unknown length $L$. Simplifying this relation, $L^2$ equals 72, and the square root of 72 provides a length of approximately 8.48 cm, which rounds to 8.5 cm.

In horizontal circular motion, the tension in the string provides the necessary centripetal force to maintain the path, governed by the formula $T = m\ell\omega^2$, where $m$ represents mass, $\ell$ is the radius, and $\omega$ is the angular speed. Because the radius remains constant, the tension is directly proportional to the square of the angular velocity. When the speed is doubled to $2\omega$, the new tension $T'$ is determined by:

$T' = m\ell(2\omega)^2 = 4m\ell\omega^2$

This calculation shows that the new tension is equivalent to $4T$.

The moment of inertia of a solid sphere about an axis passing through its center is $\frac{2}{5}MR^2$. Applying the parallel axis theorem for an axis tangent to the sphere, where the distance between the two axes equals the radius $R$, the moment of inertia becomes:
$I_{XY} = \frac{2}{5}MR^2 + MR^2 = \frac{7}{5}MR^2$

By definition, the radius of gyration $K$ is related to the moment of inertia by $I_{XY} = MK^2$. Substituting the values $M = 5\text{ kg}$ and $K = 5\text{ m}$ into the equation $MK^2 = \frac{7}{5}MR^2$, we have:
$5 \times 5^2 = \frac{7}{5} \times 5 \times R^2$

This simplifies to $125 = 7R^2$, which leads to $R = \sqrt{\frac{125}{7}} = \frac{5\sqrt{5}}{\sqrt{7}}$. Comparing this result with the given radius of $\frac{5x}{\sqrt{7}}$, the value of $x$ is $\sqrt{5}$.

Although the magnitude of the speed remains uniform, velocity is a vector quantity that depends on both magnitude and direction. In circular motion, the direction of the velocity vector $\vec{v}$ changes at every instant as the particle follows the curved path, resulting in a continuously varying velocity. This change in direction necessitates a centripetal acceleration vector that is always directed toward the center of the circle, meaning that as the particle rotates, the direction of this acceleration also shifts constantly, making the acceleration variable as well.

In a system where particles move solely due to mutual forces of attraction, the net external force is zero. Since the velocity of the center of mass remains constant in the absence of external forces, and both particles started at rest, the initial velocity of the center of mass is zero. Consequently, the speed of the center of mass remains zero regardless of the individual velocities $v$ and $3v$ of the particles.

Angular acceleration, denoted by $\vec{\alpha}$, describes how the angular velocity of a rotating body changes over time. Since the angular velocity vector is defined to point along the axis of rotation according to the right-hand rule, any change in the rotational motion necessarily inherits this same axial direction. As a result, the angular acceleration vector is always directed along the axis of rotation, perpendicular to the plane of the circular path.

The radius of gyration is defined as $k = \sqrt{\frac{I}{M}}$, where $I$ represents the moment of inertia and $M$ is the mass. A solid sphere has a moment of inertia of $\frac{2}{5}MR^2$ about its axis, resulting in $k_s = \sqrt{\frac{2}{5}}R$, while a thin hollow sphere has an inertia of $\frac{2}{3}MR^2$, leading to $k_h = \sqrt{\frac{2}{3}}R$.

Taking the ratio $k_s : k_h$ requires dividing these values to obtain $\frac{\sqrt{2/5}R}{\sqrt{2/3}R}$, which simplifies to $\sqrt{\frac{2}{5}} \times \sqrt{\frac{3}{2}} = \sqrt{\frac{3}{5}}$. Since this result does not match any of the provided alternatives, the required value falls outside the given choices.

The rotation of an object under a constant torque is governed by the rotational analogue of Newton's second law, $\tau = I \alpha$. We first determine the angular acceleration by dividing the torque of 100 Nm by the moment of inertia of 300 kg m, which results in an acceleration of 1/3 rad/s^2. Using the kinematic equation $\omega = \omega_i + \alpha t$ and knowing the wheel starts from an initial angular velocity of 0, we multiply the acceleration of 1/3 rad/s^2 by the time of 3 s to find the final angular velocity of 1 rad/s.

Since no external forces act on the shell during the explosion, the system obeys the principle of conservation of linear momentum, ensuring the total momentum before and after the event remains zero. The two equal fragments of mass $2m$ traveling at speed $v$ along perpendicular paths generate a combined resultant momentum magnitude of $\sqrt{(2mv)^2 + (2mv)^2} = 2\sqrt{2}mv$. To satisfy the conservation law, the third fragment of mass $m$ must carry an equal and opposite momentum, effectively balancing this resultant vector. Equating the momentum of this third fragment, $mv'$, to the magnitude of the other two gives $mv' = 2\sqrt{2}mv$, resulting in a final speed of $v' = 2\sqrt{2}v$.

The center of mass serves as the weighted average position of all the mass in a system, effectively acting as the point where the entire mass could be considered concentrated. To find the location of this point relative to the $10 \text{ kg}$ object, the simplest approach is to place the $10 \text{ kg}$ mass at the origin of a coordinate system, assigning it a position of $0 \text{ m}$. Since the rigid rod spans $10 \text{ m}$, the $20 \text{ kg}$ mass sits at the $10 \text{ m}$ mark on the same axis.

Using the center of mass formula $X_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$, we substitute the masses and their respective coordinates into the equation to get $\frac{10 \times 0 + 20 \times 10}{10 + 20}$. Simplifying this calculation yields $\frac{200}{30}$, which determines that the balance point of the system lies $\frac{20}{3} \text{ m}$ away from the $10 \text{ kg}$ mass.

The work-energy theorem states that the energy provided to a system to change its rotational speed is equal to the increase in its rotational kinetic energy. This energy, given as 484 J, relates to the moment of inertia and the angular velocities of the flywheel through the relationship between kinetic energy and angular speed.

To perform the calculation, convert the speeds from revolutions per minute to radians per second by multiplying by $2\pi/60$. This converts the initial speed of 60 rpm to $2\pi$ radians per second and the final speed of 360 rpm to $12\pi$ radians per second. Using the expression $\Delta KE = \frac{1}{2}I(\omega_f^2 - \omega_i^2)$, the balance is:

$484 = \frac{1}{2} I ( (12\pi)^2 - (2\pi)^2 )$

Evaluating the squared terms gives $144\pi^2$ and $4\pi^2$, meaning their difference is $140\pi^2$. The equation then simplifies to $484 = 70 I \pi^2$, and solving for the moment of inertia gives $I \approx 0.7$ kg-m^2.

A body's radius of gyration is defined as the distance from the axis of rotation where the entire mass could be concentrated while maintaining the same moment of inertia, expressed as $k = \sqrt{\frac{I}{M}}$. For a thin uniform disc, the moment of inertia about an axis through its center and perpendicular to its plane is
$I_{1} = \frac{MR^{2}}{2}$
which yields a radius of gyration of $k_{1} = \frac{R}{\sqrt{2}}$. Rotating the disc about its diameter results in a moment of inertia of
$I_{2} = \frac{MR^{2}}{4}$
which corresponds to a radius of gyration of $k_{2} = \frac{R}{2}$. Dividing the first radius of gyration by the second provides the ratio
$\frac{k_{1}}{k_{2}} = \frac{R/\sqrt{2}}{R/2} = \sqrt{2}:1$

Angular acceleration represents the rate of change in angular velocity, determined by the equation $\alpha = \frac{\omega_f - \omega_i}{t}$. Because rotational speeds are given in revolutions per minute, we must convert them to the standard unit of radians per second by multiplying each value by $\frac{2\pi}{60}$. The change in angular velocity is found by subtracting the initial 1200 rpm from the final 3120 rpm, which equals 1920 rpm. Substituting these values into the formula gives $\alpha = \frac{1920}{16} \times \frac{2\pi}{60}$. Performing this calculation simplifies to 120 multiplied by $\frac{2\pi}{60}$, resulting in an angular acceleration of $4\pi$ rad/s².

The behavior of a projectile during an explosion is governed by the principle that the system's Center of Mass continues along its original trajectory, as the impulsive forces causing the detonation are strictly internal. Because these internal forces do not exert a net external influence, effectively resulting in $\sum F_{ext} = 0$ specifically regarding the forces triggering the explosion, the system's overall momentum remains unchanged. Consequently, the assertion is incorrect in claiming that the individual fragments follow the original path, whereas the reason accurately identifies that the explosion is solely the result of internal forces.

The moment of inertia of a system about an axis is defined by the integral $I = \int r^2 dm$. Since a thin ring has all its mass distributed at a constant distance $R$ from the center, this expression simplifies to $I = R^2 \int dm$, where the integral represents the total mass of the object. Removing a $90^{\circ}$ arc accounts for one-quarter of the ring's circumference, which consequently removes one-quarter of its mass, leaving behind a remaining portion with mass $\frac{3M}{4}$. Substituting this remaining mass into the simplified formula gives the moment of inertia for the arc:
$I = \frac{3}{4} M R^2$
Comparing this result to the form $I = K M R^2$ establishes that the value of $K$ is $\frac{3}{4}$.