Centre of Mass MCQ

Explanation: This question examines the correct conceptual and mathematical description of angular momentum in rotational motion. It focuses on its physical nature and relation to other rotational quantities.

Angular momentum is defined as the product of moment of inertia and angular velocity for a rigid body rotating about a fixed axis: L = Iω. Here, I represents resistance to rotational motion, and ω represents angular speed. Because it originates from a Vector cross product (r × p), angular momentum has both magnitude and direction, making it a Vector quantity.

To reason properly, recall that moment of inertia plays a role similar to Mass in linear motion. Just as linear momentum is p = mv, rotational momentum depends directly on I. If I increases while ω remains constant, L increases proportionally. Also, since it involves direction determined by the right-hand rule, it cannot be scalar. Evaluating these properties helps identify its correct description.

This question tests understanding of the formula L = Iω and the Vector nature of angular momentum.

Explanation: This question investigates whether angular momentum is conserved when a disc moves on a rough surface and how reference points affect conservation.

Angular momentum conservation depends on external torque about a chosen point. On a rough surface, friction acts at the contact point. Friction is an external force and can produce torque about some reference points but not necessarily all.

To analyze, consider torque about the center of Mass. Friction generally produces torque there, so angular momentum may change. However, about the instantaneous point of contact, friction’s line of action passes through that point, producing zero torque about it. If NET external torque about a point is zero, angular momentum about that point remains conserved. Therefore, conservation depends on the selected reference point.

This highlights the importance of torque analysis and choice of reference frame in Rotational Dynamics.

Explanation: This question concerns angular momentum of a projectile relative to its launch point. Angular momentum about a point depends on both position Vector and linear momentum.

In projectile motion (neglecting air resistance), gravity is the only external force. To determine conservation, examine torque about the launch point. Torque equals r × F. Since gravity acts downward and its line of action does not generally pass through the starting point, a nonzero torque exists about that point.

Because angular momentum changes when external torque is present, it will vary over time relative to the starting point. Even though horizontal velocity remains constant, vertical velocity changes due to gravity, affecting momentum and therefore angular momentum.

This question tests understanding of torque’s effect on angular momentum in projectile motion.

Explanation: This question explores conservation of angular momentum in rotational systems. When no external torque acts, total angular momentum remains constant.

Angular momentum is given by L = Iω. If the gymnast pulls arms inward, the distribution of Mass shifts closer to the axis of rotation. This decreases the moment of inertia I. Since external torque is negligible, angular momentum must remain constant.

To maintain constant L while I decreases, angular velocity ω must increase. This explains why gymnasts spin faster when pulling their limbs inward. The effect follows directly from L = Iω and conservation principles.

It is similar to an ice skater spinning faster by pulling arms inward.

This tests application of conservation of angular momentum.

Explanation: This question asks about the Vector classification of angular momentum. Some Vectors are polar (like displacement), while others are axial (associated with rotation).

Angular momentum arises from the cross product r × p. Cross products produce axial Vectors, also called pseudovectors. These Vectors represent rotational effects and follow the right-hand rule for direction.

Unlike scalar quantities, angular momentum has direction perpendicular to the plane formed by position and momentum Vectors. Its direction reverses if the sense of rotation reverses.

Thus, understanding cross products and rotational quantities helps classify angular momentum correctly.

This evaluates knowledge of Vector types in rotational Physics.

Explanation: This question examines conservation of angular momentum when parts of a rotating system separate.

Initially, the system consists of the person and weights rotating together. Total angular momentum equals Iω. When the weights are released without additional force, they carry away some angular momentum.

Since external torque on the remaining system is negligible, the angular momentum of the person alone adjusts according to what remains after separation. Because the person’s moment of inertia does not change (arms unmoved), any change in angular momentum affects angular velocity.

Thus, analyzing transfer of angular momentum between separated parts determines the resulting rotational speed.

This tests understanding of angular momentum redistribution in rotating systems.

Explanation: This question relates linear speed to rotational motion. In pure rotation about a fixed axis, all particles share the same angular velocity ω.

However, linear speed depends on distance from the axis. It is given by v = rω. Since different particles lie at different radii r, their linear speeds differ.

Points farther from the axis move faster linearly than points closer to it. Even though ω is constant for all particles, v varies proportionally with r.

This demonstrates the distinction between angular and linear quantities in rotation.

The question tests understanding of v = rω in rigid body motion.

Explanation: This question compares linear and Rotational Dynamics. In linear motion, force causes acceleration according to F = ma.

In rotational motion, the analogous quantity is torque. Torque causes angular acceleration according to τ = Iα. Just as force changes linear momentum, torque changes angular momentum.

Torque depends on both magnitude of force and perpendicular distance from the axis. It represents the turning effect of force.

Understanding this parallel helps map linear concepts to rotational ones.

This tests recognition of torque as the rotational counterpart of force.

Explanation: This question applies conservation principles to a rotating system when Mass distribution changes.

When the boy jumps onto the spinning table, external torque about the axis is negligible. Therefore, total angular momentum of the combined system remains constant.

However, the moment of inertia increases because additional Mass is added farther from the axis. Since L = Iω remains constant, angular velocity adjusts accordingly.

Linear momentum is not necessarily conserved due to external forces from supports, but angular momentum about the axis is conserved.

This illustrates conservation of angular momentum in closed rotational systems.

Explanation: This question directly applies the conservation relation L = Iω. If no external torque acts, L remains constant.

If moment of inertia I decreases, then ω must change to keep L constant. Since L is fixed, reducing I requires ω to increase proportionally.

This explains phenomena like spinning skaters accelerating when pulling arms inward. The mathematical relation shows inverse proportionality between I and ω when L is conserved.

Thus, analyzing how changes in Mass distribution affect I helps predict rotational speed.

This tests quantitative understanding of conservation of angular momentum.

Explanation: This question concerns the shape of a liquid surface in rotational motion. When a Fluid rotates with constant angular velocity ω, each particle experiences centripetal acceleration equal to rω² directed toward the axis.

In the rotating frame, the outward centrifugal effect balances gravity and radial acceleration. The pressure distribution adjusts so that the liquid surface becomes perpendicular to the NET effective force at every point. Mathematically, the height of the liquid varies with radius as h ∝ r². This quadratic dependence indicates that the surface is not flat.

Because the height increases with the square of the radial distance, the surface forms a parabolic profile. This phenomenon is commonly observed when stirring liquids in a rotating container.

This question tests understanding of rotational motion and Fluid equilibrium under centripetal acceleration.

Explanation: This question applies Newton’s laws to systems of particles. The motion of the center of mass depends only on the NET external force acting on the system.

According to Newton’s second law for a system, F_ext = M a_cm, where M is total mass and a_cm is acceleration of the center of mass. If the total external force is zero, then a_cm = 0.

Zero acceleration means the velocity of the center of mass does not change. It may remain zero or continue moving with constant velocity, depending on initial conditions. Internal forces cancel due to Newton’s third law and do not affect center-of-mass motion.

This question tests understanding of system dynamics and the special role of external forces.

Explanation: This question examines motion of the center of mass in projectile motion involving rotation.

Even if a stick rotates while flying, its center of mass behaves as if all external forces act at that point. With gravity as the only external force (neglecting air resistance), the center of mass follows standard projectile motion.

Projectile motion under constant gravitational acceleration results in a parabolic trajectory. The rotational motion of the stick about its center of mass does not affect the path of the center of mass itself.

Thus, regardless of spinning or tumbling, the center of mass moves according to basic projectile equations.

This question tests separation of translational and rotational motion.

Explanation: This question analyzes forces in vertical circular motion. In such motion, the speed of the ball changes due to gravitational potential energy variation.

At the lowest point, speed is highest; at the highest point, speed is lowest. Tension in the string also varies because it must supply centripetal force in addition to balancing weight.

However, gravitational force acting on the ball remains constant in magnitude (mg), since mass and gravitational acceleration do not change significantly over the small vertical distance.

Thus, identifying which quantity depends only on constant physical parameters helps determine what remains unchanged.

This question tests understanding of vertical circular motion and force variation.

Explanation: This question connects stability with the position of the center of mass relative to the Base of support.

A body remains stable as long as the vertical line through its center of mass falls within its Base area. When tilted, the projection of the center of mass shifts. If it crosses beyond the Base boundary, gravitational torque causes the object to topple.

For a chair tipped backward, increasing the tilt moves the center of mass projection toward the edge of support. Once it moves outside, restoring torque no longer exists, and rotation continues until it falls.

Thus, stability depends directly on center-of-mass position relative to Base.

This question evaluates conceptual understanding of equilibrium and stability.

Explanation: This question involves two-body systems and mass distribution. The center of mass lies along the line joining two bodies and depends on their masses and separation.

For masses m₁ and m₂ separated by distance d, the center of mass distance from m₁ equals (m₂d)/(m₁ + m₂). Since Earth’s mass is much greater than the Moon’s mass, the center of mass lies closer to Earth.

Both bodies revolve about this common center of mass. In fact, this point lies inside Earth but not at its geometric center.

This question tests application of center-of-mass formula in astronomical systems.

Explanation: This question examines conservation principles during explosions.

When a projectile explodes, internal forces act between fragments, but the NET external force remains gravity (assuming no air resistance). Internal forces cannot change the motion of the center of mass.

Since gravity continues to act uniformly, the center of mass follows the same trajectory it would have followed if no explosion occurred. The explosion only redistributes internal motion among fragments.

Thus, the center of mass continues along the original projectile path determined by gravity.

This tests understanding of center-of-mass motion in systems undergoing internal interactions.

Explanation: This question applies conservation of momentum in an isolated system.

When two masses attract each other through mutual gravitational force, the forces are equal and opposite (Newton’s third law). Since no external force acts on the system, total linear momentum remains conserved.

If the system starts from rest, total momentum is zero. Therefore, the center of mass must remain at rest because motion of one mass is balanced by motion of the other.

Even though both masses move toward each other, the center of mass position does not change.

This tests application of momentum conservation to interacting systems.

Explanation: This question asks about the physical situation where angular velocity ω is the appropriate descriptive quantity.

Angular velocity measures rate of change of angular displacement. It is relevant when motion occurs about an axis, such as spinning wheels or rotating bodies.

In straight-line motion, linear velocity describes motion adequately. Angular velocity becomes meaningful only when rotation or circular motion is involved.

Thus, identifying the type of motion determines relevance of angular velocity.

This question checks conceptual understanding of rotational kinematics.

Explanation: This question addresses spatial positioning of the center of mass in rigid bodies.

The center of mass depends on mass distribution, not necessarily physical material location. For symmetric Solid objects, it often lies inside the body. However, for hollow or irregular shapes, it may lie outside the material.

For example, in a ring or hollow sphere, the center of mass lies at the geometric center, even though no material exists there.

Thus, center of mass location depends entirely on mass arrangement.

This tests conceptual clarity about center-of-mass positioning.

Explanation:

The question asks what determines the motion of the center of mass of a system of particles. It focuses on identifying the fundamental physical quantity responsible for governing this motion.

In mechanics, the center of mass behaves as if the entire mass of the system were concentrated at a single point. Newton’s Second Law for a system states that the NET external force acting on a system equals the total mass multiplied by the acceleration of its center of mass. Internal forces between particles cancel in pairs due to Newton’s Third Law and do not affect overall motion.

To understand this, consider that each particle obeys F = ma. When summing over all particles, internal forces cancel, leaving only external forces contributing to total acceleration. Therefore, the acceleration of the center of mass is given by the NET external force divided by total mass. This shows that only external influences determine its motion.

For example, in a rocket exploding mid-air, fragments scatter due to internal forces, but the center of mass continues along the same trajectory determined by gravity alone.

Thus, the motion of a system’s center of mass depends solely on the NET external force acting on the system.

Explanation:

This question examines a fundamental property of a system when calculating moments (mass × position Vectors) relative to the center of mass.

The center of mass is defined as the weighted average position of all particles in a system. Mathematically, it satisfies the condition that the sum of m_ir_i equals total mass multiplied by the center-of-mass position Vector. When positions are measured relative to the center of mass, this definition produces a key result.

If we shift the origin to the center of mass and compute the sum of m_ir’_i, where r’_i represents displacement from the center of mass, the total becomes zero. This is because positive and negative contributions balance exactly due to the weighted averaging definition.

Think of a balanced seesaw: the pivot (center of mass) is positioned such that clockwise and anticlockwise tendencies cancel out perfectly.

Therefore, when moments are taken about the center of mass, their algebraic sum always equals zero due to its defining property.

Explanation:

This question concerns the geometric position of the center of mass in a system containing two particles.

For two masses m₁ and m₂ located at positions r₁ and r₂, the center of mass is given by a weighted average: (m₁r₁ + m₂r₂) divided by total mass. This expression determines a point between the two position Vectors.

Because the formula is a linear combination of the two position Vectors, the resulting point must lie along the straight line joining them. Its exact position depends on relative masses: it shifts closer to the heavier particle.

For example, if both masses are equal, the center of mass lies exactly midway between them. If one mass is much larger, the center of mass lies very close to that heavier mass.

Thus, in any two-particle system, the center of mass always lies on the straight line joining the two particles.

Explanation:

This question asks for the location of the center of mass of three particles of equal mass placed at specified coordinates in a plane.

For equal masses, the center of mass coordinates are simply the arithmetic mean of the individual x-coordinates and y-coordinates. Since all masses are identical, weighting becomes unnecessary and the formula reduces to averaging.

To compute, add the x-coordinates (1 + 2 + 3) and divide by 3, and similarly add the y-coordinates (1 + 2 + 3) and divide by 3. This yields identical values for both coordinates because the points lie along a straight diagonal line.

This is similar to finding the average position of three equally weighted objects placed along a line; the balance point lies at their midpoint in terms of distribution.

Therefore, the center of mass is located at the average of the given coordinate values.