Question

In the List-I below, four different paths of a particle are given as functions of time. In these functions, \(\alpha\) and \(\beta\) are positive constants of appropriate dimensions and \(\alpha \neq \beta .\) In each case, the force acting on the particle is either zero or conservative. In List-II, five physical quantities of the particle are mentioned: \(\vec{p}\) is the linear momentum, \(\vec{L}\) is the angular momentum about the origin, \(K\) is the kinetic energy, \(U\) is the potential energy and \(E\) is the total energy. Match each path in List-I with those quantities in List-II, which are conserved for that path. LIST-I P. \(\vec{r}(t)=\alpha t \hat{\imath}+\beta t \hat{\jmath}\) Q. \(\vec{r}(t)=\alpha \cos \omega t \hat{\imath}+\beta \sin \omega t \hat{\jmath}\) R. \(\vec{r}(t)=\alpha(\cos \omega t \hat{\imath}+\sin \omega t \hat{\jmath})\) S. \(\vec{r}(t)=\alpha t \hat{\imath}+\frac{\beta}{2} t^{2} \hat{\jmath}\) LIST-II 1\. \(\vec{p}\) 2\. \(\vec{L}\) 3\. \(K\) 4\. \(U\) 5\. \(E\)

Short answer

P: \(\vec{p}\), \(\vec{L}\), \(E\); Q: \(U\), \(E\); R: \(\vec{p}\), \(\vec{L}\), \(K\), \(U\), \(E\); S: (None of the given quantities are strictly conserved due to the non-zero net force and resultant acceleration).

Step-by-step solution

Identify Conserved Quantities for Path P

The path \(\vec{r}(t)=\alpha t \hat{\imath}+\beta t \hat{\jmath}\) represents linear motion with constant velocity. Therefore, the linear momentum \(\vec{p}\) remains constant over time. No net force means no torque about the origin, so angular momentum \(\vec{L}\) is conserved as well. Due to constant velocity, kinetic energy \(K\) is also conserved. Since there is no mention of a potential field, we cannot ascertain the conservation of potential energy \(U\), but total energy \(E\) is conserved since the kinetic energy is constant and there are no non-conservative forces acting.

Analyze Path Q for Conserved Properties

For \(\vec{r}(t)=\alpha \cos \omega t \hat{\imath}+\beta \sin \omega t \hat{\jmath}\), the particle experiences simple harmonic motion in both \(\hat{\imath}\) and \(\hat{\jmath}\) directions independently. The forces in such a case are conservative, which means potential energy \(U\) and total energy \(E\) are conserved. Since the motion is periodic and returns to initial velocity at regular intervals, the linear momentum \(\vec{p}\) and kinetic energy \(K\) are not conserved. The motion does not conserve angular momentum \(\vec{L}\) because as the particle moves in the elliptical path, the lever arm and hence angular momentum changes over time.

Assess Conserved Quantities for Path R

The path \(\vec{r}(t)=\alpha(\cos \omega t \hat{\imath}+\sin \omega t \hat{\jmath})\) is a circular motion with radius \(\alpha\). For circular motion, the force acting is centripetal and conservative. Therefore, the potential energy \(U\) and total energy \(E\) are conserved. Since the particle is moving at a constant speed in a circular motion, the magnitude of linear momentum \(\vec{p}\) and kinetic energy \(K\) are conserved. Additionally, angular momentum \(\vec{L}\) is conserved because the perpendicular distance from the origin to the path of motion (radius of the circle) is constant.

Determine Conserved Elements for Path S

The path \(\vec{r}(t)=\alpha t \hat{\imath}+\frac{\beta}{2} t^{2} \hat{\jmath}\) suggests an accelerated motion due to the squared time dependence in the second term. Acceleration indicates a non-zero net force, hence the momentum \(\vec{p}\) is not conserved. The force is not necessarily conservative, which means potential energy \(U\) may not be conserved. The kinetic energy \(K\) changes due to acceleration, so it is not conserved. Angular momentum \(\vec{L}\) is also not conserved due to the changing lever arm as the particle's trajectory moves away from the origin linearly in \(\hat{\imath}\) and quadratically in \(\hat{\jmath}\). Total energy \(E\) may change if the forces are non-conservative.

Key concepts

Conservation of Momentum

The principle of conservation of momentum is fundamental in physics and asserts that the total momentum of an isolated system remains constant if no external forces act upon it. Momentum, denoted by \( \vec{p} \), is the product of an object's mass and its velocity. It's a vector quantity, which means it has both magnitude and direction.

In the context of the exercise, path P illustrates a particle moving with constant velocity, indicating no external forces are acting upon it. This means the linear momentum is conserved. However, for path S, we see accelerated motion due to the presence of a squared time component, suggesting a varying force, and therefore, the linear momentum is not conserved in this scenario.

Understanding the conservation of linear momentum is essential in solving many physical problems, especially in collision and explosion scenarios, where the momentum before and after the event remains unchanged.

Angular Momentum

Angular momentum, symbolized by \( \vec{L} \), is related to the rotation of an object about a point. It is conserved in a system when no external torque acts on the system. Angular momentum depends upon the distance of the object from the point of rotation (lever arm) and the linear momentum of the object.

In the given exercise, for path R, the particle is in circular motion at a constant radius \( \alpha \). Since the radius does not change, the lever arm remains the same and, consequently, the angular momentum is conserved. By contrast, in path Q, the lever arm changes as the particle traces out an elliptical path, resulting in a change in angular momentum over time. This situation leads to angular momentum not being conserved.

Kinetic and Potential Energy

Kinetic energy \( K \) is the energy associated with the movement of an object, while potential energy \( U \) refers to the energy stored due to an object's position or arrangement. In an isolated system where no non-conservative forces are present, the sum of kinetic and potential energy, known as the total mechanical energy, remains constant.

For example, the path R in the exercise depicts uniform circular motion. In such a case, even though the particle's direction changes, its speed does not. Thus, kinetic energy, which depends solely on the speed and mass of the particle, remains constant. Potential energy is also conserved if we assume centripetal force is conservative. In contrast, path S involves acceleration, which suggests that kinetic energy is not conserved because the speed of the particle changes over time.

Conservative Forces

Conservative forces are unique in that the work they do on a particle moving between two points is independent of the path taken by the particle. These forces have the property that they can fully convert kinetic energy to potential energy and vice versa without any loss of energy to other forms like heat or sound. Gravitational and electrostatic forces are prime examples of conservative forces.

In our exercise, the forces acting on paths Q and R are indeed conservative, as they represent simple harmonic motion and uniform circular motion, respectively. The forces that act on the particles in these paths are central forces with potential functions, meaning the total mechanical energy remains conserved. Path S, exhibiting non-uniformly accelerated motion, implies the presence of non-conservative forces or changing energy forms, highlighting that not all forces in mechanics are conservative.

Simple Harmonic Motion

Simple Harmonic Motion (SHM) is a type of periodic motion where the restoring force is directly proportional to the displacement and acts in the direction opposite to that displacement. A classic example is the motion of a mass attached to a spring when it's displaced from its equilibrium position. SHM is characterized by the object's oscillation about an equilibrium point, during which the potential and kinetic energies are interchanged but their sum (total mechanical energy) remains constant.

In the context of this exercise, path Q represents an example of SHM, as the motion can be decomposed into independent oscillations along the \( \hat{\imath} \) and \( \hat{\jmath} \) axes. Despite the interchange of kinetic and potential energy as the particle moves, the system's total energy \( E \) remains constant because the forces are conservative, i.e., they do not dissipate the mechanical energy.