Problem 18 - Master Resource Book in JEE Main Physics

Question

Let \(L_{1}, L_{2}\) and \(r_{1}, r_{2}\) are the angular momenta and position vectors of the particles at that instant about any arbitrary point \(O .\) Angular momentum of the particles, $$\mathbf{L}_{1}=\mathbf{r}_{1} \times m \mathbf{v} \text { and } \mathbf{L}_{2}=\mathbf{r}_{2} \times m \mathbf{v}$$ It resultant angular momentum of the system is \(\mathbf{L}\), then $$\mathbf{L}=\mathbf{L}_{1}+\mathbf{L}_{2}=\mathbf{r}_{1} \times m \mathbf{v}+\left(-\mathbf{r}_{2}+m \mathbf{v}\right)$$ Negative sign shown that both particles are moving in opposite directions. \(\begin{aligned} |\mathbf{L}| &=\left|\mathbf{L}_{1}\right|-\left|\mathbf{L}_{2}\right| \\ &=m v r_{1} \sin \theta_{1}-m v r_{2} \sin \theta_{2} \\ &=m v\left(r_{1} \sin \theta_{1}-r_{2} \sin \theta_{2}\right) \end{aligned}\) where \(\theta_{1}\) and \(\theta_{2}\) are the angles between \(\mathbf{r}_{1}, \mathbf{v}\) and \(\mathbf{r}_{2}, \mathbf{v}\) respectively. When particles changes their position with time, their direction of motion (v) remains unchanged and therefore distances \(O M=r_{1} \sin \theta_{1}\) and \(O N=r_{2} \sin \theta_{2}\) remains same. But \(\quad O M-O N=M N=d \quad\) (Given) \(\therefore \quad r_{1} \sin \theta_{1}-r_{2} \sin \theta_{2}=d\) (ii) From Eqs. (i) and (ii), we get \(|\mathbf{L}|=m v d\) It is constant with time. The direction of \(\mathbf{L}\) is perpendicular to the plane of \(\mathbf{r}\) and \(\mathbf{v}\) and is inward to the plane of paper, which also remains unchanged with time.

Step-by-Step Solution

Answer

The angular momentum magnitude is constant, \(|\mathbf{L}| = m v d\), and its direction is perpendicular and inward.

Step 1: Understanding the Angular Momentum

The angular momenta of two particles about a certain point are given by \( \mathbf{L}_{1} = \mathbf{r}_{1} \times m \mathbf{v} \) and \( \mathbf{L}_{2} = \mathbf{r}_{2} \times m \mathbf{v} \). This represents the cross product of the position vector with the linear momentum vector of each particle.

Step 2: Resultant Angular Momentum

The resultant angular momentum \( \mathbf{L} \) for the system is obtained by adding the individual angular momenta: \( \mathbf{L} = \mathbf{L}_{1} + \mathbf{L}_{2} = \mathbf{r}_{1} \times m \mathbf{v} + (-\mathbf{r}_{2} + m \mathbf{v}) \). The negative sign indicates that the particles are moving in opposite directions.

Step 3: Magnitude of Resultant Angular Momentum

The magnitude of the resultant angular momentum is calculated as \( |\mathbf{L}| = |\mathbf{L}_{1}| - |\mathbf{L}_{2}| \), resulting in the expression \( m v r_{1} \sin \theta_{1} - m v r_{2} \sin \theta_{2} \) which further simplifies to \( m v (r_{1} \sin \theta_{1} - r_{2} \sin \theta_{2}) \).

Step 4: Constancy of Distance Components

Given that the distances \( O M = r_{1} \sin \theta_{1} \) and \( O N = r_{2} \sin \theta_{2} \) are consistent over time, and \( O M - O N = M N = d \), we conclude \( r_{1} \sin \theta_{1} - r_{2} \sin \theta_{2} = d \).

Step 5: Final Expression for Angular Momentum Magnitude

Combining these results yield \( |\mathbf{L}| = m v d \). This implies that the magnitude of the angular momentum remains constant with time.

Step 6: Direction of the Angular Momentum

The direction of \( \mathbf{L} \) is perpendicular to the plane formed by \( \mathbf{r} \) and \( \mathbf{v} \) and points inward, a property that does not change with time.

Key Concepts

Position Vector

The position vector is a crucial element in understanding angular momentum. It denotes the position of a point related to another point, often serving as the reference or origin. In physics, it is typically represented by symbols such as \( \mathbf{r}_1 \) or \( \mathbf{r}_2 \). These vectors provide information regarding the spatial coordinates of particles in motion.
Understanding the position vector is key because it helps in determining the location and trajectory of moving objects in space. In the context of angular momentum, position vectors like \( \mathbf{r}_1 \) and \( \mathbf{r}_2 \) are used to calculate the angular momenta of particles. You can think of them as being the arms extending from a fixed point \( O \) to the location of each particle. This allows you to visualize how the particles move in relation to this origin, which is vital when analyzing their angular motion.

Cross Product

The cross product is a mathematical operation used to derive new vectors from two existing ones. In angular momentum, it's used to describe the product of the position vector and linear momentum, denoted as \( \mathbf{L} = \mathbf{r} \times m \mathbf{v} \).
The cross product has unique properties:

• It results in a vector that is perpendicular to the plane formed by the initial vectors.
• The magnitude of this vector depends on the sine of the angle between the two vectors.
• It allows you to determine the direction of the angular momentum vector.

The calculation involves not just the values of the vectors, but also the angle formed between these vectors. This is why you see terms like \( \sin \theta \) in expressions for angular momentum. Understanding the cross product allows you to compute how different forces combine to influence rotational movement.

Constant Angular Momentum

The concept of constant angular momentum refers to a situation where the magnitude of angular momentum does not change over time. This happens under certain conditions, particularly when external forces do not induce a torque on the system.
In the exercise, we see that the angular momentum expression simplifies to \( |\mathbf{L}| = m v d \), indicating a constant value with respect to time. This constant nature is crucial since it suggests a closed system where no net external forces are acting to alter the angular momentum.
Situations with constant angular momentum can be significant because they often simplify calculations and predictions, allowing one to focus on other dynamic aspects. It is this property of constancy that often aids in preserving the initial state of rotational systems, making them extremely useful in fields like mechanics and astronomy.

Resultant Angular Momentum

Resultant angular momentum refers to the combined effect of angular momentum from multiple objects or particles within a system. In the context of the exercise, it is derived by summing the angular momenta of individual particles.
Calculated as \( \mathbf{L} = \mathbf{L}_1 + \mathbf{L}_2 \), this process takes into account the various directions and magnitudes of each component's angular momentum. For systems where particles move in opposite directions, the signs in these equations reflect those movements, as indicated by the negative sign in the expression \( -\mathbf{r}_2 + m \mathbf{v} \).
The resultant angular momentum gives an overall picture of how the system is behaving rotationally. It helps in understanding the collective rotational influence of all particles when evaluated about a common point like \( O \). This is particularly useful in multi-body systems where interactions can become complex, and a unified perspective is needed.