Problem 579 - JEE Physics

Question

If the earth is treated as a sphere of radius \(R\) and mass \(M\). Its angular momentum about the axis of rotation with period \(\mathrm{T}\) is..... \(\\{\mathrm{A}\\}\left(\pi \mathrm{MR}^{3} / \mathrm{T}\right)\) \(\\{\mathrm{B}\\}\left(\operatorname{MR}^{2} \pi / \mathrm{T}\right)\) \(\\{C\\}\left(2 \pi \mathrm{MR}^{2} / 5 \mathrm{~T}\right)\) \(\\{\mathrm{D}\\}\left(4 \pi \mathrm{MR}^{2} / 5 \mathrm{~T}\right)\)

Step-by-Step Solution

Answer

The short answer is: The angular momentum of the Earth with given conditions is \(L = \dfrac{4\pi MR^2}{5T}\). Thus, the correct option is D.

Step 1: Recall the formula for angular momentum of a solid sphere

The formula for angular momentum (L) of a solid sphere of mass M and radius R, rotating about an axis through its center with an angular velocity ω is: \(L = \dfrac{2}{5}MR^2 ω\)

Step 2: Find the relationship between period and angular velocity

The period (T) of rotation of a rotating object is the time taken for one complete revolution. The angular velocity (ω) is the rate of change of angular displacement. The relationship between angular velocity (ω) and period (T) can be written as: \(ω = \dfrac{2\pi}{T}\)

Step 3: Substitute the relationship between ω and T in the angular momentum formula

Now, let's substitute the relationship between ω and T in the formula for the angular momentum of a solid sphere: \(L = \dfrac{2}{5}MR^2 \cdot \dfrac{2\pi}{T}\)

Step 4: Simplify the expression

Simplify the expression for the angular momentum: \(L = \dfrac{4\pi MR^2}{5T}\)

Step 5: Compare the simplified expression with the given options

Now, we compare our simplified expression with the given options: \(L = \dfrac{4\pi MR^2}{5T}\) matches with option D. So the correct answer is: \(\\{\mathrm{D}\\}\left(4 \pi \mathrm{MR}^{2} / 5 \mathrm{~T}\right)\).

Key Concepts

solid sphere

A solid sphere is a three-dimensional object that is perfectly symmetrical around its center. All points on the surface are equidistant from the center, which is known as the radius (R). The mass (M) of a solid sphere is distributed uniformly throughout its volume. This symmetry makes it an ideal model for calculating properties related to rotational motion, such as angular momentum.
When considering rotational dynamics, the moment of inertia plays a crucial role. For a solid sphere, the moment of inertia, which is a measure of an object's resistance to changes in its rotational motion, is calculated using the formula \(I = \frac{2}{5}MR^2\). This specific moment of inertia formula is due to the uniform distribution of mass and the spherical shape.
Understanding how to calculate the moment of inertia allows us to later determine other characteristics, such as angular momentum, when the sphere is rotating about its axis.

angular velocity

Angular velocity refers to how fast an object rotates or revolves relative to another point, often the center of rotation. It is denoted by the Greek letter omega (\omega). In the context of the Earth's rotation, angular velocity describes how quickly the Earth spins around its axis.
To find the angular velocity, one considers the angle that has been traversed over a certain period of time. In mathematical terms, it's given by the expression \(\omega = \frac{2\pi}{T}\), where \(T\) is the period of rotation. This formula highlights the relationship between the angular velocity and the time it takes for one complete revolution.
Understanding angular velocity is essential for calculating other rotational properties, such as angular momentum. It essentially tells us how many turns, in radians, an object makes in one second.

period of rotation

The period of rotation is the amount of time it takes for a rotating object to make one complete revolution around its axis. For the Earth, this period is approximately 24 hours, as it completes one full spin each day.
It is closely related to angular velocity, as they are inversely proportional. The formula \(\omega = \frac{2\pi}{T}\) can be rearranged to express the period in terms of angular velocity: \(T = \frac{2\pi}{\omega}\). This highlights that as the angular velocity increases, the period of rotation decreases, and vice versa.
Understanding the period of rotation is vital for predicting the movement and behavior of rotating objects, and is fundamental when working with rotational physics problems.

rotational motion

Rotational motion occurs when an object spins around an internal axis. It is a type of motion that occurs frequently and is observed in planets, wheels, and even everyday objects like tops or yo-yos. This type of motion is characterized by quantities such as angular displacement, angular velocity, and angular momentum.
The fundamental difference between linear and rotational motion is that instead of distance and velocity, we focus on angles and the rate of angular change. Key formulas involve quantities like the moment of inertia (for resistance to changes in motion) and angular momentum, which depends on both the inertia and the angular velocity of the object.
Analyzing rotational motion involves understanding how these concepts interact with forces and torques to result in the rotational movement of objects. Grasping the basics of rotational motion allows us to solve complex problems, predict outcomes, and design systems that efficiently utilize this fundamental type of motion.