Problem 685 - JEE Physics

Question

The radius of the earth is \(6400 \mathrm{~km}\) and \(\mathrm{g}=10 \mathrm{~ms}^{-2} .\) In order that a body of \(5 \mathrm{~kg}\) weights zero at the equator, the angular speed of the earth is \(=\ldots \ldots \ldots \mathrm{rad} / \mathrm{sec}\) (A) \((1 / 80)\) (B) \([1 /(400)]\) (C) \([1 /(800)]\) (D) \([1 /(600)]\)

Step-by-Step Solution

Answer

The angular speed of the Earth for a body of 5 kg to weigh zero at the equator is (C) \(\frac{1}{800} ~rad/s\).

Step 1: Gravitational Force

The gravitational force acting on the body can be calculated using the formula: \[F_g = m * g\] where \(F_g\) is the gravitational force, \(m\) is the mass of the body, and \(g\) is the gravitational acceleration.

Step 2: Centrifugal Force

The centrifugal force acting on the body can be calculated using the formula: \[F_c = m * R * \omega^2\] where \(F_c\) is the centrifugal force, \(m\) is the mass of the body, \(R\) is the radius of the Earth, and \(\omega\) is the angular speed.

Step 3: Balancing the forces

For the body to experience zero weight, the gravitational force should be equal to the centrifugal force: \[F_g = F_c\] \[m * g = m * R * \omega^2\]

Step 4: Solve for the angular speed (\(\omega\))

To find the angular speed \(\omega\), we can rearrange the equation: \[\omega^2 = \frac{g}{R}\] \[\omega = \sqrt{\frac{g}{R}}\] Now, we can substitute the given values to find the angular speed \(\omega\): \[\omega = \sqrt{\frac{10~m/s^2}{6400 * 10^3~m}}\] \[\omega = \sqrt{\frac{1}{640 * 10^3}}\] \[\omega = \frac{1}{\sqrt{640 * 10^3}} ~rad/s\] Now, comparing our answer with the given choices, let's check which choice matches our result: (A) \(\frac{1}{80} = \boxed{\frac{1}{\sqrt{6400}} \neq \omega}\) (B) \(\frac{1}{400} = \frac{1}{\sqrt{160000}} \neq \omega\) (C) \(\frac{1}{800} = \boxed{\frac{1}{\sqrt{640 * 10^3}} \Large\checkmark = \omega}\) (D) \(\frac{1}{600} = \frac{1}{\sqrt{360000}} \neq \omega\) So, the answer is (C) \(\frac{1}{800}\)

Key Concepts

Gravitational Acceleration

Gravitational acceleration, denoted as \(g\), is the acceleration that the Earth imparts to objects on or near its surface due to gravity. It is a crucial concept in physics and plays a significant role in determining the weight of an object. The standard gravitational acceleration on Earth is approximately \(9.81 \, \text{m/s}^2\), but for simplicity in many physics problems, it is rounded to \(10 \, \text{m/s}^2\). This value represents the constant acceleration experienced by any object in freefall near the earth's surface, assuming air resistance is negligible.

Gravitational acceleration is the reason that objects fall to the ground when dropped and why they weigh a certain amount. It is calculated using Newton's law of universal gravitation, factoring in the mass of the Earth and the radius from its center. Its equation is often simplified to

• \(F = m \cdot g\)

where \(F\) is the force due to gravity (or weight) and \(m\) is the mass of the object. This formula is also key in understanding how forces balance with centrifugal force in rotational systems.

Angular Speed

Angular speed, often denoted by the Greek letter \( \omega \), is a measure of how quickly an object rotates or revolves around an axis. It is essential when considering the motion of objects like planets or wheels, including the rotation of Earth itself.

Mathematically, angular speed is defined as the rate of change of angular displacement per unit time. In simpler terms, it tells us how much angle has been covered in a particular amount of time. Its unit is usually in radians per second (rad/s), which makes it convenient when dealing with circular motion.

For the Earth to make an object at the equator weightless through its rotation, the gravitational force must be balanced by the centrifugal force. The equation to find angular speed in such a scenario is given by:

• \(\omega = \sqrt{\frac{g}{R}}\)

where \(g\) is gravitational acceleration and \(R\) is Earth's radius. This equation arises from setting the gravitational force equal to the centrifugal force resulting from the Earth's rotation.

Earth's Radius

Earth's radius is a vital measurement when solving problems related to Earth's gravitational and rotational dynamics. It is roughly \(6400\) kilometers from the center of the Earth to its surface, though this can vary slightly due to Earth's oblate shape, being wider at the equator than pole-to-pole.

Knowing the Earth's radius allows us to connect linear and angular quantities, which is essential in our example problem. It helps calculate the centrifugal force facing a body on the Earth's surface due to the rotation. In physics, using the formula for centrifugal force:

• \(F_c = m \cdot R \cdot \omega^2\)

where \(m\) is the mass and \(\omega\) is the angular speed, Earth’s radius \(R\) is pivotal. It ensures we understand how much this force impacts objects at the equator as the Earth rotates, balancing it with gravitational force to explore concepts like weightlessness at high rotational speeds.