Problem 44 - Master Resource Book in JEE Main Physics

Question

The radius of the earth is \(6400 \mathrm{~km}\) and \(g=10 \mathrm{~m} / \mathrm{s}^{2}\) in order that a body of \(5 \mathrm{~kg}\) weight zero at the equator the angular speed of the earth is (a) \(1 / 80 \mathrm{rad} / \mathrm{s}\) (b) \(1 / 400 \mathrm{rad} / \mathrm{s}\) (c) \(1 / 800 \mathrm{rad} / \mathrm{s}\) (d) \(1 / 1600 \mathrm{rad} / \mathrm{s}\)

Step-by-Step Solution

Answer

(c) \(\frac{1}{800} \text{ rad/s}\).

Step 1: Convert the Radius to Meters

The radius of the Earth is given as 6400 km. We need to convert this into meters for consistency in units. Since 1 km is equal to 1000 meters, we have:\[\text{Radius in meters} = 6400 \text{ km} \times 1000 = 6400000 \text{ m}\]

Step 2: Establish the Condition for Weightlessness

A body experiences weightlessness when the gravitational force is equal to the centrifugal force. The gravitational force acting on the body is defined as:\[F_g = mg = 5 \times 10 = 50 \text{ N}\]

Step 3: Explain Centrifugal Force Concept

For a rotating body at the equator, centrifugal force is given by:\[F_c = m \omega^2 R\]Here, \(\omega\) is the angular speed, \(R\) is the radius of the Earth, and \(m\) is the mass of the body.

Step 4: Set Gravitational Force Equal to Centrifugal Force

Set the gravitational force equal to the centrifugal force to find the angular speed where the weight of the body is zero:\[mg = m \omega^2 R\]

Step 5: Cancel Mass and Solve for Angular Speed

Cancel the mass \(m\) from both sides:\[g = \omega^2 R\]Now solve for \(\omega\):\[\omega^2 = \frac{g}{R} \rightarrow \omega = \sqrt{\frac{g}{R}}\]

Step 6: Substitute Known Values

Substitute the known values of \(g = 10 \text{ m/s}^2\) and \(R = 6400000 \text{ m}\):\[\omega = \sqrt{\frac{10}{6400000}}\]

Step 7: Calculate Angular Speed

Calculate the value:\[\omega = \sqrt{\frac{10}{6400000}} = \sqrt{\frac{1}{640000}} = \frac{1}{800} \text{ rad/s}\]

Step 8: Select the Correct Option

Convert the calculation into the given options: \(\omega = \frac{1}{800} \text{ rad/s}\). Therefore, the correct answer is option (c): \(\frac{1}{800} \text{ rad/s}\).

Key Concepts

Earth's Rotation

The rotation of the Earth is a fascinating phenomenon that affects various aspects of life on our planet. Earth completes one full rotation around its axis approximately every 24 hours. This rotation is responsible for the day and night cycle. As Earth spins, different parts of its surface are exposed to sunlight or darkness, thereby creating daytime and nighttime.

Rotation also influences time zones and contributes greatly to weather patterns. Furthermore, the Earth's rotation leads to the Coriolis effect, which affects wind and ocean currents. This results in unique weather phenomena and climate systems globally.

It's important to note that the Earth's rotation isn't perfectly stable. Minute changes, such as the slowing down of rotation by milliseconds due to tidal forces, occur over long time periods. Despite these minor changes, the predictable rotation remains a critical component in the study of physics and astronomy.

Angular Speed

Angular speed is a measure of how fast an object rotates or revolves relative to another point. In the case of Earth's rotation, it refers to how quickly the Earth spins on its axis.

Angular speed is mathematically represented by the symbol \(\omega\) and expressed in radians per second (rad/s). It's calculated by considering the rate of rotation and the time taken to complete a revolution. For the Earth's rotation, angular speed can be computed by dividing the total angle of rotation, 2π radians for a full circle, by the rotational period, roughly 86400 seconds (24 hours).

• Angular speed formula: \(\omega = \frac{2\pi}{T}\), where T is the period of rotation.
• At the equator, Earth's angular speed aids in creating centrifugal force, impacting the sensation of gravity slightly.

The understanding of angular speed is essential when calculating forces acting on rotating bodies like Earth. It helps explain phenomena like why objects might feel lighter at the equator compared to the poles.

Gravitational Force

Gravitational force is a fundamental interaction in physics that attracts two bodies towards each other. The force is dependent on the masses of the objects and the distance between their centers. Given by Newton's law of universal gravitation, the formula is: \(F = G\frac{m_1 m_2}{r^2}\), where G is the gravitational constant.

On Earth, this force provides us with the sensation of "weight." When objects are in free fall or in orbit, they experience weightlessness because gravitational forces are essentially balanced by centrifugal forces or the trajectory of fall.

• On Earth's surface, gravity gives us an acceleration of approximately \(9.81 \text{ m/s}^2\), often approximated as 10 m/s² for simplicity in calculations.
• Gravitational force is paramount in maintaining the orbits of planets and satellites.

Gravitational force, while weak when considered between small everyday objects, becomes significant when large masses like planets are involved. In physics problems, understanding gravitational force is key for analyzing the motion of bodies under its influence.