Question

What should be the angular velocity of earth due to rotation about its own axis so that the weight at equator becomes \((\frac{3}{5})^{th}\) of initial value? (\(g =\) acceleration due to gravity, R = radius of earth)

• A. \((\frac{R}{3g})^{\frac{1}{2}}\)
• B. \((\frac{2g}{5R})^{\frac{1}{2}}\)
• C. \((\frac{5g}{9R})^{\frac{1}{2}}\)
• D. \((\frac{3g}{2R})^{\frac{1}{2}}\)

Hint:

Apparent g at the equator is g - omega^2 R.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
At the equator, the effective gravity is reduced by the centrifugal effect of rotation: \(g'=g-\omega^2R\).

Step 2: Weight condition:
Weight becomes \(\dfrac35\) of the initial, so \(g'=\dfrac35g\).

Step 3: Solve:
\[ g-\omega^2R=\frac35g\ \Rightarrow\ \omega^2R=\frac25g\ \Rightarrow\ \omega=\left(\frac{2g}{5R}\right)^{1/2} \]

Step 4: Choose:
Option (B).

Final Answer:
omega = sqrt(2g/5R). \[ \boxed{\omega=\left(\frac{2g}{5R}\right)^{1/2}} \]