Question

The angular velocity of the earth with which it has to rotate so that acceleration due to gravity on 60° latitude becomes zero is (Radius of earth = 6400 km. At the poles g = 10 ms^–2)

• A. 2.5×10^–3rad/sec
• B. 5.0×10^–1rad/sec
• C. $10 \times 10 ^ { 1 } \mathrm { rad } / \mathrm { sec }$
• D. $7.8 \times 10 ^ { - 2 } \mathrm { rad } / \mathrm { sec }$

Answer 1

Answer: (A)

2.5×10^–3rad/sec

Answer 2

Effective acceleration due to gravity due to rotation of earth$g ^ { \prime } = g - \omega ^ { 2 } R \cos ^ { 2 } \lambda$

⇒ $0 = g - \omega ^ { 2 } R \cos ^ { 2 } 60 ^ { \circ }$ ⇒ $\frac { \omega ^ { 2 } R } { 4 } = g$

⇒ $\omega = \sqrt { \frac { 4 g } { R } } = 2 \sqrt { \frac { g } { R } } = \frac { 2 } { 800 } \frac { \mathrm { rad } } { \mathrm { sec } }$

[As $g ^ { \prime } = 0$ and $\lambda = 60 ^ { \circ }$]

⇒ $\omega = \frac { 1 } { 400 } = 2.5 \times 10 ^ { - 3 } \frac { \mathrm { rad } } { \mathrm { sec } }$.