If the Earth suddenly stops rotating,then the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The apparent weight $w'$ of an object of mass $m$ at the equator is given by the formula: $w' = mg - m \omega^2 R$,where $g$ is the acceleration d

Vedclass · Accepted Answer

Increase by $m \omega^2 R$

Vedclass · Answer

(B) The apparent weight $w'$ of an object of mass $m$ at the equator is given by the formula: $w' = mg - m \omega^2 R$,where $g$ is the acceleration due to gravity,$\omega$ is the angular speed of the Earth,and $R$ is the radius of the Earth.When the Earth is rotating,the effective weight is reduced by the centrifugal force $m \omega^2 R$.If the Earth suddenly stops rotating,the angular speed $\omega$ becomes $0$.Consequently,the new apparent weight $w''$ becomes $w'' = mg$.Comparing the two,the weight increases by the magnitude of the centrifugal force,which is $m \omega^2 R$.Therefore,the weight of the object increases by $m \omega^2 R$.

If the Earth suddenly stops rotating,then the weight of an object of mass $m$ at the equator will $[\omega$ is the angular speed of the Earth and $R$ is its radius]

Similar Questions

Acceleration due to gravity on the surface of Earth is $g$. If the diameter of Earth is reduced to one-third of its original value and mass remains unchanged,then the acceleration due to gravity on the surface of the Earth is . . . . . . $g$.

The value of the acceleration due to gravity is $g_{1}$ at a height $h = \frac{R}{2}$ ($R$ = radius of the earth) from the surface of the earth. It is again equal to $g_{1}$ at a depth $d$ below the surface of the earth. The ratio $\left(\frac{d}{R}\right)$ equals

$Assertion$ : In a free fall,the weight of a body becomes effectively zero.
$Reason$ : The acceleration due to gravity acting on a body in free fall is zero.

At what height above the Earth's surface does the acceleration due to gravity become $1\%$ of its value at the Earth's surface (in $R$)? (Radius of Earth $= R$)

The height at which the weight of a body becomes $\frac{1}{9}$ of its weight on the surface of the earth (radius of the earth is $R$) is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module