With what angular velocity the earth should spin in order that a body lying at ${37^{th}}$ latitude may become weightless?
A. $\dfrac{5}{4}\sqrt {\dfrac{g}{R}} $
B. $\dfrac{{25}}{{16}}\sqrt {\dfrac{g}{R}} $
C. $\dfrac{5}{3}\sqrt {\dfrac{g}{R}} $
D. $\dfrac{{25}}{9}\sqrt {\dfrac{g}{R}} $
Answer
573.6k+ views
Hint: In order to solve this question we need to understand angular speed and centripetal force. So angular velocity is defined as velocity with which the body rotates, it is mathematically defined as the division of linear velocity with the radius in which it rotates. Centripetal force is defined as the pseudo force which acts in the opposite direction of radius, this force is the reason why a body moving on a circle is in equilibrium.
Complete step by step answer:
Latitude is defined as the angle that a body or object subtends with the equator axis; it is used to denote the position of an object on earth along with longitude.Consider earth rotates with angular velocity $\omega $ and let the mass of body be $m$.Also let the angle that it makes with equator axis be $\theta $ and the radius of earth be $R$.So the centripetal force that acts in outward direction is given by,
$f = m{R^2}\omega $
If the earth is not rotating then weight of body acts in downward direction and it is equal to, ${f_1} = mg$
Since the earth is rotating so let the weight of the body denoted by, ${f_2} = mg'$.
Since we know variation of acceleration due to gravity $g$ with $\theta $ as,
$g' = g - R{\omega ^2}{\cos ^2}\theta $
So the force ${f_2}$ becomes after putting value,
${f_2} = m(g - R{\omega ^2}{\cos ^2}\theta )$
$\Rightarrow {f_2} = mg - mR{\omega ^2}{\cos ^2}\theta $
So for the body to be weightless we know, ${f_2} = 0$.
Putting values we get,
$m(g - R{\omega ^2}{\cos ^2}\theta ) = 0$
$\Rightarrow {\omega ^2} = \dfrac{g}{{R{{\cos }^2}\theta }}$
$\Rightarrow {\omega ^2} = \dfrac{g}{{R{{\cos }^2}37}}$
Since, $\cos 37 = \dfrac{4}{5}$
So, ${\omega ^2} = \dfrac{{25g}}{{9R}}$
$\therefore \omega = \dfrac{5}{4}\sqrt {\dfrac{g}{R}} $
So the correct option is A.
Note: It should be remembered that earth is also tilted at its own axis and due to this climate changes on earth. Also it should be noted here that with some angular velocity we can feel weightlessness or in other words we can eliminate gravity, so gravity is not a force it’s merely due to space time curvature.
Complete step by step answer:
Latitude is defined as the angle that a body or object subtends with the equator axis; it is used to denote the position of an object on earth along with longitude.Consider earth rotates with angular velocity $\omega $ and let the mass of body be $m$.Also let the angle that it makes with equator axis be $\theta $ and the radius of earth be $R$.So the centripetal force that acts in outward direction is given by,
$f = m{R^2}\omega $
If the earth is not rotating then weight of body acts in downward direction and it is equal to, ${f_1} = mg$
Since the earth is rotating so let the weight of the body denoted by, ${f_2} = mg'$.
Since we know variation of acceleration due to gravity $g$ with $\theta $ as,
$g' = g - R{\omega ^2}{\cos ^2}\theta $
So the force ${f_2}$ becomes after putting value,
${f_2} = m(g - R{\omega ^2}{\cos ^2}\theta )$
$\Rightarrow {f_2} = mg - mR{\omega ^2}{\cos ^2}\theta $
So for the body to be weightless we know, ${f_2} = 0$.
Putting values we get,
$m(g - R{\omega ^2}{\cos ^2}\theta ) = 0$
$\Rightarrow {\omega ^2} = \dfrac{g}{{R{{\cos }^2}\theta }}$
$\Rightarrow {\omega ^2} = \dfrac{g}{{R{{\cos }^2}37}}$
Since, $\cos 37 = \dfrac{4}{5}$
So, ${\omega ^2} = \dfrac{{25g}}{{9R}}$
$\therefore \omega = \dfrac{5}{4}\sqrt {\dfrac{g}{R}} $
So the correct option is A.
Note: It should be remembered that earth is also tilted at its own axis and due to this climate changes on earth. Also it should be noted here that with some angular velocity we can feel weightlessness or in other words we can eliminate gravity, so gravity is not a force it’s merely due to space time curvature.
Recently Updated Pages
Write structures of the following compounds i 2 Chloro3methylpentane class 11 chemistry CBSE

What is BLO What is the full form of BLO class 8 social science CBSE

Explain the Treaty of Vienna of 1815 class 10 social science CBSE

A Paragraph on Pollution in about 100-150 Words

XIX+XXX A 49 B 51 C 55 D 44 class 5 maths CBSE

If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

Bond order ofO2 O2+ O2 and O22 is in order A O2 langle class 11 chemistry CBSE

Draw a labelled diagram of the neuron and describe class 11 biology CBSE

