Question

The angular velocity of earth so that a body at equator experiences weightlessness is omega not.Then omega is is

Answer 1

Answer:

We know that acceleration due to gravity at equator is,

ge =g−Rω^2

where, ω=0

is the angular velocity of the earth.

Angular velocity of Earth= 2π/T = 2π/86400

If the acceleration due to gravity at the equator becomes zero, the body at equator experiences weightlessness.

Let ω be the angular velocity of rotation at which the acceleration due to gravity at the equator becomes zero.

That is, ge =g−Rω^2

=0

⇒g=Rω^2

⇒ω= root g/R

Now, ω/ω0 = 1/ω0 root g/R

= 86400/2π root 9.8/6.4 × 10^6

=17

⇒ω=17ω0

Thus, the earth should rotate 17 times faster than the present rate.

Rotation speed increases by a factor of 17.

∴ the duration of day decreases by factor 17.

New duration of day, T=24/17

=1.412hr is the answer.

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Answer 2

Your question is wrong dear. The correct question I write here please look carefully.

Question:

The angular velocity of the earth on rad/s, so that bodies on equator may appear weightless. Then find the angular velocity?

Answer:

${1.25 \times10}^{-3}rad/s$

Explanation:

Let the weight of the apparent body on equator is:

${W' = W - mR}{\omega}^{2}$

As we know that,

$\boxed{W = m \times g}$

$=>{mg' = mg - mR}{\omega}^{2}$

$=>{mg' = mg - mR}{\omega}^{2}$

Cancel the mass "m" on both the sides.

We get,

$=>{g' = g - R}{\omega}^{2}$

As,

$=> {g - R}{\omega}^{2}=0$

Now, "g" goes to left hand side

$=>{R}{\omega}^{2}=g$

Now, "R" goes to right hand side

$=>{\omega}^{2}=\frac{g}{R}$

$=>{\omega}=\sqrt{\frac{g}{R}}$

Now, it is given that:

Acceleration due to gravity, g = 10 m/s²

Radius of the earth = 6400 km = 6400000 m

So, by using the formula:

$=>{{\omega}=\sqrt{\frac{g}{R}}}$

Putting the values here:

$=>{\omega}=\sqrt{\frac{10}{6.4 \times {10}^{6}}}$

$=>{\omega}=\sqrt{\frac{5}{3.2 \times {10}^{6}}}$

$=>{\omega}=\sqrt{\frac{5}{32 \times {10}^{5}}}$

$=>{\omega}=\sqrt{\frac{1}{32 \times 20000}}$

$=>{\omega}=\sqrt{\frac{1}{640000}}$

$=>{{\omega}=\frac{1}{800}}$

${\boxed{\blue{=>{\omega}={1.25 \times 10}^{-3}rad/s}}}$

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