what sould be the angular speed with which the earth have to rotate on its axis so that a person on the equator would weight 3/5th as much as present..??
Solve this equation for Earth angular speed in radians per second:
Mg - M Rw^2 = 0.6 M g
Cancel out the M's
0.4 g = R w^2
w = sqrt(0.4g/R)
Use R = 6371*10^3 m
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To calculate the angular speed at which the Earth should rotate on its axis so that a person on the equator would weigh 3/5th as much as their present weight, we need to consider the relationship between weight and the centrifugal force.
The centrifugal force is the force experienced by an object in a rotating reference frame, such as a person on the rotating Earth. It acts in the opposite direction to the acceleration towards the center of rotation, which is the force of gravity.
Let's assume the current weight of a person on the equator is W. So, if the person's weight decreases to 3/5th of W, the remaining weight is (3/5)W.
The centrifugal force is given by the formula F = m * ω^2 * r, where F is the centrifugal force, m is the mass of the object, ω (omega) is the angular speed in radians per second, and r is the radius from the center of rotation.
At the equator, r is the radius of the Earth, denoted by R. The person's mass, m, cancels out in this equation. However, the gravitational force on the person is mg, where g is the acceleration due to gravity. Since weight is equal to mg, we have weight = mg.
To solve for the angular speed ω, we equate the gravitational force to the centrifugal force:
mg = (3/5)W = m * ω^2 * R
We can cancel out the mass (m) from both sides of the equation:
g = (3/5)ω^2 * R
Next, we isolate ω by dividing both sides of the equation by (3/5)R:
g / ((3/5)R) = ω^2
Finally, taking the square root of both sides gives us the angular speed:
ω = sqrt(g / ((3/5)R))
Note: Remember to use consistent units for the gravitational constant (g) and the radius of the Earth (R) in order to obtain accurate results.
By substituting the values for g (approximately 9.8 m/s^2) and R (approximately 6,371 km), you can calculate the angular speed at which the Earth should rotate on its axis so that a person on the equator would weigh 3/5th as much as their present weight.