The earth is rotating with an angular velocity of 7.3 xx 10^(-5) rad/s. What is the tangential force needed to stop the earth in one year ? Given moment of inertia of the earth about the axis of rotation = 9.3 xx 10^(37) m^(2). Radius of the earth = 6.4 xx 10^(6) m

The earth is rotating with an angular velocity of 7.3 xx 10^(-5) rad/s

1.	The earth is rotating with an angular velocity of 7.3 xx 10^(-5) rad/s. What is the tangential force needed to stop the earth in one year ? Given moment of inertia of the earth about the axis of rotation = 9.3 xx 10^(37) m^(2). Radius of the earth = 6.4 xx 10^(6) m
Answer» SOLUTION :Initial angular frequency `=omega_(0) = 7.3 xx 10^(-5)` rad/s Final angular VELOCITY `=omega =0` Time `=t = 1 "year" = 365.25 xx 24 xx 60 xx 60s` Angular acceleration `=alpha`=? `omega = omega_(0) + alpha t` `alpha = (omega - omega_(0))/t = (-7.3 xx 10^(-5))/(36.25 xx 24 xx 60 xx 60) = -2 xx 10^(-12) rad//s^(2)` Torque =`tau =laalpha = F xx R` TANGENTIAL force = F = ? Radius of the earth =`R = 6.4 xx 10^(6) m` M.I. of the earth `= I = 9.3 xx 10^(37) kg m^(2)` `F = (I alpha)/R = (9.3 xx 10^(37) xx 2 xx 10^(-12))/(6.4 xx 10^(6)) = 2.9 xx 10^(19) N`