A body is weighed by a spring balance to be 1.000kg at the north pole.How much will it weigh at the equator?Account for earth's rotation only.Ans=0.997Kg Q2=A pendulum having a bob of mass m is hanging in a ship sailing along the equator from east to west .When the ship is stationary with respect to water the tension in the String To 1=Find the difference b/w To and the earth's attarction on the bob.2=If the ship sails at a speed v,what is the tension in the String?Radius of the earth is R and teh angular speed of it is omega Ans1=m(omega)2 R Ans 2=To +2(omega)v
Answers
1) The tension in the string when the ship is at rest w.r.t water is To
The ship is rotation with the earth. So, at the equator the net force towards the center of the earth is, F = mg – mw2R= tension on the string.
Thus, To = mg – mw2R
==> To – mg = -mw squareR
Thus, the magnitude of the required difference is, mw square R
2) The speed of rotation of the earth at the equator is, u = (2 πR/T) = wR
As the ship moves from east to west that is opposite to earth's rotation, its net rotational speed is equal to (u-v)
And the angular speed of rotation is ws = ( u-v)/R
Thus, the net downward force is, Fs = mg – mws square R = mg – m[(u-v)/R] square R
===> Fs = mg – m (u square + v square – 2uv ) / R square ] R
= mg –m [(u/R) square + (v/R) square – 2 (u/R) (v/R)] R
= mg – m [w square + 0 – 2 w (v/R)] R
[(v/R) square is very small and is neglected]
= mg – mw square R+2mwv
= To + 2mwv
This is the new tension on the string.