- #1
binbagsss
- 1,259
- 11
Q1) What angular velocity would be required for your effective weight at the equator to be third of that the north pole? You may neglect all inertial forces except the centrifugal.
I understnad the majority of my books solution which is that:
At the equator, the centrifugal force is -mRw^2k, gravitational is mgk, where k is the unit vector pointing perpendicalry out of the Earth's surface.
At the north pole there is no centrifugal force, so we want to solve:
g/3=g-Rw^2**
I am struggling to see how we have reached this equation, sign-wise.
The main equation of this topiic is ma[rotating frame]=ma[inertial frame]-inertial forces *
where here inertial forces is only the centrifugal .
So here, as other inertial forces have been neglected, I think we take the north pole as the interial frame and the equator as the rotating frame?
So from the question we want to ar , where ar is the accelertatin in the rotating frame, to be -g/3k on the LHS of * . And on the RHS we have aI= -gk, and Fcentrifugal
= -Rw^2k, which gives **.
But, (assuming the above procedure is correct), I don't understand the sign choice of the centrifugal force. I thought it always points radially outward, so it would be +Rw^2k?
Q2) The direction of the centrifugal force points directly away from the axis of rotation... I can clearly see that it must be perpendicular to both w and wxr, (by F=-mwx(wxr)but I am struggling to see how this implies it must point away from the axis.
Many Thanks to anyone who can shed some light on any of this. Greatly appreciated :) !
I understnad the majority of my books solution which is that:
At the equator, the centrifugal force is -mRw^2k, gravitational is mgk, where k is the unit vector pointing perpendicalry out of the Earth's surface.
At the north pole there is no centrifugal force, so we want to solve:
g/3=g-Rw^2**
I am struggling to see how we have reached this equation, sign-wise.
The main equation of this topiic is ma[rotating frame]=ma[inertial frame]-inertial forces *
where here inertial forces is only the centrifugal .
So here, as other inertial forces have been neglected, I think we take the north pole as the interial frame and the equator as the rotating frame?
So from the question we want to ar , where ar is the accelertatin in the rotating frame, to be -g/3k on the LHS of * . And on the RHS we have aI= -gk, and Fcentrifugal
= -Rw^2k, which gives **.
But, (assuming the above procedure is correct), I don't understand the sign choice of the centrifugal force. I thought it always points radially outward, so it would be +Rw^2k?
Q2) The direction of the centrifugal force points directly away from the axis of rotation... I can clearly see that it must be perpendicular to both w and wxr, (by F=-mwx(wxr)but I am struggling to see how this implies it must point away from the axis.
Many Thanks to anyone who can shed some light on any of this. Greatly appreciated :) !
Last edited: