Recent content by pandafish

Yes, that is what I originally assumed the problem to be asking. Does this mean when the centripetal force on the point mass balances out its weight force, the coin would no longer be rolling?

Oh, I think I understand now. Thank you

When the mass is furthest from the table, the kinetic energy is lowest, and when the mass is closest to the table, the kinetic energy is greatest?

Yes, gravitational force on the mass creates a torque that increases and decreases the rotational velocity as the coin rolls.

The rotational velocity decreases until it is 0, where the point mass is at the highest point.

I reasoned that at the coin's slowest velocity, the energy it has must just be enough for it to reach the highest potential configuration: when the point mass is directly above the centre of mass of the coin, and its GPE is ##mg(R+r)##. I used this to find the minimum velocity, but I can't think...

Thank you, that explanation makes it a lot clearer what the apparent force is.

I see what you mean. Working backwards from the solution, I realised that the 'apparent gravitational acceleration' was made up from the normal force , and the component of the centrifugal force tangential to Earth's surface. However, I'm don't quite understand why this is the case.

The right side is the centripetal acceleration, $$a = \frac{v^2}{r} = \omega^2r$$ I believe r is distance from the mass to the earth's axis of rotation, so $$a = \omega^2(Rcos\lambda)$$ I made a mistake by including mass on the right.

My bad, its suppose to be: $$\\(\frac{GM}{R^2}-N)cos\lambda+Fsin\lambda=m\omega^2Rcos\lambda$$ This force balance is suppose to be perpendicular to the axis of Earth's rotation.

I began by drawing a diagram and resolving the forces. Since the question asked for 'apparent gravity' I tried to find the normal force. I started with the equations: $$\\(\frac{GM}{R^2}-N)sin\lambda-Fsin\lambda=m\omega^2Rcos\lambda$$ $$\\(\frac{GM}{R^2}-N)sin\lambda-Fcos\lambda=0$$ Solving...