Trying to wrap my head around this bizarre physics question

[phattestlewt]

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When a person stands on the surface of the earth, where do they appear to weight the most? (Assume the earth is a perfect sphere, i.e. ignore the mountains)

A. The Equator
B. The North Pole
C. Midway between the Equator and The North Pole
D. The person's weight will be the same at any point on the earth's surface







So I answered A. The Equator which wrong, acc to the soln manual. The correct answer is B. The North Pole and reasoning given is that when a person is at the equator he/she is in uniform circular motion because the earth rotates. This motion creates a force that cancels out some of the impact of gravity. When a person is standing on the Norht Pole, the are not in uniform circular motion, so the force the feel is purely gravitational.

I guess if I think about this intuitively it makes sense but when I draw a FBD (which I think im doing wrong, so please help!) it definitely does not make this any clearer to solve.

If the persons weight (mg) point to the earth and centripetal force (mv^2/r) point towards the center of earth's rotational axis, they make both the weight and centripetal force pointing in the same direction. So shouldn't they *increase* the net force acting on that person?

Please help me understand this!

Thanks

 

[phattestlewt]

Similarly, here's a related question I had on the side just about centripetal force and Force Body diagrams.

Suppose you want to find the speed of a roller coaster at the top of a loop, when the roller coaster has just enough KE to make it around.

So in a FBD depicting a roller coaster cart at the top of the loop:
gravity (mg) points down
Normal Force points down
and since the roller coaster is centripetally accelerating, shouldn't the the Centripetal force also point down?

 

[chill3]

If the persons weight (mg) point to the earth and centripetal force (mv^2/r) point towards the center of earth's rotational axis, they make both the weight and centripetal force pointing in the same direction. So shouldn't they *increase* the net force acting on that person?

Please help me understand this!

Thanks

So I think that you may be getting tripped up on the definition of centripetal force (Fc). Fc is not a force that you draw in your diagrams explicitly as its own force, rather it represents the sum of all forces towards the center of the circular motion. In the case of someone standing on Earth, that would be the gravitational force of attraction.

 

[Blueprint MCAT Tutor]

You're better off thinking about inertia with that weight question. Setting aside revolution around the sun, movement of the whole solar system, etc. the person standing at the equator has an inertia that wants to "make" her fly off the earth at a tangent to the equator. The person at the North Pole has no such inertia. I'm also pretty sure that the Earth's radius at the equator is larger - perhaps that would also decrease gravitational force.

It's a pretty funky question, and a good example of thinking conceptually rather than purely based on equations. Lots of MCAT students hit trouble when they just sort of mechanistically plug-and-chug an equation without stopping to actually picture what's going on.

I'll admit that I'm nowhere near 100% sure on that - if you asked me to explain it with vectors and equations and stuff I'd be at a loss. But intuitively, we can imagine the guy at the equator zipping around in a circle with the Earth's rotation would feel that inertial "push" trying to fling him off the planet.

For the roller coaster question, that's a little more straightforward: here, the centripetal force is pointing towards the center of the circle. That's the definition of centripetal force - the force(s) pointing towards the center of the circle that makes an object move in a circular path.

When you're trying to reason your way through circular motion problems, ask yourself the question "WHY is this thing moving in a circle?! Things WANT to move in a straight line forever!" Whatever the force is that's making the thing move in a circle is the centripetal force.

Here, there's the gravitational force between the coaster car and the Earth. The car and the earth are tugging on each other according to the gravity equation. But that force of gravity isn't the force making the car move in a circle. If there were no tracks, the car would just fall towards the Earth according to the projectile motion equations.

So what's making the car move in a circle is the tracks pushing on the car. That's the normal force. So throughout the entire loop, the centripetal force is the normal force.

Hope that helps!

 

[4563728]

I am not sure if it is 100% right but this is how I got the answer and it lead me to say that weight at the north pole is greater than that of the equator:

For a FBD, I draw the earth with a person standing at the top, like the north pole. He has gravity pulling him towards the center of the Earth and a normal force pointing the opposite direction. Since he is not accelerating up or down, N=mg.

Now at the equator the Earth is spinning so we don't use 'ma' in our equation, we use (mv^2)/r for centripetal force. Again, gravity pulls the person towards earth's center with N pointing in the opposite direction. So our equation is: (mv^2 /r)= mg-N (N is the apparent weight). To rearrange, N=mg-(mv^2)/r. So at the equator the apparent weight of the person is reduced so at the north pole they weigh more.

Hope this helps!