Circular motion, forces, magic mountain ride.

JoshuaR

Hi there. On a midterm we just had on Monday, one question asked about a magic mountain ride called the rainbow. You get in this circular apparatus and it starts to spin, you're against the wall. Then it goes from horizontal to vertical and you stick to the walls. Anyways, one question asked that I pick a coordinate system and then set up Newton's 2nd Law for the person at the top. I chose x to the right and y going down. Then I said F=ma=mg+Fn (normal force from wall down on person to keep them in the centripetal motion). Now I didn't get any points marked down but I'm still unsure of the direction of forces. I said Fn=mv^2/r-mg. Is that right? Is it minus mg, or + mg? What signs do I use? I ride stays at constant speed. I figured the normal without gravity would just be mv^2/r. When I add gravity I figure it'll allow the person not to push on the wall as much so I said mv^2/r-mg. Is that right?

Part two. I just did this for my own sake. At the bottom of the ride, I wrote in ma=mg-Fn where Fn=mg+mv^2/r. The professor wrote in a minus sign in front of mg. What happened here? I figured I'd add the weight this time instead of subtract it. Maybe I did it wrong in the first part?

Thanks, Josh

Additional question. I'm holding a pencil in my hand and turning it in a circle. I feel it weighs more on the way up, but once I reach the top it almost flies out of my hand. Likewise it weighs less on the way down. Now in my head I can't figure if it weighs more at the top or bottom. At the top, is it flying out of my hand, or is it just about to stop feeling heavy and fly out of my hand? The bottom? These forces are confusing me.

Doc Al

All good. You are finding the net force at the top of the motion. Using down as positive, you would get: net force = mg + Fn, since both weight and normal force act down. Since the acceleration is centripetal (and downward), the net force must equal mv^2/r, thus mv^2/r = mg + Fn, and thus Fn = mv^2/r - mg (as you stated).

Perhaps your professor didn't understand the sign convention you were using or just made an error. In any case, your answer is correct.

Doc Al

I'm not sure what you are describing here. What do you mean by "turning it in a circle"? Are you rotating your arm about your shoulder while holding a pencil? Realize that what you feel is the reaction force that the pencil exerts on you in response to your force on the pencil. (And that if you swing the pencil in a circle, you are providing the centripetal force.) The actual weight of the pencil (weight = the earth's gravitational pull on the pencil) doesn't change.

JoshuaR

Sorry, I meant that I open my palm flat and place a pencil in it. I then mimic a circular motion in the vertical direction, like the ride I was describing. As my hand is going upward, I feel the force of the pencil more. I meant its apparent weight was larger. And same going down. But I'm not sure if at the very top its apparent weight is larger or smaller. Now the force should point straight down, and there's no force above the pencil to hold it so inertia allows it to jump from my hand, thus lowering the apparent weight?
And opposite at bottom of circle, it's apparent weight is higher? As if you were on a fast moving ride where the floor is beneath you. Maybe a very quick ferris wheel. At the top, standing on a scale, will you feel heavier or lighter? And the bottom? I want to say you feel lighter at the top.

JoshuaR

Ah. My hand is moving in a circle. The net force for the pencil is mv^2/r. This is at the top equal to mg-Fn (taking down to be positive). Fn=mg-mv^2/r.
At the bottom, -mv^2/r=mg-Fn. Fn=mg+mv^2/r. So apparent weight is heaviest at bottom of my circle. I think.