circular motion!

[kimk2kr]

In a circular motion how do you find the apparent weight at the top and the bottom of the motion when the velocities are given?



top: N+mg=Fcent
bottom: N-mg=Fcent


is this right?

[Phy124]

If it was a roller coaster where the person was upside down in there seat at the top, then yes, as both the normal reaction force and the weight force would be acting downwards and thus the addition of the two would be equal to the centripetal force (also acting downwards).

Once again if it was a something like a roller coaster where at the bottom the person is facing upright, then yes, the normal reaction force would be upwards and the weight force would be acting downwards. Therefore the centripetal force would be given by the normal reaction force minus the weight force.

*Note:

-The apparent weight of an object is the normal force acting on it (remember to mention this in any worded questions)
-The centripetal force is given by mv²/r

[Hutchoo]

Yeah, it's right.

[max payne]

What if at the bottom of the roller coaster, the person was upside down? wod it be
Fnet= -mg-N ?

[StumbleBum]

I don't believe that is possible as there is Weight or Normal force towards the centre so there is now way there could be a net force towards the centre, as there would be no net force then the object would not stay in uniform circular motion and it would fall away.

Also for when at the top of circular motion and being right side up (so going over say a hump) then it would be:
mg-R=F_net

[My name aint bob]

Slight thread hijacking, Though its revelvent: at the top of the coaster, if the force was centripetal, wouldn't that act with gravity, rather then pushing it outwards into the track? is it better to refer to it as centrifugal force?

[StumbleBum]

Slight thread hijacking, Though its revelvent: at the top of the coaster, if the force was centripetal, wouldn't that act with gravity, rather then pushing it outwards into the track? is it better to refer to it as centrifugal force?

I'm not quite sure what your asking, but the centripetal force does act with the gravity towards the centre of circular motion when at the top of the coaster, also remember that the centripetal force does not actually apply a force to the object it is merely the sum of the other forces acting on the object (Normal and Weight force).

[My name aint bob]

Oh thanks i didn't know this That makes alot more sense now

[Phy124]

I'm not quite sure what your asking, but the centripetal force does act with the gravity towards the centre of circular motion when at the top of the coaster, also remember that the centripetal force does not actually apply a force to the object it is merely the sum of the other forces acting on the object (Normal and Weight force).

As StumbleBum noted, the centripetal force is just the net force in the radial direction (towards the centre)

As you would notice

F_net = ma

a_radial = v²/r

F_centripetal = ma_radial = mv²/r