Why Does Apparent Weight Change on a Ferris Wheel?

[mvpshaq32]

Homework Statement

Explain the shift in apparent weight at the top and bottom of a ferris wheel.

Homework Equations

The Attempt at a Solution

I am just not able to grasp centripetal acceleration. I understand at the top point, gravity AND centripetal acceleration are downwards, so then why in the equation for Normal force N=m(g - v^2/r) imply that these forces are acting in opposite directions? And if they both point downwards, I would assume normal force would also be greater at the top point than at the bottom point. Can someone explain this?
Also, in my textbook, it says that if the ride goes fast enough such that g=v^2/r, then N=0 and the passenger is about to be airborne. Why would this happen if you eliminate the upward force and have only the downward forces? Is this not counter intuitive?

 

[Doc Al]

I am just not able to grasp centripetal acceleration. I understand at the top point, gravity AND centripetal acceleration are downwards, so then why in the equation for Normal force N=m(g - v^2/r) imply that these forces are acting in opposite directions?

What forces? The only forces involved are the normal force (which is the apparent weight) and gravity. And they do point in opposite directions. v^2/r is the acceleration, not a separate force.

As always, just use Newton's 2nd law:
ΣF = ma
N - mg = -mv^2/r

 

[jambaugh]

Centripetal acceleration is the acceleration which keeps an object in circular motion.

Since the rider is moving in a circle, the net forces on him must add up to cause only this centripetal acceleration. Add up the forces acting on the rider and see how they must result in this one force toward the hub of the wheel.

Consider at any instant what would happen to the rider if the wheel blinked out of existence.

 

[cupid.callin]

keep in mind that centripatal force is not any new force .. its just a force (of Columbian or gravitational origin) which allows something to move in circle.

At the top point what is the force (in terms of gravity and normal rxn) providing centripetal force?

 

[ashishsinghal]

actually centripetal force is not a force itself. the forces acting on the particle give it the centripetal acceleration in order to make it move in circular path. so here the net downward force is :
mg - N
this is equal to to mv^2/r
from here you get : mg - mv^2/r = N
hope this helps...

 

[jambaugh]

actually centripetal force is not a force itself. [...]

Hua? Don't confuse the centripetal force which is a physical force with the centrifugal force which is an apparent force when we adopt a rotating frame of reference.

 

[ashishsinghal]

i meant that there is no additional force such as centripetal force. it is provided by other forces acting radially.
for eg. tension provides centripetal force to the pendulum.
i am not being confused...maybe the language suggested otherwise to you

 

[jambaugh]

i meant that there is no additional force such as centripetal force. it is provided by other forces acting radially.
for eg. tension provides centripetal force to the pendulum.
i am not being confused...maybe the language suggested otherwise to you

My apologies. Now I see what you meant.