I have a question about a concept

[kevinf]

i've heard that in a loop the loop environment when the object goes to the top (upside down), gravitational force (mg) will equal to centripetal force(mv^2/r) can anyone explain why this is.

 

[Chi Meson]

This is the specific situation of the car going at the slowest possible speed in order to "loop-the-loop" and not fall from away from the track at the top.

When going faster than the slowest speed, and when the car is at the very top of the loop, there are two forces on the car: the gravitational force from the earth, and the "normal" support force from the track. When the car is at the top, both these forces are pointing straight down, and consequently both forces point toward the center of the circle of the loop. Therefore, the two forces together provide the centripetal force for continuing in a circular path.

When the car slows down, less centripetal force is needed, and so the normal force decreases (while weight remains the same). At the slowest speed, the normal force has decreased all it can (zero) and then only gravity supplies a force that is centripetal.

If the car tries to "loop it" while going even slower, it needs even less centripetal force, but weight can't decrease. So there will be too much centripetal force, and the car will be pulled out of the circular path, and instead it will take a parabolic, projectile motion plunge to the bottom of the loop.

 

[kevinf]

yeah but why does centripetal force equal the gravitational force. i heard that its because at the top the normal and the gravitational force is pointing down. is that right?

 

[Grufey]

If the particle is in a loop, then, there are two forces: the gravity force and the centripetal force. The total force in any point of the loop, wil be, the sum of the gravity force and the centripetal force in the point. If you have the object in the top point, then the sum tht force gravity and the centripetal force is cero, then the gravity force is equal to the centripetal force, but the diference is the sign.

$$mg\bar{u_g}+m\frac{v^2}{r}\bar{u_r}=0$$ This is the condition in the maximun
$$mg\bar{u_g}=-m\frac{v^2}{r}\bar{u_r}$$ Look at the sign

 

[Chi Meson]

If the particle is in a loop, then, there are two forces: the gravity force and the centripetal force. The total force in any point of the loop, wil be, the sum of the gravity force and the centripetal force in the point. If you have the object in the top point, then the sum tht force gravity and the centripetal force is cero, then the gravity force is equal to the centripetal force, but the diference is the sign.

$$mg\bar{u_g}+m\frac{v^2}{r}\bar{u_r}=0$$ This is the condition in the maximun
$$mg\bar{u_g}=-m\frac{v^2}{r}\bar{u_r}$$ Look at the sign

No, not quite. That is the explanation from the "rotational frame of reference," where you should be saying "centrifugal."

From the non-rotational point of view (that is the point of view without a "centrifugal force'), the sum of all forces on an object in circular motion is the centripetal force. That is, in uniform circular motion, the net force must be centripetal.

"Centripetal" only designates the direction of a force as "pointing toward center." It is not a force in itself (such as "centrifugal" claims to be). In all cases of circular motion, the centripetal force must be supplied by some recognizable force (gravity, tension, friction, aerodynamic lift, surface 'normal' force, etc.). If any force is directed toward the center of a circle, it is "centripetal."

As stated in my first response, for the car in a loop-the-loop, at the top, the two forces on the car are gravity and the surface "normal" force from the track. Together they add up to be the net force on the car, and hence the total "centripetal" force (assuming constant speed which is not a realistic assumption).

Re read my first response at this time.

 

[Grufey]

Ok I agree, thanks