Loop de Loop Problem: Equal Forces?

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In summary, at the top of the loop, the centripetal force must be at least equal to the force of gravity, so that the car not fall (excluding any friction). However, at the bottom of the loop things are different, because the centripetal force must act upward while gravity acts downward.
  • #1
the_obs
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Just a quick question, concerning the (probably common) loop de loop problem.
At the top, why can we say that the centripetal force has to be at least equal to the force of gravity, so that the car not fall (excluding any friction)?

Both forces point towards the bottom (or the centre of the circle, more exactly), so shouldn't the centripetal force be opposite that of gravity?
 
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  • #2
Note that the term "centripetal force" refers to the net force acting towards the center. The actual forces that comprise the centripetal force are gravity and the normal force. Since both of those forces act toward the center, the centripetal force must be at least equal to gravity.

At the bottom of the loop things are different, because the centripetal force must act upward while gravity acts downward.

Make sense?
 
  • #3
Ohhhh right, I get it.

Thank you very much! I didn't really grasp the concept of centripetal force as a net force, which actually makes sense, seeing as it's calculated by sum of forces =ma, a = v^2 /r, thus sum of forces = mv^2/r...

Thanks!
 
  • #4
Hi the_obs! :smile:
the_obs said:
I didn't really grasp the concept of centripetal force as a net force …

Maybe I'm a bit obsessive about this :rolleyes:

but I think calling it centripetal force is bad and confusing …

there are only two forces on the car, the gravitational force and the normal force …

by good ol' Newton's second law, their sum equals the mass times the centripetal acceleration

there is no centripetal force. :wink:
 
  • #5
I agree with the overall point, tiny-tim, but to say that there's no centripetal force is a bit extreme. :-p That's why I refer to centripetal force as a net force, not as an actual individual force that would appear on a free body diagram. But good point!
 
  • #6
My physics teacher always say, never include centripetal force into free-body diagram, cause its the NET force. It only exists when a some force exerted on a body PROVIDES for the REQUIREMENT of centripetal force.

Take Earth orbiting the sun, the gravitational force exerted on Earth by sun PROVIDES for the centripetal force.
 
  • #7
In order for the car to stay on the track at the top of the loop, the minimum normal force is zero. The car is in free fall during the instant it is at the top of the loop. There is no force between the track of the loop and the tires of the car (or vice versa). In this special case, the only force at the top of the loop is gravity, and being the only force, centripetal force = gravitational force.

If the car were moving slower it would fall. If the car were moving faster, then there would a be a normal force exerted by the track onto the car, coexisting with an equal and opposing force exerted by the car onto the track at all times.

The force the car exerts onto the track is the sum of the gravitational and reactive (to centripetal acceleration) forces.
 

Related to Loop de Loop Problem: Equal Forces?

1. What is the Loop de Loop problem in physics?

The Loop de Loop problem, also known as the equal forces problem, is a physics puzzle where an object is placed on a circular track and must complete a full loop without falling off. The challenge is to determine the minimum speed needed for the object to successfully complete the loop without the force of gravity pulling it down.

2. What are the forces acting on an object in the Loop de Loop problem?

The two main forces acting on the object are the centripetal force, which pulls the object towards the center of the circular track, and the force of gravity, which pulls the object downwards. In order for the object to successfully complete the loop, these two forces must be equal in magnitude.

3. How do you calculate the minimum speed for an object to complete the loop?

The minimum speed can be calculated using the equation v = √(rg), where v is the minimum speed, r is the radius of the circular track, and g is the acceleration due to gravity. This equation is derived from equating the centripetal force and the force of gravity.

4. How does the radius of the track affect the minimum speed for the Loop de Loop problem?

The radius of the track directly affects the minimum speed needed to complete the loop. As the radius increases, the minimum speed decreases. This is because a larger radius means a larger distance for the object to travel, which requires less speed to maintain the same centripetal force.

5. What other factors can affect the success of an object completing the Loop de Loop?

In addition to the minimum speed and radius, other factors that can affect the success of an object completing the Loop de Loop include the mass of the object, the friction between the object and the track, and the shape of the track. A heavier object will require a greater speed to overcome the force of gravity, and a smoother track with less friction will make it easier for the object to maintain its speed.

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