Gravity and uniform circular motion involving a loop?

In summary, a toy car with a mass of 0.27 kg is held at rest against a 1796 N/m spring compressed by 8cm, and when released, it travels 112cm along a flat surface before reaching a 0.30m high loop. The friction coefficient between the car and the flat surface is 0.4, and 0.18 between the car and the loop, with the assumption that friction can be ignored in the loop. Using the work-energy equation, the velocity of the car at the base of the loop is calculated to be 6.19m/s. To solve for the velocity at the top of the loop, the conservation of energy can be used since gravity is a conservative
  • #1
Ascendant78
328
0

Homework Statement



A 0.27 kg toy car is held at rest against a 1796 N/m spring compressed a distance of 8cm. When released, the car travels a distance of 112cm along a flat surface before reaching a 0.30m high loop. The friction coefficient of the flat surface and the toy car is (0.4, 0.18). Assume friction is so small in the loop that it can be ignored.

Question 1: What is the velocity of the toy car the moment it reaches base of the loop?

Question 2: What is the velocity of the toy car when it reaches the top of the loop?


Homework Equations



Solving question 1
Evaluation of toy car from release to bottom of loop using work-energy:

EEi + W = 1/2 (0.27) vf2

1/2 (1796)(0.08)2 + (0.47628)(0.08 + 1.12) cos(180)= 0.135 vf2

5.7472 – 0.571536 = 0.135 vf2

vf = 6.19m/s when toy car reaches the base of the loop


The Attempt at a Solution



As seen in #2 above, I know how to solve for the first question. However, I am at a complete loss as to how we approach the second part? We have dealt with uniform circular motion with no gravity and uniform circular motion with movement perpendicular to gravity, but never one where the direction of movement is constantly changing with respect to the direction of gravity. I am at a loss as to what equation I could even use that would take everything I need into consideration?
 
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  • #2
Ascendant78 said:
Solving question 1
Evaluation of toy car from release to bottom of loop using work-energy:

EEi + W = 1/2 (0.27) vf2

1/2 (1796)(0.08)2 + (0.47628)(0.08 + 1.12) cos(180)= 0.135 vf2

5.7472 – 0.571536 = 0.135 vf2

vf = 6.19m/s when toy car reaches the base of the loop

Just a comment here: what you have written above is almost completely meaningless, and almost completely useless to someone else who is trying to figure out what you're doing. The reason is because you haven't included any units on your quantities. They are just meaningless numbers. 1796 whats? It could be kilowatt-hours, for all I know. You might think I'm nitpicking and being pedantic, but I'm actually not. Without the units, it's not a physically meaningful statement, and excluding units is a great way to fall prey to silly mistakes as well.

I'm assuming EEi is the initial "elastic energy?" If that's true, then it's equal to (1/2)kx2, which I think is the first term you have on the lefthand side of the next line (but I had to guess that because there are no units). That term looks okay. Let's talk about the middle term, which I've put in boldface. This term corresponds to what you called "W", by which I assume you mean the work done by friction. You're saying that initial elastic energy + frictional work = final kinetic energy, so that all makes sense by the conservation of energy. But the middle term (the work) has nothing to indicate where any of the numbers came from. The number in red, 0.47628, in particular, has no discernible origin. It could be correct, or it could not be, but I have no way of telling either way, because you didn't write down any algebraic symbols or any of the steps that you used to arrive at it.


Ascendant78 said:
As seen in #2 above, I know how to solve for the first question. However, I am at a complete loss as to how we approach the second part? We have dealt with uniform circular motion with no gravity and uniform circular motion with movement perpendicular to gravity, but never one where the direction of movement is constantly changing with respect to the direction of gravity. I am at a loss as to what equation I could even use that would take everything I need into consideration?

Just use the conservation of energy. Gravity is a conservative force: the great thing about conservative forces is that the work they do in moving an object from point a to point b depends only on the difference between those two points. It does not depend on the path taken between those two points. So it doesn't matter that the object travels along a circular loop. To figure out its change in potential energy, all you need to do is figure out the change in height of the object. Potential energy gained tells you kinetic energy lost.

EDIT: this is why conservation of energy is so powerful a concept. It's incredibly useful.
 
  • #3
cepheid said:
Just a comment here: what you have written above is almost completely meaningless, and almost completely useless to someone else who is trying to figure out what you're doing. The reason is because you haven't included any units on your quantities. They are just meaningless numbers. 1796 whats? It could be kilowatt-hours, for all I know. You might think I'm nitpicking and being pedantic, but I'm actually not. Without the units, it's not a physically meaningful statement, and excluding units is a great way to fall prey to silly mistakes as well.

I'm assuming EEi is the initial "elastic energy?" If that's true, then it's equal to (1/2)kx2, which I think is the first term you have on the lefthand side of the next line (but I had to guess that because there are no units). That term looks okay. Let's talk about the middle term, which I've put in boldface. This term corresponds to what you called "W", by which I assume you mean the work done by friction. You're saying that initial elastic energy + frictional work = final kinetic energy, so that all makes sense by the conservation of energy. But the middle term (the work) has nothing to indicate where any of the numbers came from. The number in red, 0.47628, in particular, has no discernible origin. It could be correct, or it could not be, but I have no way of telling either way, because you didn't write down any algebraic symbols or any of the steps that you used to arrive at it.




Just use the conservation of energy. Gravity is a conservative force: the great thing about conservative forces is that the work they do in moving an object from point a to point b depends only on the difference between those two points. It does not depend on the path taken between those two points. So it doesn't matter that the object travels along a circular loop. To figure out its change in potential energy, all you need to do is figure out the change in height of the object. Potential energy gained tells you kinetic energy lost.

EDIT: this is why conservation of energy is so powerful a concept. It's incredibly useful.

Apologies for not including the units and simplifying it all. I didn't want to make a clutter with all the units and figured the velocity I obtained could be checked relatively quickly. Besides, I wasn't really concerned with question 1 anyway, just question 2.

Thanks for the information about using conservation of energy. The only reason I wasn't sure about using that one is because of the force caused by the centripetal when it goes through the circular motion. I figured if it wasn't somehow included in the equation, it would come out wrong. I guess I'm just looking at it wrong conceptually.
 
  • #4
Still looking for help on this. Can anyone clarify how you can use conservation of energy on this problem when the centripetal force will affect how high the toy car can travel?
 
  • #5
Ascendant78 said:
Still looking for help on this. Can anyone clarify how you can use conservation of energy on this problem when the centripetal force will affect how high the toy car can travel?

Hi Ascendant78,

I replied to your PM regarding this question. Sorry for not getting back to you sooner. For the benefit of other thread goers, the gist of what I said was:

Only two forces act on the object: gravity and the normal force. The normal force does no work.
 

1. What is gravity?

Gravity is a natural phenomenon by which all objects with mass are brought towards each other. It is the force that keeps planets in orbit around the sun and causes objects to fall towards the ground.

2. How does gravity affect objects in uniform circular motion?

In uniform circular motion, an object is constantly changing its direction, but its speed remains constant. Gravity acts as the centripetal force, pulling the object towards the center of the circle and keeping it in a circular path.

3. How does gravity affect an object in a loop?

In a loop, gravity pulls the object towards the center of the loop, providing the centripetal force needed to keep the object moving in a circular path. However, if the object's speed is too slow, it may not have enough centripetal force to complete the loop, causing it to fall off.

4. What factors affect the force of gravity in a loop?

The force of gravity in a loop is affected by the mass of the object, the radius of the loop, and the speed of the object. The greater the mass and speed, and the smaller the radius, the greater the force of gravity and thus the tighter the loop.

5. How does the force of gravity change as an object moves through a loop?

As an object moves through a loop, the force of gravity changes direction and magnitude. At the bottom of the loop, the force of gravity is at its maximum, providing the centripetal force needed to keep the object moving in a circular path. As the object moves up the loop, the force of gravity decreases and may even become zero at the top of the loop if the speed is sufficient to keep the object in motion.

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