Vertical Circular Motion

In summary, the conversation discusses the concept of centripetal force and its role in keeping a roller-coaster train from falling out of a circular loop. The minimum velocity required for the train to remain at the top of the loop is calculated using the formula v = √(rg), where r is the radius and g is the gravitational acceleration. The conversation also touches on the relationship between velocity and normal reaction force, and how external factors can affect the theoretical calculations. Finally, an example problem is discussed and solved using the formula v = √(rg).
  • #1
Delzac
389
0
Hi all,

A roller-coaster train is upside down when it is at the top of a circular vertical structure which is 20 m high. In order to get to the top without falling out, the velocity of the roller-coaster train at the top of the circle has to be more than
a 25 m s-1
b 14 m s-1
c 9.9 m s-1
d zero

Ans : C

The question didn't give me any mass of anything at all to work with except for height. Is mass not required for this question? i couldn't get C, in fact i chose D, zero instead.

Any help will be greatly appreciated. ( probably need clarification of concepts.)
 
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  • #2
Think about centripetal force. For anybody undergoing circular motion the sum of all the forces acting must be equal to the centripetal force. In this case there are only two forces acting (ignoring friction); the normal reaction force R and gravity mg, therefore we can write;

[tex]\frac{mv^2}{r} = R + mg[/tex]

Do you follow? Now, what will the reaction force if the coaster only just makes the loop?
 
  • #3
R = 0 i guess?

For R to be non-zero the roller coaster must have some velocity so as to be able to "press" on to the track is that right?
 
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  • #4
Delzac said:
R = 0 i guess?

For R to be non-zero the roller coaster must have some velocity so as to be able to "press" on to the track is that right?
This is correct. So can you solve the problem from here?
 
  • #5
so,

[tex] V = \sqrt{rg}

=\sqrt{(10)(9.81)}= 9.9[/tex]

But i have a qns, if there is velocity by the time i reached the top, wouldn't there be normal reaction force already? then R will be non-zero.
 
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  • #6
Delzac said:
so,

[tex] V = \sqrt{rg}

=\sqrt{(10)(9.81)}= 9.9[/tex]

But i have a qns, if there is velocity by the time i reached the top, wouldn't there be normal reaction force already? then R will be non-zero.
Sorry, for the delay in replying but I have been increasingly busy of late. Why does velocity imply a normal reaction force? Do not forget that the reaction force (if any) is not the only force that is acting, HINT: Gravity.
 
  • #7
I have a question: If R were to drop to zero, the coaster would fall, right? So, for v=9.9 m/s, it would actually lose contact with the track and fall. But for any value even minutely larger than R=0 (and thus v=9.9 m/s) the coaster would just barely "stick" to the track? Even for a super small value of R?
 
  • #8
(mv2/r) = T top + mg

The mass from the (mv2)/r and mg cancel each other out thus giving

v2/r = T top + g

where g = 10 or 9.81 whichever you are using.
For the coaster to remain in contact with the tracks, the force T top has to be greater then zero, so that the resultant force (mv2/r) is greater than the weight of the coaster pulling it downwards.

From there on it is relatively simple. Your problem, I think, was more of mathematics than physics.

r is the radius, and since it is at a height of 20m this is the diameter.

To continue your problem :

v2/10 = 0 + 9.81
v2 = 98.1
v = square root of 98.1
v = 9.9m/s

Sorry for the format, but I don't really know how to use this thing well. I hope that you understand that v2 means v squared etc
 
  • #9
Exactly yes, in theory that is what should happen, so that the centrepetal force forcing the coaster outwards are greater than the forces pulling it down. Remember though that in reality there are other external factors which may change this.

I think that even if R = 0, the coaster should not fall, because this would just mean that the forces pushing it up are equal to the forces pulling it down. I am not sure of this though
 
  • #10
Hi,
How
 
  • #11
Hi,
How con I find the velocity of an object at the top of the loop, if it has a radius of 0.4m, a mass of 0.025kg, and an energy value of 9J? This was an exam question and I couldn't work it out. The lecturer has now given it to us for homework and I still cannot work it out. I hace become an expert in these problems trying to solve this, but perhaps there is something I am not seeing. Thanks
 
  • #12
Hey Mate

This question is relatively easy, and to solve this you don't need mass. Take into account this formula v=√(rg), this is the critical speed or the minimum speed allowed to prevent the passenger from falling over at the top. Remember this is not the actual speed that the rollercaster is travelling. The 'r' stands for radius, and the 'g' stands for gravitaional acceleration (9.8). The questions states that the height is 20m, therefore the radius is 10m. Substitute 10m to r and 9.8 to g (g is constant). You will get v as 9.99ms^-1. Thus this is the critical speed. Therefore the speed of the rollercosater has to greater than 9.99ms^-1. Hope this helps :)
 

1. What is vertical circular motion?

Vertical circular motion is a type of motion in which an object moves in a circular path while also experiencing the force of gravity pulling it downward. This type of motion can be observed in objects such as roller coasters and Ferris wheels.

2. What is the centripetal force in vertical circular motion?

The centripetal force in vertical circular motion is the force that acts towards the center of the circular path, keeping the object moving in a circular motion. In the case of vertical circular motion, this force is provided by the tension in the string or track that is holding the object in place.

3. How is the speed in vertical circular motion related to the radius of the circle?

The speed of an object in vertical circular motion is directly proportional to the radius of the circle. This means that as the radius increases, the speed also increases, and as the radius decreases, the speed decreases. This relationship is described by the equation v = √(g*r), where v is the speed, g is the acceleration due to gravity, and r is the radius of the circle.

4. What is the difference between uniform circular motion and non-uniform circular motion?

Uniform circular motion is when an object moves in a circular path at a constant speed. Non-uniform circular motion is when an object moves in a circular path at a changing speed. In vertical circular motion, the speed is usually not constant, as the object experiences changes in speed due to the force of gravity.

5. How does vertical circular motion affect the weight of an object?

In vertical circular motion, the weight of an object can change as it moves along the circular path. As the object moves upwards, it experiences a decrease in weight due to the decrease in the force of gravity. As it moves downwards, it experiences an increase in weight due to the increase in the force of gravity. At the top of the circle, the weight may be equal to zero, while at the bottom of the circle, the weight may be double the normal weight of the object.

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