Riding a Loop the Loop (velocity at the bottom of the loop)

In summary, the problem involves finding the minimum height required for a car to complete a loop-the-loop without falling off at the top. This is determined by setting the normal force equal to zero and using the centripetal acceleration formula to solve for the minimum velocity needed. In the second part of the problem, the speed of the car at a specific point is calculated using the conservation of energy equation. The tangential acceleration is found by multiplying the acceleration of gravity by the cosine of the angle with respect to the vertical. The centripetal force is not a separate force, but rather the name given to the net force that produces the centripetal acceleration. It is important to be consistent with signs in calculations involving centripetal acceleration.
  • #1
clope023
992
131
[SOLVED] Riding a Loop the Loop (velocity at the bottom of the loop)

Homework Statement



A car in an amusement park ride rolls without friction around the track shown in the figure. It starts from rest at point A at a height h above the bottom of the loop. Treat the car as a particle.

YF-07-32.jpg


part A - What is the minimum value of h (in terms of R) such that the car moves around the loop without falling off at the top (point B)?

found this to be 5/2R, since:

sigma Fy = Fc - N - mg = 0

N = 0 and Fc = mv^2/R

so Fy = Fc = mg and mv^2/R = mg and v^2 = GR

h = (1/2mgR + 2mgR)/mg = 1/2R + 2R = 5/2R

now I'm stuck on the velocity at the point C.

part B - If the car starts at height h = 4.50 R and the radius is R = 22.0m , compute the speed of the passengers when the car is at point C, which is at the end of a horizontal diameter.

Homework Equations



mgR = 1/2mv^2

v^2 = [tex]\sqrt{2mgR}[/tex]

The Attempt at a Solution



got v = 29 which was wrong, I might be using the wrong formulas or the wrong radius, or the formulas in the wrong way; this is an energy conservation problem.

any help is appreciated.
 
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  • #2
another method I tried, was that:

h = 9/2R = 99m

so the conservation of energy equation would turn out to be:

mgh = 1/2mv^2

v^2 = 2mgh/m

so v = [tex]\sqrt{2gh}[/tex] = 44m/s, which also turned out to be wrong.
 
  • #3
Your h=2.5R looks good for part A. For part B, you are neglecting the fact that the coaster has speed at the top of the loop in your first attempt,; in your second attempt, you neglected it has PE at point C, if you used h=4.5R.
 
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  • #4
bump to this, I figured out how to solve the velocity, got 38.8m/s and found the radial acceleration to be 68.6m/s^2.

now I want to find the tangential acceleration, I don't know angular velocity formulas at this point, so not sure where to take the problem to find the tangential acceleration, the only thing I know would be that it would be pointing outside the loop, but not sure how to write that mathematically.
 
  • #5
forget it, lol, I solved it,

atan = v^2/R = gR/R = g
 
  • #6


erm...I just wanted to say that the reply above mine solving for the tangential acceleration is not proper... v^2/r is for *centripetal* acceleration, or radial acceleration (pointing towards the center of the circle). The tangential acceleration is "g" because the vector is pointed straight down and the only force to accelerate the object of our interest is gravity. To find the tangential accel. in a situation like this, you must multiply the acceleration of gravity (g) by the cos(angle respect to vertical).
 
  • #7
clope023 said:
1. The problem statement, all variables and given/known

sigma Fy = Fc - N - mg = 0

N = 0 and Fc = mv^2/R

so Fy = Fc = mg and mv^2/R = mg and v^2 = GR

h = (1/2mgR + 2mgR)/mg = 1/2R + 2R = 5/2R



I hope it is ok to bring up an old thread because I don't understand these calculations. I cannot understand what the force F_{c} represents and why it's direction is in the positive y-direction.

I do see that the loop exerts a normal force which is downward but why is this force equal to 0?
 
  • #8
Should not the force F_{c} be directed along the trajectory since it is just the mass(scalar)*acceleration(vector) and acceleration has two components; One to the center of the orbit and one along the trajectory.
 
  • #9
zeralda21 said:
I hope it is ok to bring up an old thread because I don't understand these calculations. I cannot understand what the force F_{c} represents and why it's direction is in the positive y-direction.
That post is a bit confused. Fc stands for the centripetal force, which acts downward (at the top of the loop).

The analysis should be:
ƩFy = Fc = -N -mg
I do see that the loop exerts a normal force which is downward but why is this force equal to 0?
When finding the minimum speed to maintain contact, you want the normal force to just equal zero.
 
  • #10
Doc Al said:
That post is a bit confused. Fc stands for the centripetal force, which acts downward (at the top of the loop).

The analysis should be:
ƩFy = Fc = -N -mg

When finding the minimum speed to maintain contact, you want the normal force to just equal zero.

I understand now why N has to go to zero. Good.But is Fc an actual force? Shouldn't it therefore be included in the analysis as ƩFy=-Fc-N-mg? Or is it just a representation of the forces acting?

It is something with the signs..If we proceed with ƩFy=Fc=-N-mg=-mg and use that Fc=mv^2/R we obtain that v^2=-gR which is wrong since (g,R)>0.
 
  • #11
zeralda21 said:
But is Fc an actual force? Shouldn't it therefore be included in the analysis as ƩFy=-Fc-N-mg? Or is it just a representation of the forces acting?
Fc is not a separate force and should not be included in ƩFy. Centripetal force is just the name we give to the net force that produces the centripetal acceleration. I advise not even using the term centripetal force and just stick to centripetal acceleration.

It is something with the signs..If we proceed with ƩFy=Fc=-N-mg=-mg and use that Fc=mv^2/R we obtain that v^2=-gR which is wrong since (g,R)>0.
You just need to be consistent with signs. v^2/R is just the magnitude of the centripetal acceleration; The direction is downward so the sign will be negative:
ƩFy = ma
-N-mg = -mv^2/R
 
  • #12
Doc Al said:
Fc is not a separate force and should not be included in ƩFy. Centripetal force is just the name we give to the net force that produces the centripetal acceleration. I advise not even using the term centripetal force and just stick to centripetal acceleration.You just need to be consistent with signs. v^2/R is just the magnitude of the centripetal acceleration; The direction is downward so the sign will be negative:
ƩFy = ma
-N-mg = -mv^2/R

Well put, I understand. One last concern: In the following step:

h=(1/2mgR + 2mgR)/mg is clearly equivalent to mgh=(1/2mgR + 2mgR) which means that the potential energy at beginning (mgh) is equal to mg2R(top of the circle)+mgR/2 (the difference from in energy from initial position and on top of the circle).

How do we know that difference is mgR/2?
 
  • #13
zeralda21 said:
Well put, I understand. One last concern: In the following step:

h=(1/2mgR + 2mgR)/mg is clearly equivalent to mgh=(1/2mgR + 2mgR) which means that the potential energy at beginning (mgh) is equal to mg2R(top of the circle)+mgR/2 (the difference from in energy from initial position and on top of the circle).

How do we know that difference is mgR/2?
What you know is that the initial PE must equal the PE (2mgR) + KE at the top.

And since you just determined that mv^2/R = mg at the top, you know that the KE = 1/2mv^2 = mgR/2.
 
  • #14
Doc Al said:
What you know is that the initial PE must equal the PE (2mgR) + KE at the top.

And since you just determined that mv^2/R = mg at the top, you know that the KE = 1/2mv^2 = mgR/2.

I see. Thank you DocAl. You have been of great help.
 

Related to Riding a Loop the Loop (velocity at the bottom of the loop)

1. What is the minimum velocity required to safely ride a loop the loop?

The minimum velocity required to safely ride a loop the loop depends on the size and shape of the loop. Generally, it is recommended to have a minimum velocity of at least 8 meters per second to complete a circular loop without falling. However, for larger or more elliptical loops, a higher velocity may be required.

2. How does the velocity at the bottom of the loop affect the rider?

The velocity at the bottom of the loop is crucial for the rider's safety and experience. If the velocity is too low, the rider may not have enough momentum to complete the loop and may fall. On the other hand, if the velocity is too high, the rider may experience excessive G-forces and may feel discomfort or even lose consciousness.

3. Can a rider control their velocity at the bottom of the loop?

Generally, the velocity at the bottom of the loop is determined by the initial velocity and the force of gravity. However, a rider can slightly adjust their velocity by changing their position and posture during the ride. For example, leaning forward can increase the velocity, while leaning back can decrease it.

4. How does the weight of the rider affect the velocity at the bottom of the loop?

The weight of the rider does not directly affect the velocity at the bottom of the loop. However, a heavier rider may experience slightly higher G-forces due to their larger mass. It is important for riders to follow height and weight restrictions set by the ride operators to ensure their safety.

5. Is there a maximum velocity that a rider can safely ride a loop the loop?

The maximum safe velocity for riding a loop the loop depends on the design and safety measures of the ride. In general, most amusement park rides are designed with a maximum velocity in mind to ensure the safety and comfort of riders. It is important for riders to follow all safety instructions and guidelines provided by the ride operators to minimize the risk of injury.

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