Optimizing Loop-the-Loop Height for Marble Shoot Tracks

  • Thread starter topgun08
  • Start date
  • Tags
    Marble
In summary, the largest theoretically possible value for H/D is found when the ball is at the top of the D ramp and has all KE.
  • #1
topgun08
15
0

Homework Statement


A popular pastime used to be to construct helter-skelters: tracks along which to
roll (without slipping) a ball (a marble say, or ball bearing). I made one that mimics a
fairground loop-the-loop (see attached .jpg). If the drop from start to base of the loop is D, and the top of the loop is a height H above the bottom, then what is the largest theoretically
possible value for H/D?

Homework Equations


?? - I am very confused.

Also, if my picture is bad, the height D is obviously supposed to be larger than the height H.
Thanks in advanced
 

Attachments

  • phys.jpg
    phys.jpg
    5 KB · Views: 480
Physics news on Phys.org
  • #2
topgun08 said:

Homework Statement


A popular pastime used to be to construct helter-skelters: tracks along which to
roll (without slipping) a ball (a marble say, or ball bearing). I made one that mimics a
fairground loop-the-loop (see attached .jpg). If the drop from start to base of the loop is D, and the top of the loop is a height H above the bottom, then what is the largest theoretically
possible value for H/D?


Homework Equations


?? - I am very confused.

Also, if my picture is bad, the height D is obviously supposed to be larger than the height H.
Thanks in advanced

Welcome to the PF. The easiest way to solve this type of problem is to use energy considerations. When the ball is at the top of the D ramp, it has all potential energy (PE) and no kinetic energy (KE) because it is not rolling yet. Are you familiar with how to calculate the PE of an object of mass m raised by some height D?

At the bottom of the ramp, the ball has all KE, although part of that KE is from its linear motion (what is the equation for KE as a function of an object's mass and velocity?), and part is from its rotation (are you familiar with a ball's Moment of Inertia and the energy associated with rolling?).

At the top of the loop, the ball has to be moving fast enough not to fall off the track. Are you familiar with how to figure that out?
 
  • #3
Thank you for the help! I should have realized this was an energy problem.

PE = mgD
KE (at the beginning of the ramp) = .5mv2 + .5Iw2

Now I don't know how fast the ball has to move at the top of the loop to not fall off...

Thank you so far though!
 
  • #4
I guess the marble is supposed to complete one round in the loop? Or it just has to go and stop at the highest point of the loop?
Anyway, think of contact. If it loses contact with the loop, then what happens is the normal force = ... And when it comes to force, you can calculate it, since you know speed by the energy conservation law and weight, right? :wink:
 
  • #5
you can use kinematics to do it that would take you a long time and its tedious, so use gravitational and kinetic energy, basically conservation of energy to solve it. which would be very fast
 
  • #6
hikaru1221 said:
I guess the marble is supposed to complete one round in the loop? Or it just has to go and stop at the highest point of the loop?
Anyway, think of contact. If it loses contact with the loop, then what happens is the normal force = ... And when it comes to force, you can calculate it, since you know speed by the energy conservation law and weight, right? :wink:

So at the top of the loop-the-loop,
mgD = mgH + .5Iw2 + .5mvH2.

The marble will not fall with the same force it will fall. So at the top of the loop, the marble would fall if both rotational and kinetic energy were not up to par?

But I can't seem to figure out when the normal force of the track will be less than the force of gravity pulling the marble down. I can't seem to relate force to the energy equations.
 
  • #7
i don't think you can relate force to energy..
 
  • #8
seto6 said:
i don't think you can relate force to energy..

hmmm...What should my next step be? At height H on the loop-the-loop what will cause the marble to fall/not fall?
 
  • #9
do you have to finish this by today.
cuz I am kinda busy got physics finals tomorrow morning.
 
  • #10
seto6 said:
do you have to finish this by today.
cuz I am kinda busy got physics finals tomorrow morning.

oh don't worry about it. Its not actually homework or anything. Its just to help me practice for the midterm which is on Thursday for me.

Goodluck
 
  • #11
seto6 said:
you can use kinematics to do it that would take you a long time and its tedious, so use gravitational and kinetic energy, basically conservation of energy to solve it. which would be very fast

Well, in this problem, dynamics is inevitable.

@topgun08: Think of my hint :wink: The key is the normal force. The energy conservation law is just for calculating speed.
 
  • #12
hikaru1221 said:
Well, in this problem, dynamics is inevitable.

@topgun08: Think of my hint :wink: The key is the normal force. The energy conservation law is just for calculating speed.

Okay here is what I have After checking out the equations and realizing what your normal force hint was all about.

PEtotal=mgD
KEat any point = .5mv2 + .5Iw2

Using the known formulas for a sphere's moment of inertia and for angular acceleration...
I = 2/5mr2
w = v/r
I substitute these formulas into my KE equation and compute:
KE = .5mv2 + .5(2/5mr2)(v/r)2
KE = .5mv2 + .2mv2
KE = .7mv2

Now once inside the loop-the-loop the marble will have centripetal acceleration. For this problem we want the normal force at the top of the loop-the-loop to be 0. So the only force acting on it will be the centripetal force whose acceleration is equal to gravity's acceleration. Thus the Fc = mac where ac = g

Using the formula that ac = v2/R where R is the radius of the loop-the-loop, I can solve for v2
ac = g = v2/R ==> v2 = Rg

Plug this back into the KE equation:
KE = .7mv2
v2 = Rg
KE = .7mRg

Applying the conservation of Energy Law at the top of the loop-the-loop:
PEtotal = KEtop + PEtop
mgD = .7mRg + mgH
Realize that H = 2R
D = .7R + 2R
D = 2.7R
(1/2.7)D = R
.74 = 2R/D
.74 = H/D

Is my logic and math right? Any mistakes or is this correct?
 
  • #13
It's correct :wink:
 
  • #14
hikaru1221 said:
It's correct :wink:

thanks for the help. much appreciated!
 

FAQ: Optimizing Loop-the-Loop Height for Marble Shoot Tracks

What is "Marble Shoot Loop-the-Loop"?

"Marble Shoot Loop-the-Loop" is a scientific experiment that involves setting up a track with a loop at the end and launching a marble through the loop to observe its trajectory.

How does "Marble Shoot Loop-the-Loop" work?

The marble is launched with a specific velocity and angle to ensure it has enough momentum to make it through the loop. The loop is designed with a specific radius and angle to keep the marble moving in a circular path. Gravity and the marble's inertia allow it to complete the loop and continue along the track.

What can we learn from "Marble Shoot Loop-the-Loop"?

By observing the marble's trajectory through the loop, we can learn about the principles of motion, such as momentum, centripetal force, and gravity. We can also use this experiment to study the effects of changing variables, such as the velocity and angle of launch, on the marble's movement.

What materials are needed for "Marble Shoot Loop-the-Loop"?

The materials needed for this experiment include a track, a loop, a marble, and a launching mechanism, such as a ramp or a catapult. Optional materials may include a stopwatch or video recording device to measure the marble's speed and trajectory.

Can "Marble Shoot Loop-the-Loop" be done at home?

Yes, "Marble Shoot Loop-the-Loop" can be easily set up and conducted at home using simple materials like cardboard, paper cups, and marbles. It is a fun and educational experiment for children and adults alike to explore the laws of motion.

Back
Top