What is the correct formula for solving part b of this problem?

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Homework Help Overview

The discussion revolves around a physics problem involving a solid brass ball rolling along a loop-the-loop track. The original poster has successfully solved part (a) of the problem, which concerns the minimum height from which the ball must be released. Part (b) asks for the magnitude of the horizontal component of the force acting on the ball at a specific point on the track after being released from a height of 6R.

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Approaches and Questions Raised

  • The original poster attempts to apply conservation of energy principles and relates various physical quantities, including mass, height, and velocity. They express confusion about the correct application of formulas and the resulting calculations. Other participants suggest using energy conservation equations and provide alternative formulations to approach the problem.

Discussion Status

Participants are exploring different methods to solve part (b) of the problem. Some have provided guidance on using conservation of energy and have shared specific equations, while others express uncertainty or seek further clarification on related concepts.

Contextual Notes

There is mention of a potential discrepancy in available resources, as one participant notes that the problem is not found in their physics book, indicating possible variations in curriculum or textbook content.

HobieDude16
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ok, here's a problem that you have to work out with all variables, and i frankly have trouble doing that... i got part a, and i did a lot of work on part b, but can't seem to come up with the right answer... any help appreciated...

In Figure 11-32, a solid brass ball of mass m and radius r will roll without slipping along the loop-the-loop track when released from rest along the straight section. For the following answers use g for the acceleration due to gravity, and m, r, and R, as appropriate, where all quantities are in SI units.

Fig. 11-32
hrw7_11-32.gif


(a) From what minimum height h above the bottom of the track must the marble be released to ensure that it does not leave the track at the top of the loop? (The radius of the loop-the-loop is R. Assume R r.)

(b) If the marble is released from height 6R above the bottom of the track, what is the magnitude of the horizontal component of the force acting on it at point Q?

ok, for part a, i got 2.7R whichi is correct, so then i moved on to part b, and i was told by my TA to start with the formula mg6R = .5mv^2 + .5Iomega^2 + mgR... so i did, and tried to solve for v. i used 2/5mR^2 for I and for omega, used v/R... so if i do that, and solve for v, then i use a=v^2/r and then when i have that, F=ma so i have the idea, i just can't get it right. anybody have any ideas? thanks in advance
John
 
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anybody have any ideas?
 
maybe i can find some help on another forum or something
 
Ok you work with conservation of energy and do, mgh=mgR+7mv^2/10. So at point Q you have F=mv^2/R=(50/7)mg
 
awesome, can't believe i didnt think of that, lol
 
btw u can find this one in almost every physics book it a classic


mg6R = 1/2mv^2+1/2Iw^2 + mgR

mg6R = 1/2mv^2 + 1/2 (2/5mr^2)w^2 + mgR

6gR = 1/2v^2 + 1/5r^2(v/r)^2+gR

5gR = 7/10 v^2

50/7gR = v^2

F = 50/7mg
 
Last edited:
ah, really? its not in ours (since its asked as a question in ours... lol)
 
Help...

Hey HobieDude16,

Can you show me how you got part A? I'm learning angular momentum too and I have a similar problem except the ball is not on a ramp. Thanks.
 
Hobie sent you a message hope you can help :)
 
  • #10
sqrt(9.8*r)=sqrt(10/7*9.8*(h-2R))
basically, i looked through the book, found examples of a guy riding a bike around a loop, and then found an example of a ball rolling down a hill, and combined and adjusted to work... maybe that might help
 

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