Rotational motion down an incline (ring/hoop)

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Homework Help Overview

The problem involves a 24kg metal ring rolling down a 30-degree incline from a height of 3.4 m, with questions related to its linear speed at the bottom of the ramp, the effects of its moment of inertia, and its average linear acceleration down the incline.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply the conservation of energy principle to find the linear speed, while also considering the moment of inertia for further calculations. Some participants question the wording of the problem regarding friction and its implications on the motion of the ring.

Discussion Status

Participants are exploring different interpretations of the problem, particularly regarding the assumptions about friction. Some guidance has been offered regarding the equations linking initial speed, final speed, distance, and acceleration, but no consensus has been reached on the correct approach to part three.

Contextual Notes

There is some ambiguity in the problem statement regarding whether the ring rolls without slipping or if it is frictionless, which affects the interpretation of the motion. Additionally, the original poster has noted a correction in terminology related to acceleration.

floraha
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Homework Statement


A 24kg metal ring with 24cm diameter rolls without slipping down a 30 degree incline from a height of 3.4 m.

1) According to the law of conservation of energy what should be the linear speed of the ring at the bottom of the ramp
2) if the ring has a moment of inertia of I=mr^2 what will its linear speed be at the base of the incline?

3) what is the avg linear acc of this ring down the incline?

Homework Equations

3. The Attempt at a Solution [/B]
For part 1, I got the math down to the square root of 2gh, because I assumed they were just asking for Vfrictionless. I did this through the equation PE=KE(translational) + KE (rotational) from which I got mgh = 1/2mv^2+1/2iw^2 where i=moment of inertia and w=angular velocity. In this case (i) would just be zero, getting me 2gh

For part 2 I did the same thing, except I substituted the moment of inertia in for (i) getting the math down to the square root of gh.

I had the most difficult time with part three. I wasn't sure where to start so I just worked out a(tangental) = r * angular acceleration. My teacher told me the answer is gsin(x) but I have no idea how he got there.
 
Last edited:
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floraha said:
rolls without stopping
Are you sure that is the correct wording and it wasn't supposed to say, "rolls without slipping". Additionally, because it says "rolls", that suggests that it is not frictionless. If it was frictionless, it would not rotate at all; it would just slide.

Edit: Sorry I forgot to welcome you to Physics Forums. Welcome!
Edit2: Hmm. I think I can see your position that part 1 maybe is supposed to assume frictionless due to the wording of part 2. However, for part 1 the ring had the same moment of inertia as specified in part 2.
 
@TomHart Thank you for replying! You were right about the wording, it was actually "slipping." Also, for part 3 it autocorrected to arc when it should have been acc (as in acceleration).
 
floraha said:
what is the avg linear acc of this ring down the incline?
You know the initial speed, the final speed and the distance (which is...?). What equation do you know linking those three to acceleration?
 

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