Time hollow cylinder spends on ramp

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Homework Help Overview

The discussion revolves around a hollow cylinder rolling up an incline, focusing on the distance it travels and the time spent on the incline. The problem involves concepts from energy conservation and kinematics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the application of energy conservation principles and question the treatment of the cylinder's moment of inertia. There is discussion about the forces acting on the cylinder, including friction, and how these affect the calculations for distance and time.

Discussion Status

Some participants have recalculated the height and distance based on the correct treatment of the cylinder as a thin hoop. There is ongoing exploration of how to determine the time spent on the incline, with suggestions to consider average speed.

Contextual Notes

Participants are navigating through assumptions about the cylinder's properties and the effects of friction. There is a noted discrepancy between calculated values and those provided in the textbook.

cashmoney805
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Homework Statement


A hollow cylinder (hoop) is rolling on a horizontal surface at speed v= 3.3 m/s when it reaches a 15 degree incline. (a) How far up the incline will it go? (b) How long will it be on the incline before it arrives back at the bottom?


Homework Equations


Energy conservation:
1/2 Iw^2 + 1/2mv^2 = mgh

v^2 = vo^2 +2ax
x=volt + .5at^2
v=vo+at


The Attempt at a Solution


I for this problem is (I think) 1/2 m (r1^2 + r2^2), but the r's cancel because w^2 = v^2/(r1^2 + r2^2)

So I believe the energy equation simplifies to .25v^2 + .5v^2 = gh

I solve and get h = .833 and so the length = .833/sin15 = 3.22
However, the book says the length is 4.29.

Also, after that I don't know where to go.

For time, I tried v=vo+at, for a I used 9.8sin 15 and t (up) I got 1.3s
However, the total t= 5.2 s

Thank you so much!
 
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cashmoney805 said:

The Attempt at a Solution


I for this problem is (I think) 1/2 m (r1^2 + r2^2), but the r's cancel because w^2 = v^2/(r1^2 + r2^2)
Careful. Treat this as a thin hoop--with a single radius. (You're using a formula for a thick hollow cylinder. You'd have to set r1 = r2 to use that one.)

So I believe the energy equation simplifies to .25v^2 + .5v^2 = gh
You'll need to redo this, after fixing the above.

For time, I tried v=vo+at, for a I used 9.8sin 15 and t (up) I got 1.3s
However, the total t= 5.2 s
Careful--gravity is not the only force acting. A simpler way would be to figure out the average speed as it goes up the ramp.
 
What other force is acting on it?

Ok, I treated the hoop as a thin one and now my equation is:

v^2 = gh
h = 1.1 m, so length is 1.1/sin15 = 4.29m woo!

Now for t... I don't really know what to use anymore :/
 
cashmoney805 said:
What other force is acting on it?
Friction.

Ok, I treated the hoop as a thin one and now my equation is:

v^2 = gh
h = 1.1 m, so length is 1.1/sin15 = 4.29m woo!
Good.

Now for t... I don't really know what to use anymore
You know the distance. What's the average speed up the incline?
 
Doc Al said:
Friction.


Good.


You know the distance. What's the average speed up the incline?
vo/2

Ah then do x/v = t. How do you know when to use average velocity though? I never seem to use that.
 

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