Angular Acceleration of toy train

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Homework Help Overview

The discussion revolves around a toy train rolling on a horizontal track, specifically focusing on calculating the angular acceleration and the time it takes to stop from a given angular speed. The subject area includes concepts of angular motion and friction.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between linear and angular quantities, questioning how to apply the equations of motion given the lack of mass in the problem. There is discussion on using the force equation F = ma to derive angular acceleration.

Discussion Status

Some participants have provided guidance on how to approach the problem, suggesting to treat mass as a variable that will cancel out. There is an indication that calculations have been attempted, with one participant reporting a calculated value for angular acceleration.

Contextual Notes

The problem does not provide the mass of the train, which has led to questions about how to apply the equations effectively. The initial angular speed is given in revolutions per minute, which may require conversion for calculations.

carleon
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Hi, thanks for taking a look!

A toy train rolls around a horizontal 1.4-m-diameter track. The coefficient of rolling friction is 0.12.
(a) What is the magnitude of the train's angular acceleration after it is released?
(b) How long does it take the train to stop if it's released with an angular speed of 16 rpm?

I think I know which equations to use: α = ω^2*r and a = rα, but there is no mention of velocity in the problem so I am stuck.
 
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welcome to pf!

hi carleon! welcome to pf! :smile:
carleon said:
(a) What is the magnitude of the train's angular acceleration after it is released?

find a (from F = ma), then divide by r to get α :wink:
 


Thanks, but since mass isn't given, how would I find F=ma?
 
call the mass "m" … it'll cancel :smile:
 
I got it! α=1.68rad/s^2

Thanks a bunch. :biggrin:
 

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