What is the acceleration of a cylindrical shell rolling down an inclined plane?

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Homework Help Overview

The discussion revolves around a hollow cylindrical shell rolling down an inclined plane, specifically focusing on its acceleration and speed under various conditions. The problem involves concepts from dynamics and rotational motion, including the moment of inertia and forces acting on the shell.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster attempts to apply energy conservation principles and equations of motion to find the speed and acceleration of the shell. Some participants question the correctness of the moment of inertia used and the role of friction in the analysis.

Discussion Status

Participants are actively engaging with the problem, with some providing corrections and suggestions for reevaluation of the moment of inertia and the forces involved. There is acknowledgment of potential errors in the original poster's reasoning, particularly regarding the effects of friction and the implications of the moment of inertia on the results.

Contextual Notes

There is a mention of the initial conditions, such as the shell starting from rest and the angle of the incline. The discussion also hints at the differences in behavior when comparing a hollow cylindrical shell to a solid cylinder, though specifics are not resolved.

Amlung
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Homework Statement



A hollow cylindrical shell with mass M = 100 g and radius R = 5 cm rolls without
slipping down an inclined plane making an angle [tex]\alpha = 30[/tex]° with the horizontal.

(a) If the initial speed of the shell is zero, what will be the speed of its center of

(c) Calculate the linear acceleration of the center of mass of the shell. How long
does it take the shell to roll 1:5 meters along the plane with zero initial velocity?

(d) If the shell is replaced with a solid cylinder what will be the answer to the
previous question?


Homework Equations



[tex]I =\frac{2}{3}MR^{2}[/tex]

[tex]K = \frac{1}{2}Mv^{2} + \frac{1}{2}I\omega^{2}[/tex]

[tex]\omega = \frac{v}{r}[/tex]


The Attempt at a Solution



(a)

[tex]mgh = \frac{1}{2}Mv^{2} + \frac{1}{2}I\omega^{2}[/tex]

[tex]gh = \frac{1}{2}v^{2} + \frac{1}{3}R^{2}\omega^{2}[/tex]

[tex]gh = \frac{1}{2}v^{2} + \frac{1}{3}R^{2}\frac{v^{2}}{R^{2}}[/tex]

[tex]gh = v^{2}(\frac{1}{2}+\frac{1}{3})[/tex]

[tex]v = \sqrt{\frac{6gh}{5}}[/tex]


(b)

[tex]\sum F_{x} = ma = mgsin(\alpha)[/tex]

[tex]a = gsin(\alpha)[/tex]



(d)

According to my equation it shouldn't change anything...


I don't think b/c are correct though

Thanks for any help.
 
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Last edited by a moderator:
Amlung said:

Homework Equations



[tex]I =\frac{2}{3}MR^{2}[/tex]
That's not the rotational inertia of a thin cylindrical shell.


(a)

[tex]mgh = \frac{1}{2}Mv^{2} + \frac{1}{2}I\omega^{2}[/tex]

[tex]gh = \frac{1}{2}v^{2} + \frac{1}{3}R^{2}\omega^{2}[/tex]

[tex]gh = \frac{1}{2}v^{2} + \frac{1}{3}R^{2}\frac{v^{2}}{R^{2}}[/tex]

[tex]gh = v^{2}(\frac{1}{2}+\frac{1}{3})[/tex]

[tex]v = \sqrt{\frac{6gh}{5}}[/tex]
Correct the moment of inertia and redo.


(b)

[tex]\sum F_{x} = ma = mgsin(\alpha)[/tex]

[tex]a = gsin(\alpha)[/tex]
Gravity is not the only force acting parallel to the incline. What about friction?

(d)

According to my equation it shouldn't change anything...
But in part a you found that the speed does depend on the moment of inertia. Which means that you made a mistake in your thinking somewhere.
 
totally forgot about friction thanks xD
and wrong moment of inertia...

thx ^^
 

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