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

AI Thread Summary
The discussion focuses on calculating the acceleration and speed of a hollow cylindrical shell rolling down an inclined plane. The initial calculations for speed and acceleration are based on energy conservation and the moment of inertia, but there are errors in the moment of inertia used and the consideration of friction. The correct moment of inertia for a hollow cylinder is highlighted, and the impact of friction on the forces acting parallel to the incline is emphasized. Additionally, the implications of replacing the hollow shell with a solid cylinder are considered, noting that the moment of inertia affects the speed. Overall, the importance of accurately applying physics principles to solve the problem is underscored.
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 \alpha = 30° 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



I =\frac{2}{3}MR^{2}

K = \frac{1}{2}Mv^{2} + \frac{1}{2}I\omega^{2}

\omega = \frac{v}{r}


The Attempt at a Solution



(a)

mgh = \frac{1}{2}Mv^{2} + \frac{1}{2}I\omega^{2}

gh = \frac{1}{2}v^{2} + \frac{1}{3}R^{2}\omega^{2}

gh = \frac{1}{2}v^{2} + \frac{1}{3}R^{2}\frac{v^{2}}{R^{2}}

gh = v^{2}(\frac{1}{2}+\frac{1}{3})

v = \sqrt{\frac{6gh}{5}}


(b)

\sum F_{x} = ma = mgsin(\alpha)

a = gsin(\alpha)



(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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Amlung said:

Homework Equations



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


(a)

mgh = \frac{1}{2}Mv^{2} + \frac{1}{2}I\omega^{2}

gh = \frac{1}{2}v^{2} + \frac{1}{3}R^{2}\omega^{2}

gh = \frac{1}{2}v^{2} + \frac{1}{3}R^{2}\frac{v^{2}}{R^{2}}

gh = v^{2}(\frac{1}{2}+\frac{1}{3})

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


(b)

\sum F_{x} = ma = mgsin(\alpha)

a = gsin(\alpha)
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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