Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I've been trying to solve this problem:

A cylinder is rolling down an inclined plane (angle between plane and horizon α). Coefficient of friction is µ. What is the translational and angular speed of the cylinder when it's traveled distance is l( at the beginning v = 0)? Assume that it rolls without slipping.

So basically I started with energy conservation:
$$ E_p=mgh=mgl \sin(\alpha) $$ $$ E_p = E_r + E_t = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 $$ $$ v^2 = 4gl \sin(\alpha)/3 $$ Now the tricky part is that we need to find angular speed, but we don't have the radius of cylinder... is there a way to find the radius? I was thinking of maybe momentum conservation law(though I don't understand it completely)?

Any help appreciated!

share|improve this question
    
the radius should be given to you. There is no way to find radius from the data you have specified. Maybe you are missing something. Is the moment of inertia given? –  udiboy Jul 10 '13 at 19:05
add comment

3 Answers

up vote 1 down vote accepted

Actually you proved, with the third equation, that the translational speed does not depend on the radius $r$ of the cylinder, or on its mass $m$, but only on $g$ and $\alpha$. Indeed it reads like a differential equation defining the motion of the cylinder.

If the translational speed of the cylinder axis does not depend on the radius, then the angular speed has to depend on it, since they are related by the relation $v=\omega r$.

I do not see how that fact can be overcome by any computation including the total momentum of the system. This momentum can be computed from the speed at distance $l$, but will depend on the radius $r$, because it is squared in the moment of inertia, but not in the angular speed. Getting the total momentum by other means, such as resolving the differential equation will not do any better. Imho (but I have not practiced this sport for a long time).

But having $r$ in the expression of the angular speed is not a crime.

BTW What are you supposed to do with the coefficient of friction $\mu$ ? Do you have to check the cylinder does not slip?

share|improve this answer
    
I'm not sure what are you supposed to do with the coefficient in this case, it's given that it rolls without slipping –  Rugilena Jul 15 '13 at 14:28
add comment

The moment of inertia for a cylinder has a factor of $r^2$ in it, while we can change $\omega$ to $v$ using the equation $\omega = v/r$. Thus, $\omega^2 = v^2/r^2$. Plugging this and the moment of inertia in to your conversation of energy equation you'll find that the $r$'s cancel so you don't have to worry about the radius any more.

It seems to me that the third equation you have written is the result of what I described above. The final step I still haven't quite figured out. However, the cylinder's momentum is only conserved if there is no force acting on it, which is not the case here, so I don't think that will be useful.

share|improve this answer
add comment

The cylinder is laid down on it's side and rolls. Thus, the axis of rotation is the one along the height (z-axis) of the cylinder. Along this axis, the cylinder has a moment of inertia of $$I = \frac{1}{2}MR^2$$

which you can easily calculate with a bit of experience. Now, if the cylinder rolls without slipping, the angular velocity of the cylinder is linked to the speed of the center of mass ($v$) and the radius of the cylinder ($R$) by $$ \omega = \frac{v}{R} $$

as was pointed out in the post of Joshua. Putting all this together,

$$ E_p = E_r + E_t = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2 =\frac{3}{4}Mv^2 $$

Which after inverting for $v^2$ yields

$$ v^2 = \frac{4}{3}gl\sin(\alpha) $$

PS: If the cylinder indeed rolls without slipping, enery conservation still holds as you wrote it down, because in that case, the friction force does now work, but simply allows the cylinder to roll, and not slide.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.