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I have two ways of calculating the final speed of a solid cylinder going down an incline with friction that are apparently contradicting.

The first way is using conservation of energy. $mgh = \frac{1}{2}Iw^2 + \frac{1}{2}mv^2$ which after some work gives $mgh = \frac{3}{4}mv^2$. Therefore $v = \sqrt{\frac{4}{3}gh}$.

The second way is to use gravity and friction to determine the acceleration of the cylinder to be $\frac{2}{3}g\sin{\theta}$ (where $\theta$ is the angle between the horizontal and incline). Now if there was no friction, the cylinder would accelerate at $g\sin{\theta}$. Thus the cylinder with friction is accelerating at two-thirds the rate of a frictionless one. So the final speed of the cylinder with friction must also be two-thirds of the frictionless cylinder. We know a frictionless cylinder will have final speed $\sqrt{2gh}$ (since $mgh = \frac{1}{2}mv^2$). So by our two-thirds velocity logic, the cylinder with friction must have final speed $\frac{2}{3}\sqrt{2gh} = \sqrt{\frac{8}{9}gh}$.

So I have two ways of calculating the final speed of a cylinder with friction going down an incline and they do not match. Which is correct and why?

This link contains the derivations for both approaches: https://farside.ph.utexas.edu/teaching/301/lectures/node108.html but does not reconcile the apparent contradiction.

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2 Answers 2

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Thus the cylinder with friction is accelerating at two-thirds the rate of a frictionless one. So the final speed of the cylinder with friction must also be two-thirds of the frictionless cylinder.

This would only be true if they were accelerating for the same amount of time, since acceleration is the change of velocity per unit time. However, since the cylinder with friction accelerates at a lower rate, but for the same distance, this lower acceleration is applied for a longer time than in the frictionless case. Therefore the logic you were applying gives a final velocity that is too low.

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    $\begingroup$ I agree! So you accelerate at two-thirds the rate, but that acceleration lasts $\sqrt{\frac{3}{2}}$ times longer so your final speed is $\frac{2}{3}\sqrt{\frac{3}{2}}\sqrt{2gh}$ which gives back exactly the same answer as the first approach. Cheers! $\endgroup$ Commented Apr 22, 2022 at 0:41
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The problem assumes the cylinder rolls without slipping. v is the final velocity of the center of mass (CM). The energy solution is correct.

For the second approach the constant acceleration down the incline is ${2 \over 3} g sin(\theta)$. The acceleration of the center of mass (CM) is given by: $\vec F_{net} = M\vec a_{CM}$ where $\vec F_{net}$ is the net external force, $M$ is the total mass, and $\vec a_{CM}$ is the acceleration of the CM. Using the relationships for constant acceleration $v^2 = 2as$ where $s$ is the distance down the incline. $sin(\theta) = {h \over s}$, So $v = {\sqrt {4 \over 3}gh}$.

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