# Contradicting answers for the speed of a solid cylinder down an incline

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.

• 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! Commented Apr 22, 2022 at 0:41
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 $$v = {\sqrt {4 \over 3}gh}$$.