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A uniform, solid cylinder of mass $m$ and radius $r$ rolls without slipping down a plane, inclined at an angle $θ$ to the horizontal. You may assume, without proof, that the moment of inertia of the cylinder is given by $I=\frac{1}{2}mr^2$.

Taking the axis parallel to the inclined plane going down as the positive $x$ axis and the perpendicular going up as the $y$ axis ($z$ axis going into the plane of paper).

a) Write down an expression for the angular acceleration about $z$.

b) Show that the acceleration along $x$ is given by $a_x=B\sin\theta$, where $B$ is a constant to determine.

c) Show that the coefficient of friction must be at least $\frac{1}{3}\tan\theta$.

What I did first was write down all forces acting on the object. ($N$ is the normal reaction force and $F_F$ the frictional force, $W_x$ and $W_y$ are the components of weight in each axis).

$$\sum F_y=W_y - N = ma_y=0 \implies W_y=N, a_y=0$$

$$\sum F_x=W_x-F_F=ma_x$$ $$W_y=Wcos\theta\implies F_F=\mu N=\mu mg\cos\theta$$ $$W_x=W\sin\theta\implies mg\sin\theta - \mu mg\cos\theta=ma \implies g(\sin\theta-\mu\cos\theta) = a_x(linear)$$ Then (a) $$\vec{\tau}=\vec{r} \times \vec{F} = \left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 0 & -|r| & 0 \\ -|F_F| & 0 & 0 \end{array} \right|=-|F_F||r|\hat{k} = I\alpha_z$$ $$\implies -|\mu mg\cos\theta||r|\hat{k}=\frac{1}{2}mr^2\alpha_z\implies\alpha_z=-\frac{2}{r}\mu|g\cos\theta|$$ For (b): (minus sign from accelerating in the other direction) $$a_x = -\alpha_z r+g(\sin\theta-\mu\cos\theta)$$ $$= 2\mu |g\cos\theta|+g\sin\theta-g\mu\cos\theta = \mu g\cos\theta + g\sin\theta$$ $$\implies g(\frac{\mu \cos\theta}{\sin\theta}+1)\sin\theta = g(\mu\cot\theta+1)\sin\theta$$ Have I done something wrong in my calculations? NowIGetToLearnWhatAHeadIs says the frictional force I calculated has $\cos\theta$ instead of what probably should be $\sin\theta$, but as far as I'm concerned, $N=F_y=W\cos\theta$, so where does the $\sin\theta$ come from?

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closed as off-topic by ACuriousMind, Kyle Kanos, Brandon Enright, Ali, John Rennie Nov 17 '14 at 6:54

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  • $\begingroup$ You say the friction force is $\mu m g \cos \theta$. Does this make sense for $\theta =0$? $\endgroup$ – Brian Moths Nov 17 '14 at 1:52