A uniform cylinder of mass \(M\) sits on a fixed plane inclined at an angle \(\theta\). A string is tied to the cylinder’s rightmost point, and a mass \(m\) hangs from the string, as shown in Figure above. Assume that the coefficient of friction between the cylinder and the plane is sufficiently large to prevent slipping.

What is \(m\), in terms of \(M\) and \(\theta\), if the setup is static?

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In the above image, I have marked out the forces on the cylinder, and also, I have removed the mass \(\displaystyle m\).

Firstly, if the system is static, then the tension in the string must be \(\displaystyle mg\), so that there is equilibrium for the mass \(\displaystyle m\).

For attaining rotational equilibrium,Considering the torque about the contact point of the cylinder and the inclined plane,

\[Mg(R\sin\theta) = T(R-R\sin\theta)\]

The torque due to the Normal Reaction and the Frictional force, is zero, since those forces pass through the contact point (the point which we are considering the torque about)

\[\Rightarrow Mg(R\sin\theta) = mgR(1-\sin\theta)\]

\[\Rightarrow M\sin\theta = m(1-\sin\theta)\]

\[\Rightarrow \boxed{m = \frac{M\sin\theta}{1-\sin\theta}}\] – Anish Puthuraya · 3 years, 1 month ago

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– Hafizh Ahsan Permana · 3 years ago

nice explanationLog in to reply

– Karthik Kannan · 3 years, 1 month ago

Anish, which software do you use to draw diagrams?Log in to reply

– Anish Puthuraya · 3 years, 1 month ago

I use Photoshop CS5Log in to reply

– Karthik Kannan · 3 years, 1 month ago

Oh, I don't have it. Could you recommend something else with which I can put a decent diagram?Log in to reply

– Anish Puthuraya · 3 years, 1 month ago

Paint is not bad, if you just draw simple diagrams.Log in to reply

– Karthik Kannan · 3 years, 1 month ago

Thanks, Anish! : )Log in to reply