# Exercise 2.31 cylinder and hanging mass

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?

Note by Beakal Tiliksew
4 years, 4 months ago

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alt text

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)

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

$\Rightarrow M\sin\theta = m(1-\sin\theta)$

$\Rightarrow \boxed{m = \frac{M\sin\theta}{1-\sin\theta}}$

- 4 years, 4 months ago

nice explanation

- 4 years, 3 months ago

Anish, which software do you use to draw diagrams?

- 4 years, 4 months ago

I use Photoshop CS5

- 4 years, 4 months ago

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

- 4 years, 4 months ago

Paint is not bad, if you just draw simple diagrams.

- 4 years, 4 months ago

Thanks, Anish! : )

- 4 years, 4 months ago