A uniform cylinder of mass \(M\) lies on a fixed plane inclined at an angle \(\theta\) with horizontal. A…

A uniform cylinder of mass \(M\) lies on a fixed plane inclined at an angle \(\theta\) with horizontal. A light string is tied to the cylinder at the right most point, and a mass \(m\) hangs from the string, as shown. Assume that the coefficient of friction between the cylinder and the incline plane is sufficiently large to prevent slipping. For the cylinder the remain static, the value of \(m\) is

1. \(\frac{M \sin \theta}{1-\sin \theta}\)
2. \(\frac{M \cos \theta}{1+\sin \theta}\)
3. \(\frac{M \sin \theta}{1+\sin \theta}\)
4. \(\frac{M \cos \theta}{1-\sin \theta}\)

Solution

According to diag.
\(\begin{array}{l}
\mathrm{T}=\mathrm{mg} \\
\mathrm{f}=\mathrm{Mg} \sin \theta+\mathrm{T} \sin \theta \\
=\mathrm{Mg} \sin \theta+\mathrm{mg} \sin \theta
\end{array}\)
Balancing torque about C;
\(\mathrm{Tr}=\mathrm{fr}\)
\(\Rightarrow \mathrm{mgr}=(\mathrm{Mg} \sin \theta+\mathrm{mg} \sin \theta) \mathrm{r}\)
\(\Rightarrow \mathrm{m}=\frac{\mathrm{M} \sin \theta}{1-\sin \theta}\)

Asked in: JEE Mains - Rotational Motion - Chapter Test

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