Question

A uniform cylinder of radius $R$ and mass $M$ can rotate freely about a stationary horizontal axis $O$ (figure shown above). A thin cord of length $l$ and mass $m$ is wound on the cylinder in a single layer. Find the angular acceleration of the cylinder as a function of the length $x$ of the hanging part of the cord. The wound part of the cord is supposed to have its centre of gravity on the cylinder axis.

• A. ${\beta }_z=\frac{2mgx}{lR(M+2m)}$
• B. ${\beta }_z=\frac{4mgx}{lR(M+2m)}$
• C. ${\beta }_z=\frac{2mgx}{lR(3M-2m)}$
• D. None of these

Answer 1

The correct option is A ${\beta }_z=\frac{2mgx}{lR(M+2m)}$

Let us use the equation $\frac{d{M}_z}{dt}=N_z$ relative to the axis through $O$ (1)
For this purpose, let us find the angular momentum of the system $M_z$ about the given rotation axis and the corresponding torque $N_z$. The angular momentum is $M_z=I\omega +mvR=(\frac{m_0}{2}+m)R^2\omega$
[where $I=\frac{m_0}{2}{R}^2$ and $v=\omega R$ (no cord slipping)]
So, $\frac{d{M}_z}{dt}=(\frac{M{R}^2}{2}+m{R}^2){\beta }_z$ (2)
The downward pull of gravity on the overhanging part is the only external force, which exerts a torque about the z - axis, passing through $O$ and is given by, $N_z=\left(\frac{m}{l}\right)xgR$
Hence from the equation $\frac{d{M}_z}{dt}=N_z$
$(\frac{M{R}^2}{2}+m{R}^2){\beta }_z=\frac{m}{l}xgR$
Thus, ${\beta }_z=\frac{2mgx}{lR(M+2m)}>0$