Question

A right circular cylinder of radius r and weight W is suspended by a cord that is wound around its surface. If the cylinder is allowed to fall, find the tension in the cord.

• A. $\frac{W}{3}$
• B. $\frac{2W}{3}$
• C. $\frac{W}{2}$
• D. W

Answer 1

The correct option is A $\frac{W}{3}$

Moment of inertia of the cylinder about the perpendicular axis through C

$I_c=\frac{W}{g}\frac{r^2}{2}$

Since C is moving downward with an acceleration ${\ddot{x}}_c$, an inertial force $\frac{W}{g}{\ddot{x}}_c$ is incorporated opposite to that.

Also the cylinder rotates clockwise, an inertial torque $I_c\ddot{\theta }$ is incorporated in the opposite direction $\ddot{\theta }$ in the free body diagram.

$\sum F_Y=0$

$T+\frac{W}{g}\ddot{x}-W=\mathrm{0....}\left(1\right)$

$\sum M_C=0$

$\text{And}Tr-I_C\ddot{\theta }=0......\left(2\right)$

$\left(1\right)\Rightarrow T=W-\frac{W}{g}\ddot{x}$

$\therefore \left(2\right)\Rightarrow (W-\frac{W}{g}\ddot{x})r-\frac{W}{g}\frac{r^2}{2}\ddot{\theta }=0$

$(W-\frac{W}{g}\ddot{x})r=\frac{W}{g}\frac{r^2}{2}.\frac{\ddot{x}}{r}$


$W=(\frac{W}{2g}+\frac{W}{g})\ddot{x}$

$\ddot{x}=\frac{2g}{3}$

$\therefore T=W-\frac{W}{g}.\frac{2g}{3}=\frac{W}{3}$