Question

a solid cylinder of mass $m=4kg$ and radius $R=10cm$ has two ropes wrapped around it, one near each end. The cylinder is held horizontally by fixing the two free ends of the cords to the hooks on the ceiling such that both the cords are exactly vertical. The cylinder is released to fall under gravity.
Find the tension along the strings.

• A. $6.53$N
• B. $5.23$N
• C. $3.23$N
• D. $4.43$N

Answer 1

The correct option is A $6.53$N

Let a be the acceleration when the cylinder is falling freely$.$

$m\times a=m\times g-2\times T--------\left(1\right)$

Let r be the radius of cylinder$.$ If torque acting on both the ends are similar then net torque is given by$,$

$T=T_1+T_2=2\times r\times T=1\times \alpha ------\left(2\right)$

Where I is moment of inertai and α is angular acceleration due to rod rotation$.$

if we substitute for I and α in the above equation we get$,$

$\begin{array}{c}2\times r\times T=\frac{1}{2}m{r}^2\left(\frac{a}{r}\right)=\frac{1}{2}m\times r\times a\\ hence,T=\frac{1}{4}m\times a\end{array}$

Substituting the above equation for $T$ in $e{q}^n.\left(1\right),$ we get $a=\left(2/3\right)g$

$So,T=\frac{1}{4}\times 4\times \frac{2}{3}\times 9.8=6.53$

Hence,

option $\left(A\right)$ is correct answer.