Question

A fly wheel of radius $10cm$ and moment of inertia of $0.1kg{m}^2$ can be rotated in vertical plane without friction about its natural axis. A light string is wound around circumference of wheel and a body of mass $2.5kg$ is suspended at lower end of the string. If this suspended body is released, its acceleration is : ($g$ is the acceleration due to gravity)

• A. $\frac{g}{2}$
• B. $\frac{g}{3}$
• C. $\frac{g}{5}$
• D. $\frac{g}{10}$

Answer 1

The correct option is C $\frac{g}{5}$

$I=0.1kg{m}^2$
$R=10cm$
$2.5g-T=2.5a$ ---------- (1)
$a=\alpha R$ [no string slip condition]
and $RT=I\alpha$
using above $e{q}_n$ $\Rightarrow RT=I\frac{a}{R}$
$\Rightarrow T=I\frac{a}{R^2}=\frac{\left(0.1\right)a}{.01}=10a$
putting it in $e{q}^n$ (1)
$2.5g=12.5a$
$a=\frac{1}{5}g$