Question

The pulley shown in the figure has radius $20\text{cm}$ and MOI $0.2{\text{kg m}}^2$. Spring used has force constant $50{\text{N m}}^{-1}$. The system is released from rest. Find the velocity of $1\text{kg}$ block when it has descended $10\text{cm}$.(Take $g=10{\text{ms}}^{-2}$)

• A. $\frac{1}{2}{\text{ms}}^{-1}$
• B. $\frac{1}{\sqrt{2}}{\text{ms}}^{-1}$
• C. $\frac{1}{\sqrt{3}}{\text{ms}}^{-1}$
• D. none

Answer 1

The correct option is A $\frac{1}{2}{\text{ms}}^{-1}$

Given : Moment of inertia of the pulley $I=0.2kg-m^2$
Radius of the pulley $r=0.2m$
Spring constant $k=50N/m$, mass of the block $m=1\text{kg}$ and the descend of the block $x=10\text{cm}=0.1\text{m}$.

Applying the law of conservation of energy

$mgx=\frac{1}{2}k{x}^2+\frac{1}{2}I{\omega }^2+\frac{1}{2}m{v}^2$

Also $\omega =\frac{v}{r}$

$\Rightarrow 1\left(10\right)\left(.1\right)=\frac{1}{2}\left[50\right(.1{)}^2+\left(.2\right){\left(\frac{v}{.2}\right)}^2+1\left(1.v^2\right)]$
$\Rightarrow 2=0.5+6v^2$
or $v=\frac{1}{2}{\text{ms}}^{-1}$

hence option A is the correct answer.