Question

The pulley shown in the figure has radius $20\text{cm}$ and moment of inertia $0.2{\text{kg m}}^2$. 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 $10\text{cm}$

• 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 $\frac{1}{2}{\text{ms}}^{-1}$

Applying the conservation of mechanical energy,
Decrease in P.E. = Increase in K.E.
$mgx-\frac{1}{2}k{x}^2=\frac{1}{2}I{\omega }^2+\frac{1}{2}m{v}^2$
$1\left(10\right)\left(0.1\right)=\frac{1}{2}\left[50\right(0.1{)}^2+\left(0.2\right){\left(\frac{v}{0.2}\right)}^2+1\left(v^2\right)]$
$\Rightarrow 2=0.5+6v^2$
or $v=\frac{1}{2}{\text{ms}}^{-1}$