Question

The pulley shown in figure has a radius of $20cm$ and moment of inertia $0.2kg-m^2$. The string going over it is attached at one end to a vertical spring of spring constant $50N/m$ fixed from below, and supports a $1kg$ mass at the other end. The system is released form rest with the spring at its natural length. Find the speed of the blocks when it has descended through $10cm$. Take $=10m/s^2$

Answer 1

Given:

The inertia of the pulley is $I=0.2kg-m^2$

The radius of the pulley is $r=0.2m$

The spring constant of the spring is $K=50N/m$,

The mass of the block connected to the spring is $m=1kg$

Take $g=10m{s}^2$ and the height through which the block decends is $h=0.1m$

The sum of the kinetic energy attained by the block, rotational energy of pulley and the spring potential energy will be equal to the loss in the potential energy of the block.

The angular velocity of the pulley is:

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

$=\frac{v}{0.2}$

Therefore applying laws of conservation of energy

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

$\Rightarrow 1\times 10\times 0.1=\frac{1}{2}\times 1\times v^2+\frac{1}{2}\times 0.2\times \frac{v^2}{0.04}+\frac{1}{2}\times 50\times 0.01$ (x=h)

$\Rightarrow 1=0.5v^2+2.5v^2+1/4$

$\Rightarrow 3v^2=\frac{3}{4}$

$\Rightarrow v=\frac{1}{2}=0.5m/s$