# Solve the following :

Question:

Two blocks of masses $400 \mathrm{~g}$ and $200 \mathrm{~g}$ are connected through a light string going over a pulley which is free to rotate about its axis. The pulley has a moment of inertia $1.6^{\times 10^{-4}} \mathrm{~kg}-\mathrm{m}^{2}$ and a radius $2.0 \mathrm{~cm}$. Find (a) the kinetic energy of the system as the $400 \mathrm{~g}$ block falls through $50 \mathrm{~cm}$, (b) the speed of the blocks at this instant.

Solution:

Translatory Motion Equation

$0.4 \mathrm{~g}_{-} T_{1}=0.4 \mathrm{a}-(\mathrm{i})$

$T_{2}-0.2 \mathrm{~g}=0.2 \mathrm{a}-(\mathrm{ii})$

Rotational Motion Equation, $\tau=I \omega$

$T_{1} R-T_{2} R=\left(1.6 \times 10^{-4}\right)\left(\frac{a}{R}\right)$-(iii)

Solving (i),(ii) and (iii)

$a=g / 5$.

Speed of block $=\sqrt{2 \times a c c . \times h e i g h t}$

$=\sqrt{2 \times \frac{g}{5} \times \frac{1}{2}}$

$=1.4 \mathrm{~m} / \mathrm{s}$

Total kinetic energy of system

$=\frac{1}{2} m_{1} v^{2}+\frac{1}{2} m_{2} v^{2}+\frac{1}{2} I \omega^{2}$

$=\frac{1}{2}(0.4)(1.4)^{2}+\frac{1}{2}(0.2)(1.4)^{2}+\frac{1}{2}\left(1.6 \times 10^{-4}\right)\left(\frac{1.4}{0.02}\right)^{2}$

$=0.98 \mathrm{~J}$