Question

Two blocks of masses 400 g and 200 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 $\times {10}^{-4}$ ${\text{kg-m}}^2$ and a radius 2.0 cm. Find (a) the kinetic energy of the system as the 400 g block falls through 50 cm, (b) the speed of the blocks at this instant.

Answer 1

According to the question 0 Ag - $T_1$ = 0.4a ...(1)

$T_2$-0.2g = 0.2a ...(2)

$T_1-T_2$ r = $\frac{la}{r}$ ...(3)

From eq. 1,2 and 3,

$\Rightarrow a=\frac{(0.4-0.2)g}{(0.4+0.2+\frac{1.6}{0.4})}=\frac{g}{5}$

Therefore (b) $V=\sqrt{\left(2gh\right)}=\sqrt{2\times g\times \frac{1}{5}}$

$\Rightarrow \sqrt{\left(\frac{g}{5}\right)}=\sqrt{\left(\frac{9.8}{5}\right)}=1.4m/s$

(a) Total kinetic energy of the system

$=\frac{1}{2}{m}_1{V}^2+\frac{1}{2}{m}_2{V}^2+\frac{1}{2}{18}^2$

=$(\frac{1}{2}\times 0.4\times {1.4}^2)+\{\frac{1}{2}\times (\frac{1.6}{4}\times {1.4}^2)\}$

= $(0.2+0.1+0.2)(1.4{)}^2$

= $0.5\times 1.96=0.98$