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 $1.6\times {10}^{-4}kg-m^2$ and a radius $2.0$cm. Find the kinetic energy of the system as the $400$ g block falls through $50$cm.

Answer 1

According to equation,

$0.4g-T_1=0.4a⟹\left(1\right)$

$T_2-0.2g=0.2a⟹\left(2\right)$

$(T_1-T_2)r=\frac{Ia}{r}⟹\left(3\right)$

From equation 1,2 and 3

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

$V=\sqrt{2ah}=\sqrt{(2\ast {gl}^5\ast 0.5)}$

$\sqrt{\frac{g}{5}}=\sqrt{\frac{9.8}{5}}=1.4\frac{m}{s}$

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

$=(\frac{1}{2}\ast 0.4\ast {1.4}^2)+(\frac{1}{2}\ast 0.2\ast {1.4}^2)+(\frac{1}{2}\ast \frac{1.6}{4}\ast {1.4}^2)$

$=0.98Joules$