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 $1·6\times {10}^{-4}\mathrm{kg}-{\mathrm{m}}^2$ 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

From the free body diagram, we have:
$$\begin{gathered} 0.4g-T_1=0.4a...\left(i\right) \\ {T}_2-0.2g=0.2a...\left(ii\right) \\ \left(T_1-T_2\right)r=\frac{la}{r}...\left(iii\right) \end{gathered}$$

On solving the above equations, we get:
$a=\frac{\left(0.4-0.2\right)g}{\left(0.4+0.2+{\displaystyle \frac{1.6}{0.4}}\right)}=\frac{g}{5}$
On solving the (b) part of the question first, we have:
Speed of the blocks = $v=\sqrt{\left(2ah\right)}=\sqrt{2\times \frac{g}{5}\times 0.5}=\sqrt{\left(\frac{9.8}{5}\right)}=1.4\mathrm{m}/\mathrm{s}$

(a) Total kinetic energy of the system
$$\begin{gathered} =\frac{1}{2}{m}_1{v}^2+\frac{1}{2}{m}_2{v}^2+\frac{1}{2}I{\omega }^2 \\ =\frac{1}{2}{m}_1{v}^2+\frac{1}{2}{m}_2{v}^2+\frac{1}{2}I{\left(\frac{v}{r}\right)}^2 \\ =\left(\frac{1}{2}\times 0.4\times {\left(1.4\right)}^2\right)+\left(\frac{1}{2}\times 0.2\times {\left(1.4\right)}^2\right)+\left(\frac{1}{2}\times 1.6\times {10}^{-4}\times {\left(\frac{1.4}{2\times {10}^{-2}}\right)}^2\right) \\ =\left(0.2+0.1+0.2\right){\left(1.4\right)}^2 \\ =0.5\times 1.96=0.98\mathrm{joule} \end{gathered}$$