Question

Two blocks of masses $400g$ and $200g$ 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 $2.0cm$. Find
(a) the kinetic energy of the system as the $400g$ block falls through $50cm$,
(b) the speed of the blocks at this instant.

Answer 1

The tension developed in the string connected to $400g$ block is $T_1$ and that in the string of $200g$ block is $T_2$.

Balancing the force on the block of mass $400g$.

$m_{1}g-T_1=m_{1}a$

$0.4g-T_1=0.4a$ ...(1)

Balancing the forces on the block of mass $200g$.

$m_{2}g-T_2=m_{2}a$

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

Write the torque balancing equation for the pulley:

$\tau =I\alpha$

$(T_1-T_2)r=I\times \frac{a}{r}$ ... (3)

From equation $1,2$ and $3$

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

b) Therefore $V=\sqrt{2ah}=\sqrt{(2\times g/5\times 0.5)}$$\Rightarrow \sqrt{\left(g/5\right)}=\sqrt{\left(9.8/5\right)}=1.4m/s$

a) Total kinetic energy of the system$=\frac{1}{2}{m}_1{v}^2+\frac{1}{2}{m}_2{V}^2+\frac{1}{2}I{\omega }^2$

$=(\frac{1}{2}\times 0.4\times {1.4}^2)+(\frac{1}{2}\times 0.2\times {1.4}^2)+(\frac{1}{2}\times (1.6\times {10}^{-4})\times {\left(\frac{1.4}{4\times {10}^{-4}}\right)}^2)=0.98$ Joule.