Question:
The pulley shown in figure has a radius $10 \mathrm{~cm}$ and moment of inertia $0.5 \mathrm{~kg}-\mathrm{m}^{2}$ about its axis. Assuming the inclined planes to be frictionless, calculate the acceleration of the $4.0 \mathrm{~kg}$ block.
Solution:
Here,
$I=0.5 \mathrm{~kg}_{-} \mathrm{m}^{2}$
$\mathrm{R}=0.1 \mathrm{~m}$
Translatory Motion Equation
$4 g^{\sin 45-T_{2}=4 a}-(\mathrm{i})$
$T_{1}-2 g \sin 45=2 a$-(ii)
Rotational Motion Equation $\tau=\mathrm{I} \alpha$
$T_{2} R-T_{1} R=I\left(\frac{a}{R}\right)_{-\text {(iii) }}$
Using (i),(ii) and (iii)
$\mathrm{a}=0.25 \mathrm{~m} / \mathrm{s}^{2}$