Question

The pulley shown in figure (10-E8) has a radius $10cm$ and moment of inertia $0.5kg-m^2$ about its axis. Assuming the inclined planes to be frictionless, calculate the acceleration of the $4.0kg$ block.

Answer 1

Neutron's law equation

for the blocks

$2\sqrt{2}g-T_2=4a$ ...... (1)

$T_1-g\sqrt{2}=2a$ ....... (2)

Then, equation for the pulley

$T=I\alpha$ $[a=R\alpha ]$

$(T_{2}R-T_{1}R)=\left(0.5\right)\frac{a}{R}$

$T_2-T_1=\frac{\left(0.5\right)a}{R^2}$

$T_2-T_1=\frac{\left(0.5\right)a}{(0.1{)}^2}$

$T_2-T_1=50a\to$ ...... (3)

Adding $e{q}^n$ (1) and (2) we have

$T_1-T_2=6a+g\sqrt{2}-2\sqrt{2}g$

$T_1-T_2=6a-g\sqrt{2}$ ...... (4)

Adding $e{q}^n$ (3) and (4) we have

$0=50a+6a-g\sqrt{2}$

$g\sqrt{2}=56a$

$a=\frac{g\sqrt{2}}{56}$

$a=\frac{9.8\times \sqrt{2}}{56}$

$a=0.24748m/s^2$