Question

The pulley shown in figure has a moment of inertia I about its axis and its radius is $r$. Calculate the magnitude of the acceleration of the two blocks. Assume that the string is light and does not slip on the pulley.

Answer 1

Suppose the tension in the left string is $T_1$ and that in the right string in $T_2$. Suppose the block of mass $m_1$ goes, down with an acceleration $\alpha$ and the other block moves up with the same acceleration. This is also the tanqential acceleratlon of the rim of the wheel as the string does not slip over the rim. The angular acceleration of the wheel is, therefore, $\alpha =a/r$. The equations .of motion for the mass $m_1$, the mass $m_2$ and the pulley are as follows :
$m_{1}g-T_1=m_{1}a$ .......... (i)
$T_2-m_{2}g=m_{2}a$ .......... (ii)
$T_{1}r-T_{2}r=l\alpha =l\alpha /r$ .......... (iii)
Putting $T_{1}and{T}_2$ from (i), and (ii) into (iii),
$\left[\right(m_{1}g-a)-m_2(g+a\left)\right]rI\frac{a}{r}$
Which gives a $=\frac{(m_1-m_2)g{r}^3}{I+(m_1+m_2)r^2}$