Question

A string is wrapped on a wheel of moment of inertia 0⋅20 kg-m² and radius 10 cm and goes through a light pulley to support a block of mass 2⋅0 kg as shown in figure (10-E5). Find the acceleration of the block.

Figure

Answer 1

Moment of inertia of the bigger pulley, I = 0.20 kg-m²
r = 10 cm = 0.1 m,
Smaller pulley is light. Therefore, on neglecting its moment of inertia, we have:
Mass of the block, m = 2 kg


From the free body diagram, we get:
$mg-T=ma...\left(i\right)$
$Tr=I\alpha$ And $a=\alpha r$
$\Rightarrow T=\frac{Ia}{r^2}...\left(ii\right)$
Using equations (i) and (ii), we get:
$$\begin{gathered} mg=\left(m+\frac{I}{r^2}\right)a \\ \Rightarrow a=\frac{mg}{m+{\displaystyle \frac{I}{r^2}}} \\ =\frac{2\times 9.8}{2+\left({\displaystyle \frac{0.2}{0.01}}\right)} \\ =\frac{19.6}{22}=0.89\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$