Question

The pulley shown in figure (10-E8) has a radius 10 cm and moment of inertia 0⋅5 kg-m² about its axis. Assuming the inclined planes to be frictionless, calculate the acceleration of the 4⋅0 kg block.

Figure

Answer 1

From the figure, we have:
$$\begin{gathered} T_1-m_{1}g\sin \theta =m_{1}a...\left(i\right) \\ {T}_2-T_1=I\frac{a}{r^2}...\left(ii\right) \\ {m}_{2}g\sin \theta -T_2=m_{2}a...\left(iii\right) \end{gathered}$$
Adding equations (i) and (iii), we get:
$m_{2}g\sin \theta +\left(-T_2+T_1\right)-m_{1}g\sin \theta =\left(m_1+m_2\right)a$

$$\begin{gathered} \Rightarrow \left(m_2-m_1\right)g\sin \theta +\left(-\frac{I}{r^2}\right)a=(m_1+m_2)a \\ \Rightarrow \left(m_2-m_1\right)g\sin \theta =(m_1+m_2)a+\left(\frac{I}{r^2}\right)a \\ \Rightarrow a=\frac{\left(m_2-m_1\right)g\sin \theta }{\left(m_2+m_1+{\displaystyle \frac{I}{r^2}}\right)} \\ =\frac{\left(4-2\right)\times 10\times \left(1\sqrt{2}\right)}{\left(4+2\right)+\left(0.5/0.01\right)} \\ =\frac{\left(2\times 10\times {\displaystyle \frac{1}{\sqrt{2}}}\right)}{\left(6+50\right)} \\ =0.248=0.25\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$