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## Q. 1.5

Determine the angular acceleration of the pulley of Figure 1.17.

## Verified Solution

Consider the system of rigid bodies composed of the pulley and the two blocks. If $\alpha$ is the counterclockwise angular acceleration of the pulley, then, assuming no slip between the pulley and the cables, block A has a downward acceleration of $r_{A}\alpha$ and block B has an upward acceleration of $r_{B}\alpha$.

Summing moments about the center of the pulley, neglecting axle friction in the pulley, and using the free-body diagrams of Figure 1.17(b) assuming moments are positive counterclockwise yields

$\sum{M_{\omicron _{ext} } } =\sum{M_{\omicron _{eff} } }$

$m_{A}gr_{A} – m_{B}gr_{B}= I_{P} \alpha +m_{B}r^{2}_{A}\alpha +m_{B}r^{2}_{B}\alpha$

Substituting given values leads to $\alpha =7.55 rad/ s^{2}.$