Solution :
The angular acceleration, [tex]$\alpha$[/tex] is obtained from the equation of the [tex]$\text{Newton's second law}$[/tex] of rotational motion,
Thus,
[tex]$\tau = F \times d$[/tex]
or [tex]$\tau = I \times \alpha$[/tex]
where [tex]$\tau$[/tex] is torque, F is force, d is moment arm distance, I is the moment of inertia
Thus, [tex]$\alpha=\frac{(F\times d)}{I}$[/tex]
Now if the force and the moment arm distance are constant, then the [tex]\text{angular acceleration is inversely proportional to the moment of inertia.}[/tex]
That is when, F = d = constant, then [tex]$\alpha \propto \frac{1}{I}$[/tex] .
Thus, moment of inertia, I is proportional to mass of the bar.
The mass is less for the bar in case (1) in comparison with that with the bar in case (2) due to the holes that is made in the bar.
Therefore, the bar in case (1), has less moment of inertia and a greater angular acceleration.
