The way I think about it, since the pulley have mass, they are a bit harder to turn? I see it as if it is friction and it increases the work required to make it move.
For example, for a block with 10 kg that is attached to a pulley with radius 1 meter (assume that the disc is a frictionless pulley), then we drop the block at some height 5 meters from the ground. Let's say that the pulley's mass is still unknown and right before the block hits the ground, the angular velocity is 9 rad/s. How do we find its angular acceleration with what we know?
Here's where I am at right now:
I need $F = ma$ and its rotational equivalent $\tau = I\alpha$. Also, $I = \dfrac{1}{2}MR^2$
In the problem I gave, the torque is caused by the tension in the string acting in the pulley from the block. We are going to use the fact that the tangential speed of the pulley is equal to the speed of the block.
$\alpha$ can be found by the relation:
\begin{align*}
a &= R\alpha \\\\
\alpha &= \frac{a}{R}
\end{align*}
Hence,
\begin{align*}
T \cdot R &= \dfrac{1}{2}MR^2\frac{a}{R}\\
T &= \frac{1}{2}Ma
\end{align*}
For the block, $mg$ acts on the block, pulling it downward and this is being opposed by tension $T$:
\begin{align*}
mg – T &= ma
\end{align*}
Recall $T = \frac{1}{2}Ma$, hence we can substitute:
\begin{align*}
ma &= mg – \frac{1}{2}Ma\\
mg &= ma + \frac{1}{2}Ma\\
mg &= a(m+\frac{1}{2}M)\\
a &= \frac{mg}{m + \dfrac{1}{2}M}
\end{align*}
Since $M$ is unknown, how do I find $M$?
Best Answer
By rotating, the pulleys acquires kinetic energy at the expense of the other energies present in the system. If at the beginning the pulley is not moving and you have a total energy $E_0$, at the end you get $$E_0=E_{block}+E_{pulley}$$ where $E_{pulley}$ is the kinetic energy of the pulley. If the pulley is massless ($M=0$), than $E_{pulley}=0$ (because $E_{pulley}\sim M=0 $ so the problem is easier.
I won't solve your homework unless you put some effort into trying on your own, but the steps are: