For a, I did:
$width = 4m$ ; $F = 19 N$ ;$ work = 205 rev/min$ or $3.42 rev/sec$.
Then, $m=F/g = 19/9.8m/s^2 = 1.94 kg$
Then, $a = 1/12mr^2$
Then $1/12 (1.94 kg)(4m^2)$
So, $I = 2.6 kgm^2$
For b, I did:
momentum = Iw
= 2.6kgm^2 * 3.42 rev/s = 56kg/m^2/s
Are those right?
Let $E$ be a rod of length $L$ and radius $r_1$. The rod is rotating about one end, the moment of inertia is given by the definition:
$$I=\iiint_{E} r^2 dm$$
Where $r$ is the distance from the axis of rotation, and $m$ is referring to mass. But assuming constant density, density is $\frac{dm}{dV}=\delta=\text{const}$ and so $dm=\delta dV$ so we have,
$$I=\iiint_{E} r^2 \delta dV$$
$$=\delta \iiint_{E} r^2 dV$$
Let the rod be around the $z$ axis and start at $z=0$ then to $z=L$, define the $z=0$ end to be the one we are rotating about. Then we have:
$$I=\delta \iiint_{E} z^2 dV$$
Now we switch to polar, do not confuse the $r$ here below to the previous $r$ in the formula for moment of inertia.
$$=\delta \int_{0}^{2\pi} \int_{0}^{r_1} \int_{0}^{L} r z^2 dz dr d\theta$$
$$=(\delta)(2\pi)(\frac{r_1^2}{2})(\frac{L^3}{3})$$
$$=\frac{\delta (\pi r_1^2 L)}{3}L^2$$
$$=\frac{(\delta)(V(E))}{3}L^2$$
Here $V(E)$ denotes the volume of $E$ and note volume•density=mass for constant density.
$$=\frac{M}{3}L^2$$
This result does not agree with the formula you are using.
You have used the formula for rotational inertia of a thin rod rotating about an axis perpendicular to the rod through the center of mass of the rod. But the rod in the question is rotating around an axis that passes across one end of the rod, not through the center.
In the formula for rotational momentum, $I\omega,$ the correct units for $\omega$ in this question would be radians per second, not revolutions per second.
The formula for moment of inertia you're using is wrong.
The moment of inertia of a uniform rod about an axis through its centre is $I_1 = \frac{1}{12}ml^2$
whereas the moment of inertia of a uniform rod about an axis through one end is $I_2 = \frac{1}{3}ml^2$
You can work one out from the other by a trivial application of the parallel axis theorem.
In this case, you are to use the second formula. So $I = \frac 13 ml^2$.
Convert all values to SI units. Here $m = \frac Wg \approx \frac{19}{9.81} \approx 1.94 \mathrm{kg}$ So the moment of inertia is $\frac 13 (1.94)(4^2) = 10.33 \mathrm{kg.m^2}$ (to 3 significant figures, which is the expected level of precision).
You might get a slightly different answer depending on the value of $g$ you're given or supposed to take.
For the angular momentum, use $L = I\omega$ (note that large $L$ here is not length but angular momentum). Convert $\omega$ to SI units by finding $2\pi f$, where $f$ is the frequency of revolution in Hertz. So in this case, $\omega = 2\pi\frac{205}{60} \approx 21.5 \mathrm{rad/s}$. Hence the final answer is $10.33*21.5$ (taking more figures in the intermediate calculation), which should give you: $222 \mathrm{kgm^2/s}$
