Let's look at the given informations. We have an initial angular velocity, and we know how many revolutions it takes to get to zero angular velocity. So we have $\omega_i$, $\omega_f$, and $\theta$. Looking at that list of givens I would guess to use
$$
\omega_f = \alpha t+\omega_i \quad\text{and}\quad \theta_f=\frac{1}{2}\alpha t^2 + \omega_i t + \theta_i
$$
This has all the items we are interested in, and all the givens, and only time as an extra piece. Then we have two equations and two unknowns. Solving for $t$ in the first one we get
$$
t=\frac{\omega_f -\omega_i}{\alpha}
$$
Then the second equations becomes after substituting in
$$
\theta_f = \frac{1}{2}\alpha \left( \frac{\omega_f -\omega_i}{\alpha} \right)^2 + \omega_i \left( \frac{\omega_f -\omega_i}{\alpha} \right)
$$
$$
\alpha=\frac{1}{2\theta_f}(\omega_f -\omega_i)^2+\frac{\omega_i}{\theta_f}(\omega_f -\omega_i)
$$
now we know that the final speed is zero, so put that in and simplify.
$$
\alpha=\frac{1}{2\theta_f}(\omega_i)^2-\frac{\omega_{i}^{2}}{\theta_f}=-\frac{1}{2}\frac{\omega_{i}^{2}}{\theta_f}
$$
Hope this helps.
Hint : Friction opposes tendency to move. Tension is produced if the string is stretched $very$ slightly. So, increase friction to maximum and then tension will act if necessary.
Ironically, you are thinking absolutely right. Give yourself a cookie.
From part $a$, we know that the blocks will be at rest at all angles below that.
You are also right as at very small angles there is no need of tension and we can ignore it to solve for, again, an angle condition. You have done excellent work. Congrats.
Now we come to the middle angles. Oh... they drive you insane, don't they?
Let's start. We can start our analysis from 2 blocks, 1 will give a contradiction and other will give a result, but I will start with the one giving contradiction. This will help you.
All angles are in degrees :
$\theta=35 $
Lets start by analysing Block A (No racism intended)
Gravity is trying to pull it down : $5*10*\sin(35)N=28.67N$
Friction comes to the rescue(up) : $50*\cos(35)N=16.38N$ // read my hint to know why friction is put max here
As it is at rest, $T=12.29N$
Now Block B is also at rest,
Weight = $114.71N$
max f= $81.92N$
$16.38+114.71=12.29+f$
$f=118.8N$
OOPS, it exceed max value. So, Lets start by analysing Block B. (I love alliteration)
Gravity trying : $114.71N$
Friction comes to the rescue(up) : $81.92N$
You can take from here I guess. calculate tension. Note that you have to revise your calculation for tension again as reaction friction force will be provided by A. Better assume it $f$ from starting FBD of B.
This will yield the correct answer. Friction will be less than max value for upper block. In most cases, You should start analysing with heavier block(my experience). Hope your doubts are cleared.
Best Answer
1) The first thing I notice is that you have stated that the velocity at the end of the ramp is $2\textrm{ m/s}$. Remember that the can is accelerating as it rolls down the ramp, so the equation $v=\textrm{d}s/\textrm{d}t$ is not applicable here for finding the instantaneous velocity at the bottom. The can does indeed average $2\textrm{ m/s}$ during its trip, but this is not the final velocity of the can. Use this new corrected value to calculate angular frequency.
2) I find this problem simpler to solve using energy analysis. Take the can's initial potential energy:
$$ E_\textrm{pot} = m g h = 3.234\textrm{ J}\quad.$$
We also know that the final kinetic energy of the can must equal this due to the conservation of energy, but the final energy of the can must be broken into translational kinetic energy (due to the can's movement) and rotational kinetic energy (due to the rotation). (This is why your solution above was giving incorrect answers as it didn't take translational kinetic energy into account.) Thus, we also know that:
$$ E_\textrm{pot}(t=0) = E_\textrm{kin,trans}(t=1.5\textrm{ s}) + E_\textrm{kin,rot}(t=1.5\textrm{ s})\quad,$$
which, for our case, is
$$ 3.234\textrm{ J} = \frac{1}{2} mv^2 + \frac{1}{2}Iω^2\quad.$$
Plugging in known values to this equation with the correct value for angular velocity $\vec \omega = \vec r \times \vec v$ gives the accepted answer:
$$ I = 0.000187\textrm{ kgm}^2\quad.$$
As for part B, the height of the can is irrelevant because as long as we know the mass and radius of the can, we can solve the problem. The ‘extra mass’ resulting from lengthening the can would be centered about the can's original center of mass, and as such the moment of inertia would not be affected for this problem.