Angular velocity tank receiving cereal

AI Thread Summary
The discussion revolves around calculating the angular velocity of a cylindrical container receiving cereal while rotating. The initial moment of inertia is given as 2.0 kgm², with the container rotating at 20 RPM. The cereal falls at a rate of 0.5 kg/s for 10 seconds, adding a total of 5 kg to the container's mass. Participants emphasize using the conservation of angular momentum to find the final angular velocity, noting that the total moment of inertia will be the sum of the container's inertia and that of the added cereal. Clarity on terminology, specifically using "moment of inertia" instead of "rotational inertia," is also discussed.
Granger
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Homework Statement


The cylindrical container, when empty, has the moment of inertia of 2.0 Kgm^2 around the axis of rotation. It is rotating freely on its axis 20 revolutions per minute, when it begins to receive cereal, which falls vertically along its axis at the rate of 0.5 kg-1. The radius of the container is 0.40m. Calculate the angular velocity after 10 seconds.

Homework Equations



$$ \tau = I \alpha $$
$$ L = I \omega$$
Conservation of angular momentum

The Attempt at a Solution


I am a bit lost with this problem.
At first I thought about using conservation of angular momentum
$$L_i=L_f $$

$$I_i= 2.0 kgm^2$$
$$\omega _i = 2\pi /3$$

My problem is how can I get the final rotational inertia. The only thing I know is that the new mass is the first one + 5 kg (0.5 * 10s) and that the radius is 0.40 m.
But I don't know the expression of the rotational inertia...
 
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After it has settled, you should be able to consider the grain as a uniform cylindrical disc, rotating with the bin.
 
scottdave said:
After it has settled, you should be able to consider the grain as a uniform cylindrical disc, rotating with the bin.

And then the total rotational inertia should be the sum of the container one and the disk one?
 
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Granger said:
And then the total rotational inertia should be the sum of the container one and the disk one?
Yes. Though I would be more comfortable if you had said "moment of inertia" rather than "rotational inertia".
 
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