Angular Momentum: Uniform Cylinder Drum Rotation

AI Thread Summary
A uniform solid cylindrical drum with a mass of 1.5 kg and radius of 0.5 m rotates about a fixed axis at an initial angular speed of 4 rad/s. When a 0.3 kg block attached to a string is jerked into motion, the problem involves calculating the new angular speed of the drum, which is expected to be 2.86 rad/s. The discussion emphasizes the conservation of angular momentum, noting that the moment of inertia changes as the block pulls on the drum. Participants clarify that the block can be treated as a point mass at a distance of 0.5 m from the axis of rotation, which is crucial for accurate calculations. The conversation concludes with a confirmation of the approach and understanding of the dynamics involved.
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A uniform solid cylindrical drum of mass 1.5kg and radius 0.5m is free to rotate about a fixed, smooth, horizontal axis which coincides with the axis of the cylinder. The axis is at a height of 2m above a horizontal table, and a light string of AB of length 4m has one end attached to the heighest point of the cylinder. A block of mass 0.3kg is attached to end B of the string and rests on the table. The drum begins to rotate at a constant angular speed of 4 rad s-1 in a clockwise direction. Calculate the angular speed of the drum immediately after the block is jerked into motion.

I have no idea how to approach this question! The answer is supposed to be 2.86 rad s-1.

Any help is greatly appreciated. :smile:
 

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I'm looking at this and the only thing i can think of is to find the angular acceleration of the drum due to the block pulling it down.
 
I think the way I'm supposed to approach this question is by equating the sums of initial and final angular momentum for both the drum and the block. I can do the drum just fine, but all my attemps of finding the angular momentum of the block didn't produce the required answer.

Here's what I get for the drum:
initial angular momentum = moment of inertia * angular speed = 3/16 * 4 = 0.75 Nms
final angular momentum = 3/16 w, where w is the new angular speed
 
Ohhh right. Ofcourse.

The final moment of inertia has changed, because its now pulling the block. You can simplify the block as a point mass at point A, and find the new MOI.
 
I tried that:
final angular momentum = [3/16 + 0.3*(2^2 + 0.5^2)] w = 1.4625w
conservation of angular momentum => w = 0.75/1.4625 = 20/39

Which isn't the right answer. :frown: Am I doing something wrong?
 
final angular momentum = [3/16 + 0.3*(2^2 + 0.5^2)] w = 1.4625w

Your radius isn't right. How far is the point A from the axis of rotation?
 
If I take it to be at point A then I get the right answer! But.. Why should it be at point A?
 
I thought about it a bit and this is what I concluded:
Initial jerk turns the drum 4 rad, and so the block is pulled up 4*0.5 m, because A turns an arc length r \theta. So the block adheres to the circumference of the circle and can be considered as a point mass 0.5m away from the axis of rotation.

Is this correct?
 
Exactly, the block is 'pulling' or resisting motion AT the radius 0.5m, thus you can treat it as a particle in motion at the edge of the drum. Good thinking.
 
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Thanks for the help -- and good night!
 
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