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Torque and acceleration

  1. Aug 7, 2008 #1
    Question: Assume that the Titanic is drifting in Southampton harbor before its fateful journey and the captain wishes to stop the drift by dropping an anchor. The iron anchor has a mass 4000 kg. It is attached to a massless rope. The rope is wrapped around a reel in the form of a solid disk of radius 0.150 m and mass 400 kg, that rotates on a frictionless axle.

    a) While the anchor continues to drop through the water, the water exerts a drag force of 2500 N on it. With what acceleration does the anchor move through the water?

    b) While the anchor drops through the water, what torque is exerted on the reel?

    There is a couple of previous parts to this question, the first part asked what was the acceleration of anchor as it fell through the air and it came out to be 9.33 m/s2. To solve that I substitued for moment of inertia with 1/2MR^2 and moved things around on the equation ma = mg -T. For part a here I simply used my equation of combining torque and free body diagram and subtracted 2500N from the total weight of the anchor but that does not give me the correct answer. Part B should be easy if part A is solved. Help please? Thanks
  2. jcsd
  3. Aug 7, 2008 #2


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    If the anchor is accelerating through the water then energy is also going into accelerating the 400 kg reel that the line is unspooling off. I don't sense that you have accounted for that.

    Edit: Sorry I misunderstood your first first statement. That's how you arrived at the 9.33m/s^2. So yes it seems that the drag should be subtracted from the weight and of the same sign as the torque of the anchor line spool. But I think you should show your arithmetic as in looking at you 9.33 number I believe that you didn't account for the angular momentum and angular acceleration correctly.
    Last edited: Aug 7, 2008
  4. Aug 8, 2008 #3
    Basically what I did to solve the first part of the question (which asked what was the acceleration as it feel through the air, neglecting air resistance) was ma = mg - T. To solve to tension, I also used torque, which is torque = F x r = I times angular acceleration. If you substitute for I by using 1/2MR^2 and plug that in for tension, I came to the correct solution of 9.33 m/s2. Now using that same set up equation, I simply subtracted the 2500N from the total weight to account for the force of the water and I assumed that that would give me the correct acceleration but it is not. Anyone have ideas on what else to do next?
  5. Aug 8, 2008 #4


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    But angular acceleration is different from the "a" in F = ma is it not?
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