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Homework Help: Acceleration of a falling tower

  1. Dec 8, 2008 #1
    This is problem 67 in Chapter 10 of Halliday/Resnick/Walker 8th Edition. This is one of their three star (= hard) problems. I can get the beginning but then get stuck.

    A cylindrical chimney of height L=55m falls over. Treat it as a thin rod. Get the radial acceleration and tangential acceleration of the top when it makes an angle of 35 degrees with the vertical. Then find the angle where the tangential acceleration will equal g (= the acceleration of gravity).

    Okay, I can get the radial acceleration. Here's the workings: Using the parallel-axis theorem, the rotational inertia of the rod is I=(1/12) m L + m (L/2)^2=(1/3) m L^2 (with m=mass of rod). The first term is the I of the rod around its center of mass (com), which is at L/2, and the second term is the mass of the rod times the distance^2 between the com and the base of the rod.

    The com of the rod (at L/2) starts with a potential energy of U= m g (L/2). When it is leaning at an angle of 35 degrees from the vertical, the height of the com is now (L/2) cos(35)=22.52m. So the change in U is m g ( L/2) (1-cos(35)). This change equals the kinetic energy which is K=(1/2) I w^2 where w=the angular velocity. But we know I is (1/3) m L^2. Plugging this in and solving for w gives w=SQR( 3 g (1-cos(35))/L)= .311 rad/s.

    The velocity of the end of the rod is v=w L=17.1m/s. The radial acceleration is ar=v^2/L=5.32m/s^2. This is the answer in the book.

    Now, for the tangential acceleration, where I'm stuck. At first I figured that the vector sum of the radial acceleration (ar) and the tangential acceleration (at) should equal the acceleration of gravity, since the tower is falling. So this would be g^2=ar^2+at^2, and solving for (at) gives at=8.23m/s^2. But the book says 8.43m/s^2. I've done it using all the significant figures in the calculator and I don't think its a rounding problem. So then I started wondering why I thought the end of the rod should fall at g. Maybe the center of mass (com) at L/2 should fall at g, and the end of the rod is being "whipped" to a higher acceleration. I tried g^2=ar^2+at^2 for the midpoint, which is easy since w is the same. This give ar(com)=2.658m/s^2 and at(com)=9.43m/s^2. I thought there would be a proportion between this and the (at) at the distance L, and the change in distance, but this doesn't seem to work either.

    Any guidance much appreciated.
  2. jcsd
  3. Dec 8, 2008 #2
    I got 8.43 m/s2 by calculating the torque at the base caused by mg at the center of mass at 35o, then use the moment of inertia to get the acceleration at the top.
  4. Dec 8, 2008 #3
    Clever! I like it. The problem says as a hint not to use torque, which is why I went the way I did. Is there a way to get the answer with my approach? What I mean is, from ar, without using torque?
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