This question is from HRW's Fundamentals of Physics. It goes:

A tall, cylindrical chimney falls over when its base is ruptured. Treat the chimney as a thin rod of length $55.0\;m$. At the instant it makes an angle of $35.0^\circ$ with the vertical as it falls, what is the radial acceleration of the top?

What I did is this:

The energy of the top-most point at the chimney must've been conserved. Therefore: $$K_0 + U_0 = K + U$$ $$\Rightarrow mgl = \frac{1}{2}mv^2 + mgl\cos{35^\circ}$$ $$\Rightarrow \frac{v^2}{l} = 2g(1 - \cos{35^\circ})$$

Now $\frac{v^2}{l}$ is the radial acceleration, considering $l$ to be the length of the chimney, in other words the radius of the circle which is the path that the top of the chimney takes. Calculating that expression gives us $\frac{v^2}{l} = a_r = 3.54\;m/s^2$. However the correct answer according to the book is $a_r = 5.32\;m/s^2$.

Where did I make a wrong assumption?


closed as off-topic by John Rennie, David Hammen, Jon Custer, sammy gerbil, ZeroTheHero Aug 22 '17 at 0:56

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – John Rennie, David Hammen, Jon Custer, sammy gerbil, ZeroTheHero
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ Hi and welcome to the Physics SE! Please note that we don't answer homework or worked example type questions. Please see this Meta post on asking homework/exercise questions and this Meta post for "check my work" problems. $\endgroup$ – John Rennie Aug 20 '17 at 4:21
  • 1
    $\begingroup$ Doesn't that post say that homework questions are allowed? And even gives a guideline for good ones: "A good homework question states the problem clearly, shows an attempt to work through it, and identifies the specific issue that is giving the questioner trouble. [...]" Forgive me if I missed some part of the post saying that it is indeed not allowed, if that's the case I'll promptly delete my question. $\endgroup$ – Deathkamp Drone Aug 20 '17 at 4:57
  • $\begingroup$ Since the rest of the chimney is exerting both radial and tangential forces on the top-most point, its sum of potential and kinetic energy need not be constant... $\endgroup$ – DJohnM Aug 20 '17 at 8:12
  • 3
    $\begingroup$ The question shows an attempt and what assumptions went in, also the OP only asks for a pointer into the right direction. I think that this is okay as-is. $\endgroup$ – Martin Ueding Aug 20 '17 at 8:12
  • $\begingroup$ As the question hints use energy conservation for the rod to find the angular speed of the rod remembering that you have both rotational and translational kinetic energy in play. $\endgroup$ – Farcher Aug 20 '17 at 8:24

You made the wrong assumption when you said:

The energy of the top-most point at the chimney must've been conserved.

Energy is only conserved for isolated systems. Your assumption would be correct if the top of the chimney was disconnected from the remainder. However, there are forces within the chimney that distribute energy from one part to the other.

You will need to use the moment of inertia of the solid rod (therefore the hint in the problem) and use angular momentum and torque.

  • $\begingroup$ Could the downvoter please tell me what the problem with this answer is? Did I give away too much? $\endgroup$ – Martin Ueding Aug 20 '17 at 9:14
  • $\begingroup$ I see. The chimney as a whole however is isolated since it left the ground, so we should apply conservation of energy to that instead, that was my mistake. $\endgroup$ – Deathkamp Drone Aug 20 '17 at 18:29

For completeness, I shall post the actual solution. The energy of the entire chimney is conserved, therefore:

$$K_{0_{chimney}} + U_{0_{chimney}} = K_{chimney} + U_{chimney}$$ $$\Rightarrow mg\frac{l}{2} = \frac{1}{2}I\omega^2 + mg\frac{l}{2}\cos35^\circ$$ $$\Rightarrow mgl = \frac{1}{3}ml^2\omega^2 + mgl\cos35^\circ$$ $$\Rightarrow \frac{1}{3}l\omega^2 = g(1 - \cos35^\circ)$$ $$l\omega^2 = 3g(1 - \cos35^\circ)$$

The angular velocity $\omega$ is the same at all points of the chimney, including at the top, and therefore we have arrived at the solution, for the radial acceleration at the top is given by $a_r = l\omega^2$. So the value of the acceleration is $l\omega^2 = 3g(1 - \cos35^\circ)$ which indeed evaluates to $5.32\;m/s^2$.


Not the answer you're looking for? Browse other questions tagged or ask your own question.