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This is from a practice exam, I've been sitting here thinking about it for over an hour and can't convince myself of an answer, or write down any relevant exact equations.

A bar of uniform density $\rho$, length $L$, and negligible width is suspended by two inextensible ropes of negligible mass.

The second rope is cut, what is the tension of the other rope shortly after cutting? What is the tension's rate of change?

I've been trying to work it out with numbers $g=10 m/s^2$, $\rho=30kg/m$ and $L=1m$ just to keep it simple. Obviously before the rope is cut the tension of both ropes is $T_1 = T_2 = 150N$ and after the bar falls down and is stationary, then $T_1 = 300N$ so there is some smooth function of the tension (I think). The angle between the first rope and the bar is $90\deg$ and increases to $180\deg$, and the force due to gravity can be divided into normal and tangential forces. The normal force increases the angular momentum, and the normal force is a maximum the moment the rope is cut. I'm having trouble putting this all together, but I think I am on the right track. Thanks in advance.

edit: As stated below, this is the force of gravity components $F_N = \rho Lg Sin(\theta )$ and $F_T = -\rho Lg Cos(\theta )$

I still have trouble applying angular momenta to this problem.

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  • $\begingroup$ Did you do a free body diagram, which includes inertial forces? $\endgroup$ – ja72 Aug 16 '13 at 0:37
  • $\begingroup$ @ja72 yes, but that doesn't illustrate the angular momentum of the bar very well. Using symmetries (note that $\theta$ goes from $90\deg$ to $180\deg$) I calculated $F_N = \rho Lg Sin(\theta )$ and $F_T = -\rho Lg Cos(\theta )$ $\endgroup$ – walczyk Aug 16 '13 at 0:51
  • $\begingroup$ The problem has two degrees of freedom. One is the rod orientation and the other is the rope angle. Solving for both is doable, but rather complex. Getting the results immediately after the cut is a much simpler problem because the rod is horizontal and the rope vertical, and nothing has velocity. $\endgroup$ – ja72 Aug 16 '13 at 13:47
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Assume horizontal Rod

Sum of forces equals acceleration of center of gravity

$$ T - m g = m \ddot{y}_C $$

Sum of moments about center of gravity equals mass moment of inertia times angular acceleration

$$ T \frac{L}{2} = I \ddot{\theta} $$

As the bar rotates by a small angle $\theta$ the center of gravity height is found by $y = -\frac{L}{2} \theta$ or by differentiating twice, $\ddot{y} = -\frac{L}{2} \ddot{\theta}$

Together it all comes as

$$ \begin{aligned} T & = m g - m \frac{L}{2} \ddot\theta \\ T \frac{L}{2} & = \left( \frac{m}{12} L^2 \right) \ddot \theta \end{aligned} $$

Solve the above for $T$ and $\ddot{\theta}$. Hint the value does not have to be more than $\frac{m g}{2}$.

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  • $\begingroup$ How do you get $T\frac{L}{2}$? I don't follow that, but I understand the rest completely. $\endgroup$ – walczyk Aug 16 '13 at 1:51
  • $\begingroup$ Solving for T I got mg/4 as an answer, which doesn't seem to make sense, why would the tension be cut in half? $\endgroup$ – walczyk Aug 16 '13 at 4:04
  • $\begingroup$ Personally I would do the moments calculation about the point of suspension by the second rope. That way you wouldn't have to worry about any issues arising from the fact that your centre of mass is not in an inertial frame (it is accelerating). $\endgroup$ – G. Paily Aug 16 '13 at 7:20
  • $\begingroup$ Are you assuming that the rope doesn't oscillate? $\endgroup$ – pppqqq Aug 16 '13 at 11:59
  • $\begingroup$ $T \frac{L}{2}$ is force times distance, for the total moment applied on the center of the rod. $\endgroup$ – ja72 Aug 16 '13 at 13:45

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