Perpendicular weight force on an object that is tipping over [closed]

I'm currently working on a problem I can't seem to find an answer to. I have an object that is hanging over a cliff. This object is exactly 12m in length, and it starts off in equilibrium (6m over the cliff, 6m on the ground). Now, the cliff is crumbling away with a velocity of 0.3 m/s. I would like to find a function that gives me the torque (of its weight force) acting on the object in dependance of time (My ultimate target would be the angular acceleration, but I have already found a working formula for the moment of inertia). If we call the meters of the object currently suspended over the cliff "b" and the current angle alpha, then the torque at any time is $$(b-6) \cdot mg \cdot \cos(\alpha)$$. The problem here is evident: We don't know alpha, since that's literally what I want to find! The solution is most probably going to involve integration, I tried finding something using the small steps method to then put it all into an integral, but I can't for the life of me find something that works. This is currently the best I can come up with:

$$\alpha_s(t)$$ is the time to angular acceleration function $$\frac{\tau}{I}$$ for the system still in equilibrium, so $$\tau = (6-(6+0,3t))mg = -0,3mg$$

$$\alpha(t) = mg \cdot (0.1 \cdot \int_0^{0.1}\alpha_s(t) dt + 0.1 \cdot \int_{0.1}^{0.2}\alpha_s(t) \cdot \cos(0.1 \cdot \int_0^{0.1}\alpha_s(t) dt) + \cdots )$$

Or in other words, $$a_1 := 0.1 \cdot \int_0^{0.1}\alpha_s(t) dt \\ a_{n+1} := 0,1 \cdot \int_{\frac{n}{10}}^{\frac{n+1}{10}}\alpha_s(t) \cdot \cos(a_n) dt \\ \alpha(t) \approx \sum_0^t a_n$$

I don't really have an idea of how to pack this mess into an integral, I'd be really grateful if someone could help me out. Thanks.

closed as off-topic by John Rennie, Jon Custer, user191954, Aaron Stevens, ZeroTheHeroNov 4 '18 at 2:13

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I'm going to put this in dimensionless form: t is relative to √(l/g), with length of rod 2l, so that in the equation l just becomes 1.

Let velocities be dimensionless - relative to √(gl), and let θ be the angle through which the rod has tipped. The mass just cancels out, as it simply multiplies both sides.

It's very unlikely it has an analytical solution; but it's orobably a good one for the Runge-Kutta numerical method - it would give you an extremely precise numerical solution in a trice.

Just incase you aren't familiar with recasting the problem in terms of these 'dimensionless' variables: it's a standard procedure in computational mathematics & physics. It might look awkward, but it colossally simplifies matters. You can then substitute any actual numbers just by scaling the abstract dimensionless solution. Well -worth getting to grips with!

If anyone does know how to solve the differential equations I come up with here analytically, I would be extremely interested!

Take care to apply torque = (Iω)' rather than Iω', if the moment of inertia is changing. This is the kind of problem that lends itself to being cast in terms of varying moment of inertia - but it's not necessary; and I think in my attempt I will take the centre of mass as the point about which rotation is defined. If you were to use the edge instead though, you would have a varying moment of inertia.

It's a very tricky problem; & I'm wondering whether in the end all those terms that make the resulting differential equation so intractable might not cancel out, and the problem pan out being actually quite tractable.

I'm also wondering whether it might not better casting the problem as that of a rod sliding over an edge with initial velocity v. I think considering rotation about the centre-of-mass might be much easier as well. but then there is a reaction force at the edge as well, this time - say $$F$$, normalised to $$mg$$, and introduce the assumption that there is no friction at the edge, so that $$F$$ acts only normally on the rod . Let the horizontal & downward vertical displacrments (normalised to $$l$$, be $$x$$ & $$y$$, respectively).

$$\dot{\dot{x}} = F\sin{\theta}$$ ...... (i)

$$\dot{\dot{y}} = 2 - F\cos{\theta}$$ ...... (ii)

$$2\dot{\dot{\theta}}/3 = F\sqrt{x^2 + y^2}$$ ...... (iii)

$$\cos{\theta} = \frac{x}{\sqrt{x^2 + y^2}}$$ ...... (iv)

$$\sin{\theta} = \frac{y}{\sqrt{x^2 + y^2}}$$ ...... (v)

Time derivatives of (iv) & (v):

$$\sin{\theta}.\dot{\theta} = \frac{xy\dot{y}-y^2\dot{x}}{(x^2 + y^2)^\frac{3}{2}}$$

$$\therefore\dot{\theta} = \frac{x\dot{y}-y\dot{x}}{(x^2 + y^2)}$$ ...... (vi)

$$\cos{\theta}.\dot{\theta} = \frac{x^2\dot{y}-xy\dot{x}}{(x^2 + y^2)^\frac{3}{2}}$$

$$\therefore\dot{\theta} = \frac{x\dot{y}-y\dot{x}}{(x^2 + y^2)}$$ ...... (vi)

$$\dot{\dot{\theta}} = \frac{(x^2+y^2)(x\dot{\dot{y}}-y\dot{\dot{x}})+2(xy.(\dot{x}^2-\dot{y}^2)-\dot{x}\dot{y}(x^2-y^2))}{(x^2 + y^2)^2}$$ ...... (viii)

Substituting these back into (i) & (ii):

$$\dot{\dot{x}} = \frac{2y((x^2+y^2)(x.\dot{\dot{y}}-y\dot{\dot{x}})+2(xy.(\dot{x}^2-\dot{y}^2)-\dot{x}\dot{y}(x^2-y^2)))}{3(x^2 + y^2)^3}$$ ...... (x)

$$\dot{\dot{y}} = 2 - \frac{2x((x^2+y^2)(x\dot{\dot{y}}-y\dot{\dot{x}})+2(xy.(\dot{x}^2-\dot{y}^2)-\dot{x}\dot{y}(x^2-y^2)))}{3(x^2 + y^2)^3}$$ ...... (xi).

I don't know whether I've done it right this time. I do make errors (as you've proably seen - I've already erased much of this answer!) ... but I think if there were some scope for simplification, I would have had some kind of clue by now, & I'm just not getting any. Usually, even if I'm beginning to make errors, I at least glimpse leads - "if that were such & such then all those would drop out ..." - that kind of thing. So I think this is a sheer number-crunching type problem; and as number-crunching problems go, it isn't too bad. The Runge-Kutte method would dispatch it with ease.

I think your tutor might possibly be showing you an example of how even a very simply defined problem, such as pitching a horizontal rod over an edge - even a frictionless one - can be hardly tractable.

Could possibly make one last attempt using polar co-ordinates & the wellknown kinematic relations for radial & tangential accelerations in terms of derivatives of $$r$$ & $$\theta$$.

$$Fr = 2\dot{\dot{\theta}}/3$$ ......(i)

$$r\dot{\dot{\theta}}+2\dot{r}\dot{\theta}=2\cos{\theta}-F$$ ......(ii)

$$\dot{\dot{r}}-r\dot{\theta}^2 = 2\sin{\theta}$$ ......(iii) .

Substituting (i) into (ii) gives

$$(r+\frac{2}{3r})\dot{\dot{\theta}}+2\dot{r}\dot{\theta}=2\cos{\theta}$$ ...... (iv).

$$(3r^2+2)\dot{\dot{\theta}}-6r(\cos{\theta}-\dot{r}\dot{\theta})= 0$$...... (v).

Equations (iii) & (v) are much more pleasant... but I still don't see a clue to an analytical solution. But they are still more pleasant for plugging into a numerical solution algorithm. The symmetry of the equations in terms of $$x$$ & $$y$$ strongly suggests that perhaps two mutually orthogonal linear combinations of the $$x$$ & $$y$$ would be a more natural choice of variables; and what more natural than distance along the rod & perpendicular to it!?

I would just like to apologise to the OP for 'thrashing-out' this problem 'in your face' so to speak, such that you had to witness the spectacle of all my dumb mistakes. It might have been better if I'd worked it out abitt more before setting it out in public. I think I'm through with it now, though!

• Welcome to Physics! Note that we have an equation editor built into the site which helps for readability of your post. – Kyle Kanos Nov 2 '18 at 9:52
• Yes thankyou! I've noticed that other people are using ssomething of that kind. I think it probably is overdue that I get to grips with it! – AmbretteOrrisey Nov 2 '18 at 9:54
• Thanks heaps for making such an effort in answering my question. I’m working on this problem for a project in high school - I found it myself by analyzing a movie scene, so it’s not from any tutor. I am suprised though that it turned out to be so difficult! I took a look at the Ringe-Kutta method (hadn’t heard of it before - sounds really interesting though, I’m definitely going to read into it). Thanks again! – MateInTwo Nov 6 '18 at 19:33
• Not heard of the Runge-Kutta method!!? Oh! you'll love it! It's a colossally powerful & ingenious method for the numerical solution of differential equations. A veritable workhorse! (quoting a line that has acquired some degree of fame from a certain textbook (Numerical Recipes)). You begin to realise, when you start trying your own problems just how 'filtered' high school problems are! Your welcolme to my contribution - it was a pleasure. Perhaps the most important bit of it though is that about putting equations in dimensionless form - I very strongly urge get to grips! – AmbretteOrrisey Nov 7 '18 at 1:22