Homework Help: Rod hanging by strings

1. Sep 29, 2015

Vibhor

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

When string joining B is burnt , the forces acting on the rod are its weight , and tension in the left string ,both being vertical ,the acceleration of its COM would be vertical . Hence option A is correct .

Now how should particle at A move ?

Thanks

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2. Sep 29, 2015

Hesch

Why don't you try it out?
http://www.buero-netshop.de/shop/images/products/main/q-connect-lineal-20cm-kunststoff-transparent.jpg
Hang it ( in the hole, using sewing thread ) and hold it ( at 20cm ). Let it go.

Watch what happens.

3. Sep 29, 2015

haruspex

This is not a valid test. Yes, over time there will be a motion at A, and the motion will initially be horizontal. But this does not prove that instantaneously there is a horizontal acceleration at A. It could be that the first and second derivatives of x are zero, but the third is positive.

4. Sep 29, 2015

Hesch

If x is the position, I thought that the acceleration ( which is the question ) is the second derivative: d2x/dt2.

5. Sep 29, 2015

haruspex

Quite so, but my point is that at t=0 it may be that both $\dot x$ and $\ddot x$ are zero. The first nonzero derivative will surely be positive, but it might be third derivative.

6. Sep 29, 2015

Hesch

At t = 0 the acceleration of A will be positive to the right because the thread at A is vertical, thus not yielding any horizontal force ( B is burned ). The rod will begin to rotate clockwise, and the center of the rod will be kept in the same horizontal position. Thus A must accelerate to the right, and B must accelerate to the left.

We are speaking of the acceleration at t = 0+ ?

7. Sep 29, 2015

Vibhor

This is a possibility . But why must A accelerate to the right only . It can accelerate upwards and B can accelerate downwards such the COM of the rod accelerates downwards ?

Can this be proven mathematically ?

Last edited: Sep 29, 2015
8. Sep 29, 2015

Staff: Mentor

Can you show us your FBD for t=0+? The sum of torques and forces are non-zero for t=0+. That should help you start to answer this question...

9. Sep 29, 2015

Vibhor

Tension acts vertically upwards and weight acts vertically downwards on the rod .

Agreed . But how does that show that particle at A accelerates horizontally ?

10. Sep 29, 2015

Staff: Mentor

11. Sep 29, 2015

Vibhor

Please see the attached picture .

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12. Sep 29, 2015

Staff: Mentor

So given that FBD, what is the sum of forces and sum of moments on the bar?

13. Sep 29, 2015

Vibhor

Please understand that merely having a net force acting and a net torque around the COM doesn't prove that the particle at A accelerates horizontally .

14. Sep 30, 2015

haruspex

Consider a particle moving in one dimension according to x=ct3. What is its acceleration at t=0?
However, rereading the question, I see that it seems to be asking about the acceleration at some time marginally greater than zero. This strikes me as strange since offering the two options as either horizontal or none would make more sense at t=0. Moreover, as soon as you allow t>0 there will be a vertically upward acceleration too, making neither answer correct.

Anyway, it is possible to prove there is a nonzero horizontal acceleration even at t=0, but it takes a little more analysis than the observational approach you posted.

Last edited: Sep 30, 2015
15. Sep 30, 2015

Hesch

The problem statement says:

So, what is meant by: "just after this"?

Well, the acceleration = d2x/dt2. Now, some physicists think that the time itself is quantified into amounts of ≈ 1E-49s. So if we write:

v(t) = dx(t)/dt

we must add: "assuming dt >> 1E-49s". Otherwise the equation doesn't make sense. So I think that the author by "just after this" means something like "within the next μs".
Should I consider what will happen within the next 1E-52s, I would get crazy. I don't want to get into that discussion due to quantified time.
I don't understand your conclusion here? As for the vertical accelerations Bacc ≈ g , centeracc ≈ ½g , Aacc = 0.

16. Sep 30, 2015

Staff: Mentor

If there is a net torque around the COM, what does that tell you about the angular acceleration around the center of mass? How is the velocity of particle A related kinematically to the velocity of the center of mass and the angular velocity (vectorially)?

17. Sep 30, 2015

Vibhor

Hello Sir ,

There is a net angular acceleration around the center of mass = Net torque / MI

$\vec{v}_A = \vec{\omega} \times \vec{r} + \vec{v}_{CM}$

18. Sep 30, 2015

Staff: Mentor

Actually, this is close, but is not quite correct. The correct relationship is:

$$\vec{v}_A = \omega r\vec{i}_θ + \vec{v}_{CM}$$
where $\vec{i}_θ$ is a unit vector in the circumferential (tangential direction). The time derivative of this is the acceleration of particle A. If θ is the angle that the unit vector in the tangential direction makes with the positive x axis, how is the unit vector in the tangential direction related to the unit vectors in the x and y directions?

Chet

19. Sep 30, 2015

Vibhor

$\vec{i}_θ = cosθ\hat{i}+sinθ\hat{j}$

20. Sep 30, 2015

Staff: Mentor

OK. Now substitute that into the equation for the velocity of particle A and take the time derivative. (You can just leave the time derivative of the center of mass in dv/dt form). What do you get?

Chet

21. Sep 30, 2015

Vibhor

$\vec{v}_A = \omega rcosθ\hat{i} + \omega r sinθ\hat{j} + \vec{v}_{CM}$

$\vec{a}_A = r[\ddot{\theta}cos\theta - {\dot{\theta}}^2sin\theta]\hat{i} + r[\ddot{\theta}sin\theta + {\dot{\theta}}^2cos\theta]\hat{j} + \dfrac{d\vec{v}_{CM}}{dt}$

22. Sep 30, 2015

Staff: Mentor

What is the value of the angular velocity of the rod at time zero? Substitute this value into your equation for a, and call the angular acceleration α. So what is the acceleration of particle A at time zero in terms of these parameters? What is the horizontal component?

Chet

23. Sep 30, 2015

haruspex

To a first approximation, the acceleration vector at A at a small time t after release is (a, bt2) for nonzero constants a, b.
Specifically, if the radius of inertia of the rod is k, the horizontal distance between A and the mass centre is w, the angle of the rod to the horizontal is theta, and the length of the string at A is r, then $a=\frac{w^2}{w^2+k^2}g\tan(\theta)$, $b=\frac{a^2t^2}r$. (The vertical acceleration is, of course, centripetal.)
It follows that at t=0 there is only a nonzero horizontal acceleration, but that at any later time there is both horizontal and vertical acceleration.

24. Sep 30, 2015

Vibhor

$\vec{a}_A = rcos\theta\alpha\hat{i} + rsin\theta\alpha\hat{j} + \dfrac{d\vec{v}_{CM}}{dt}$

$\vec{a}_{A,x} = rcos\theta\alpha$

25. Sep 30, 2015

Staff: Mentor

Good. Nice job.