Spider on the rod

1. Jul 17, 2014

Tanya Sharma

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

2. Relevant equations

3. The attempt at a solution

Applying conservation of angular momentum w.r.t center O

MvL/4 = (ML2/12+ML2/16)ω , where ω is the angular speed of (rod+spider)

This gives ω = (12v/7L)

The torque τ acting on the system when the spider is at distance 'r' and the rod has rotated by an angle θ is given by τ = 2Mgrcosθ (1)

τ = (dθ/dt)(dI/dt) (2)

where I = (ML2/12+Mr2)

dI/dt = 2Mr(dr/dt) = 2Mr(dr/dθ)(dθ/dt) (3)

From the above equations I get gcosθdθ = ω2dr . Integrating I get v = (7/6)√(gL)

I don't know the answer . Is this the correct way to approach the problem ?

I would be grateful if somebody could help me with the problem .

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Last edited: Jul 17, 2014
2. Jul 17, 2014

TSny

Tanya, your approach looks good to me. But I don't see how you get the factor of 2 in your equation (1).

3. Jul 17, 2014

Tanya Sharma

Okay.

When the spider is at distance r ,the CM should be at distance (r/2) from the center .In that case torque should be (2M)g(r/2)cosθ = Mgrcosθ .

This gives v = (7√2/12)√(gL) .

Is this correct ?

4. Jul 17, 2014

TSny

I think your expression for the answer might be right. At least that's what I got with a somewhat different approach.

But, I'm not following your analysis of the torque. If you choose to find the external torque relative to the CM of the entire system, wouldn't you need to consider the reaction force at the pivot of the rod? And I don't understand the (2M) factor. Relative to the CM of the system, does the torque due to the weight of the rod act in the same direction as the torque due to the weight of the spider?

In your first post, I thought you were taking O as the origin for applying $\tau$ = dL/dt.
For example, it looks to me that your expression I = ML2/12 + Mr2 is for O as the origin.

[I like O as the origin for doing the analysis ]

5. Jul 17, 2014

Tanya Sharma

But I have done the analysis using the center of the rod O .I am not finding external torque relative to the CM .

The CM of the system is a distance (r/2) from O .The net gravitational force acting on the CM is 2Mg(mass of rod+spider) .Hence the net torque about O = (2Mg)(r/2)cosθ = Mgrcosθ

Wouldn't the gravitational force on the system(rod+spider) be acting at the CM ?

Sorry if I am missing something very obvious .

Last edited: Jul 17, 2014
6. Jul 17, 2014

ehild

Yes, the torque about O is τ=Mgrcosθ. But your equation (1) in the OP stated that it was τ=2Mgrcosθ.

The torques add if there are separate parts of a system. No need to determine the CM.

The rod has its CM at O. The torque of gravity is zero on it.

You did the following: Rcm=( M*0+Mr)/(2M), τ=( M*0+Mr)/(2M) * (2Mg) cosθ.

ehild

Last edited: Jul 17, 2014
7. Jul 17, 2014

TSny

OK. I now see what you were doing. Your expression Mgrcosθ is correct for the torque about O.

The way I was looking at it: the weight of the rod acts through O, so it produces no torque about O. That leaves just the torque about O due to the spider, which is Mgrcosθ. Same as your expression.

Good!

[Edit: I see ehild already clarified it. Thanks.]

8. Jul 17, 2014

Tanya Sharma

Just a small doubt .

We add torques on different parts of the system , but they should be due to external forces only . The torque due to Normal force N on spider as well as rod cancel each other .

Right ?

So if we consider the spider and rod in isolation then ,

Net torque on spider about O = Mgrcosθ - Nr

Net torque on rod about O = Nr

Net torque on the system = Mgrcosθ - Nr + Nr = Mgrcosθ

Is it alright ?

9. Jul 18, 2014

TSny

Yes. That all looks good.

For this problem the rod has constant ω after the collision. What can you say about the normal force N between the spider and the rod?

10. Jul 18, 2014

ehild

You are right, but we know that the torques of interaction cancel. Also, the torque of the force exerted by a fixed axis is zero.

Torque is due to external forces. It is gravity now. The torque of gravity on the whole system is the sum of the torques of gravity on the parts.

ehild

11. Jul 18, 2014

Tanya Sharma

zero .

12. Jul 18, 2014

TSny

Yes. That was how I first approached the problem. I wondered how the spider would need to move along the rod so that it wouldn't exert a normal force on the rod. It led to the same result as your method using torque.

13. Jul 18, 2014

Tanya Sharma

What an intelligent spider . Looks to be some character of a blockbuster sci-fi movie .

Your follow up question is really nice . I never thought that normal force will be zero .

Do you mind sharing your approach ? Did you conclude that normal force is zero using the same equations I formed in post#8 ? I mean how did you proceed after concluding that normal force is zero .

Last edited: Jul 18, 2014
14. Jul 18, 2014

Tanya Sharma

Thanks a lot. Exactly what I was looking for

15. Jul 18, 2014

TSny

A very talented spider, indeed.

I considered polar coordinates (r, θ) for the spider, with O as the origin. Equation (18) here reduces to $$\ddot{\vec{r}} = (\ddot{r}-\omega^2 r)\hat{r} + 2\omega\dot{r} \hat{\theta}$$ If the normal force is zero, then the only force on the spider in the $\hat{\theta}$ direction is $mg\cos \theta$.

So, F = ma in the $\hat{\theta}$ direction gives $mg\cos \theta = m(2\omega \dot{r})$.

Or, $mg\cos \theta = m(2\omega^2 \frac{dr}{d\theta})$, etc.

16. Jul 18, 2014

ehild

Very clever! It looks polar coordinates are the natural ones for you. Anyway, we line in a rotating world

ehild

17. Jul 18, 2014

TSny

Yes. As you noted in post #8, the net torque on the rod about O is Nr. But the net torque on the rod must be zero since the rod has no angular acceleration.

18. Jul 18, 2014

Tanya Sharma

I agree . TSny relishes polar coordinates .

Beautifully done .

Thank you for teaching me an alternative way of approaching the problem .

19. Jul 18, 2014

TSny

Using the $\hat{r}$ component of $$\ddot{\vec{r}} = (\ddot{r}-\omega^2 r)\hat{r} + 2\omega\dot{r} \hat{\theta}$$ you can determine the radial force that the spider must exert on the rod as a function of θ or t.

20. Jul 18, 2014

Tanya Sharma

Thank you . It's really nice to learn better ways to approach a problem.

Last edited: Jul 18, 2014