Solving Spider on the Rod Problem Using Conservation of Angular Momentum

[Tanya Sharma]

Homework Statement

Homework Equations

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 .

 

[TSny]

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

 

[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 ?

 

[TSny]

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) .

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 ]

 

[Tanya Sharma]

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?

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 .

 

[ehild]

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 .

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

 

[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.]

 

[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 ?

 

[TSny]

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 ?

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?

 

[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

 

[Tanya Sharma]

What can you say about the normal force N between the spider and the rod?

zero .

 

[TSny]

zero .

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.

 

[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 .

 

[Tanya Sharma]

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.

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

Thanks a lot. Exactly what I was looking for

 

[TSny]

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

A very talented spider, indeed.

I mean how did you proceed after concluding that normal force is zero .

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.

 

[ehild]

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.

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

ehild

 

[TSny]

Did you conclude that normal force is zero using the same equations I formed in post#8 ?

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.

 

[Tanya Sharma]

It looks polar coordinates are the natural ones for you.
ehild

I agree . TSny relishes polar coordinates .

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.

Beautifully done .

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

 

[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.

 

[Tanya Sharma]

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