Two rods joined together, find the angle of twist

[Jahnavi]

Homework Statement

Homework Equations

The Attempt at a Solution

Honestly , I have no idea about this problem . No theory whatsoever has been given in the book . Please help me in understanding this problem and the relevant concepts involved to solve it .

Thanks

 

[haruspex]

No theory whatsoever has been given in the book

Hmm... that's a bit tough.
Try https://en.m.wikipedia.org/wiki/Polar_moment_of_inertia. I could not quickly find a simpler treatment.

 

[Jahnavi]

I am attaching the solution to this problem . Please explain this . What is ζ ? What is B ? How did they write the second step ?

 

[haruspex]

I am attaching the solution to this problem . Please explain this . What is ζ ? What is B ? How did they write the second step ?

View attachment 224873

I do not see any ζ. τ Is the torque. This will be the same in each rod, so τ=τ'.
A is the total angle of twist (i.e. θ in the question as stated). A' is the portion of that which is in the smaller rod, so the larger rod twists through A-A'.
B is some modulus, a property of the material. Since the rods are of the same material this will cancel.
The polar moment of inertia describes how well the cross-sectional geometry of the object resists twisting. For a cylinder this is ##\frac{\pi r^4}2##. But the longer the rod, the more the twists of small disc elements add up, leading to ##\tau=\frac{B\pi r^4}{2l}\theta##.
Applying that to each rod and equating the torques gives their equation.

 

[Jahnavi]

Thank you .

τ Is the torque. This will be the same in each rod, so τ=τ'.

Why would the torques be same in the two rods ?

 

[haruspex]

Thank you .
Why would the torques be same in the two rods ?

By action and reaction. The torque the first rod exerts on the second must be equal and opposite to the torque the second exerts on the first.

 

[Jahnavi]

By action and reaction. The torque the first rod exerts on the second must be equal and opposite to the torque the second exerts on the first.

Yes .

But , isn't torque , due to the force applied at the free end of the large rod . Why would this torque be same throughout the two rods ?

I might be misunderstanding things .

 

[haruspex]

Why would this torque be same throughout the two rods ?

For the reason I gave, action and reaction at the interface.
Or we can consider the rods made up of thin slices. Each slice of each rod transmits the torque to the next slice. If this were not so there would be a net torque on a slice, leading to angular acceleration.
It's like tension and compression in linear forces.

 

[Jahnavi]

Thanks . Your reasoning is very nice

A is the total angle of twist (i.e. θ in the question as stated). A' is the portion of that which is in the smaller rod, so the larger rod twists through A-A'.

I still do not understand why the difference of angle of twist at the ends of the rod appears in the expression for torque on the rod .

Please explain this .

 

[Charles Link]

Perhaps I can give an input here if may, even though I don't know how to work this problem completely yet. This problem is the rotational twisting version of what would be a linear elasticity type problem if the torque/twisting were replaced by a tension ## T ##. In the case of a vertical pulling force of ## F ## , a tension ## T =F ## would be created in the whole system, and the two parts would get stretched according to an equation of the form ## \frac{\Delta L_i}{L_i}=\alpha_i T ## where ## \alpha_i ## would depend upon some elasticity constant of the material, in the form of ## \alpha_i=\frac{b}{A_i }## where ## b ## is an elasticity property of the material, (the same for both if the material is the same) and ## A_i ## is the cross-sectional area. (Here I needed to google this to get what is common vocabulary: ## b ## here is the inverse of Young's modulus for the material). ## \\ ## Again, the twisting problem is the rotating version of the linear elasticity problem. I would need to google it to learn the details, but it is analogous to the linear elasticity problem. In the case of the linear problem, the item of interest would be the stretching of the bottom section that occurs. For this problem, the twisting that occurs on the lower section needs to be computed. ## \\ ## In this problem, they don't give ## \tau ## which would be the equivalent of ## T ## for the linear problem, but rather ## \Delta \theta_{total} ## which would be the equivalent of ## \Delta L_{total}=\Delta L_1+\Delta L_2 ## for the linear problem. ## \\ ## Editing: I think I see what their solution is, and the use of the letter ## A ## for angle is very clumsy: Basically the equation of interest is ## \Delta \theta_i=\tau (\frac{2 L_i}{\pi B_i r_i ^4}) ##. The rest is quite straightforward... How you do the algebra, to make it as simple as you can, is a matter of choice. You can solve for ## \tau ## first, but there are a couple of shortcuts that will allow you to find ## \Delta \theta_2 ##, without actually solving for ## \tau ##. In the solution presented, they set the ## \tau's ## equal. That's one possible approach, but the algebra can be done in a couple of ways.

 

[haruspex]

Thanks . Your reasoning is very nice
I still do not understand why the difference of angle of twist at the ends of the rod appears in the expression for torque on the rod .

Please explain this .

The torque, τ, puts a twist into each rod. The two twists are in the same direction, so they add up. We are given what they add up to: A in the solution, but called θ in the original question. So if one twists through angle A' then the other twists through A-A'.

 

[Jahnavi]

The two twists are in the same direction, so they add up.

I might be wrong , but I think twists at the two ends of a rod in opposite direction should add up .

Suppose there is a twist A at the end in a rod and there is also a twist A in the same direction at the other end . Wouldn't the net twist be zero ?

 

[Charles Link]

I might be wrong , but I think twists at the two ends of a rod in opposite direction should add up .

Suppose there is a twist A at the end in a rod and there is also a twist A in the same direction at the other end . Wouldn't the net twist be zero ?

What would the twist look like if the material was the same radius throughout the whole length? ## \\ ## The twist they are looking for is the angle at length ## L_2=\frac{L_{total}}{3} ## up from the bottom. The twist ## \Delta \theta ## is proportional to the length, as well as the torque ## \tau ##, but there is an additional constant as well as a factor of ## r^4 ## in the denominator of the twist angle formula See post 10 above for the formula.

 

[haruspex]

I might be wrong , but I think twists at the two ends of a rod in opposite direction should add up .

Suppose there is a twist A at the end in a rod and there is also a twist A in the same direction at the other end . Wouldn't the net twist be zero ?

I mean looking always from the same end. If you are looking from the end of rod 1 and it twists clockwise by θ up to the point where it meets rod 2, then rod 2 twists clockwise by φ from there up to its far end, then the total clockwise twist is θ+φ.

 

[Jahnavi]

then the total clockwise twist is θ+φ.

The twist given in the question is the net total twist given to the system of the two rods ?

 

[haruspex]

The twist given in the question is the net total twist given to the system of the two rods ?

Yes.

 

[Jahnavi]

Perhaps I can give an input here if may, even though I don't know how to work this problem completely yet. This problem is the rotational twisting version of what would be a linear elasticity type problem if the torque/twisting were replaced by a tension ## T ##. In the case of a vertical pulling force of ## F ## , a tension ## T =F ## would be created in the whole system, and the two parts would get stretched according to an equation of the form ## \frac{\Delta L_i}{L_i}=\alpha_i T ## where ## \alpha_i ## would depend upon some elasticity constant of the material, in the form of ## \alpha_i=\frac{b}{A_i }## where ## b ## is an elasticity property of the material, (the same for both if the material is the same) and ## A_i ## is the cross-sectional area. (Here I needed to google this to get what is common vocabulary: ## b ## here is the inverse of Young's modulus for the material). ## \\ ## Again, the twisting problem is the rotating version of the linear elasticity problem. I would need to google it to learn the details, but it is analogous to the linear elasticity problem. In the case of the linear problem, the item of interest would be the stretching of the bottom section that occurs. For this problem, the twisting that occurs on the lower section needs to be computed. ## \\ ## In this problem, they don't give ## \tau ## which would be the equivalent of ## T ## for the linear problem, but rather ## \Delta \theta_{total} ## which would be the equivalent of ## \Delta L_{total}=\Delta L_1+\Delta L_2 ## for the linear problem. ## \\ ## Editing: I think I see what their solution is, and the use of the letter ## A ## for angle is very clumsy: Basically the equation of interest is ## \Delta \theta_i=\tau (\frac{2 L_i}{\pi B_i r_i ^4}) ##. The rest is quite straightforward... How you do the algebra, to make it as simple as you can, is a matter of choice. You can solve for ## \tau ## first, but there are a couple of shortcuts that will allow you to find ## \Delta \theta_2 ##, without actually solving for ## \tau ##. In the solution presented, they set the ## \tau's ## equal. That's one possible approach, but the algebra can be done in a couple of ways.

Thank you so much !