My Elasticity Doubt

Hey guys. I had a question on elasticity which looked something like this.

We have small beads of mass say \(m\) and Mass of rod say \(M\). Now we need to find the extension in the rod after the beads collide with the the perfectly elastic rubber. Given that it rotates with angular speed \(\omega\) and

Young’s Modulus=Y\text{Young's Modulus} = Y and Area=A\text{Area}=A. [And if you need more variable you can take it. I think the given variables are enough]

No friction is there between the beads.

Now see my solution.


We have a rotating rod. So we will find out the extension for rotation (Say x1x_{1}) and for the collision say x2x_{2}.

So is it that we need to find out x1+x2x_{1}+x_{2}.

Now we can find x1x_{1} easily as follows.

Lets take a differential portion of the rod as dxdx.

Now the mass of the differential mass = dm=λdxdm = \lambda dx

As rod is uniform we have λ=ML\lambda = \dfrac{M}{L}

    dm=MLdx\implies dm= \dfrac{M}{L} dx.

Now We Stress= ForceArea\dfrac{Force}{Area}

    Stress=dmω2xA\implies Stress = \dfrac{dm \omega^2 x}{A}

    Stress=Mω2dxAL\implies Stress = \dfrac{M \omega^2 dx}{AL}

    StressStrain=Y\implies \dfrac{Stress}{Strain} = Y

    Mω2x2dxYAL=d(Δx)\implies \dfrac{M \omega^2 x^2 dx}{YAL} = d(\Delta x)

Integrating from L+L-L \rightarrow +L we get

    Δx=2Mω2L23AY=x1\implies \Delta x = \dfrac{2M \omega^2 L^2}{3AY} = x_{1}

Now we have the x2x_{2} part.

For x2x_{2}. Can we say that the force exerted by each mass on the rod is mω2Lm \omega^2 L and then

F=2mω2lF= 2 m \omega^2 l

    Δx2=2mω2L2YA\implies \Delta x_{2} = \dfrac{2m \omega^2 L^2}{YA}

So We can take x1+x2=2Mω2L2+6mω2L23YAx_{1}+x_{2} = \dfrac{2 M \omega^2 L^2 + 6m \omega^2 L^2}{3YA}

But i am confused if i can take the sum or if the beads play any role in extension or not.


Note by Md Zuhair
3 years, 5 months ago

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Zuhair.... If we take lengh of the rod to be L then why should we integrate from -L to +L(instead -L/2 to +L/2). I also think the tension in the differential portion also depend on the mass and acceleration of the all small part of the rod to the side of nearer end of the rod . So we have to integrate and find out the total tension contributed to the differential mass as a function of 'distance from axis' and substitute it in Young's equation. I do not think the bead will involve.

Hari Govind Sekhar - 3 years, 5 months ago

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Oh yes sir. Thats typo error because I was typing without any written peice. So take lenght as 2L. Or else I will just edit it sometime. Can you please provide a solution if you didnt did it this way and your issues abput thte tension. Thanks

Md Zuhair - 3 years, 5 months ago

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Otherwise, is the concept right?

Md Zuhair - 3 years, 4 months ago

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