How Does Bending Affect the Shape of a Clamped Rod with a Mass?

  • Thread starter Logistics
  • Start date
In summary: Use your initial conditions to determine any constants of integration.In summary, the problem involves a long thin rod with a mass attached to its upper end. The coordinates (x, y) of any point on the rod satisfy a given equation, and the goal is to show that x can be expressed as a function of y using the given constants. By introducing a new variable z and differentiating the equation, it can be seen that the resulting equation is similar to that of a harmonic oscillator. Using this information, the final differential equation needed to solve for z can be identified and solved using initial conditions. The result will be an expression for x as a function of y.
  • #1
Logistics
30
0
Hey guys this is my question

----------------------------------------------------------------------

A long thin rod is clamped vertically at its lower end and a mass M is attached to its upper end. The coordinates (x, y) of any point on it satisfy the equation

[tex] EI\frac{d^2x}{dy^2} = Mg(a-x)[/tex]

where E, I and a are constants. Given that x = 0 when y = 0 and x = a when y = L show that

[tex] x = a [1- \frac{sin\omega(L-y)}{sin\omega L}][/tex]

where [tex]\omega^2 = \frac{Mg}{EI}[/tex]. (Hint: Let z = a - x).

----------------------------------------------------------------------


Now I have got this here:

[tex] z = a - x \Rightarrow \frac{dz}{dy} = -\frac{dx}{dy} \Rightarrow \frac{d^2z}{dy^2} = -\frac{d^2x}{dy^2}[/tex]

Thus:

[tex] EI\left(-\frac{d^2z}{dy^2}\right) = Mgz[/tex]

[tex] \frac{d^2z}{dy^2} + \frac{Mg}{EI}z = 0[/tex]


I'm thinking that it's correct so far, How would I solve that?. ie. Continue on from here


Could someone help me out please


Thanks
 
Last edited:
Physics news on Phys.org
  • #2
Logistics said:
Hey guys this is my question

----------------------------------------------------------------------

A long thin rod is clamped vertically at its lower end and a mass M is attached to its upper end. The coordinates (x, y) of any point on it satisfy the equation

[tex] EI\frac{d^2x}{dy^2} = Mg(a-x)[/tex]

where E, I and a are constants. Given that x = 0 when y = 0 and x = a when y = L show that

[tex] x = a [1- \frac{sin\omega(L-y)}{sin\omega L}][/tex]

where [tex]\omega^2 = \frac{Mg}{EI}[/tex]. (Hint: Let z = a - x).

----------------------------------------------------------------------


Now I have got this here:

[tex] z = a - x \Rightarrow \frac{dz}{dy} = -\frac{dx}{dy} \Rightarrow \frac{d^2z}{dy^2} = -\frac{d^2x}{dy^2}[/tex]

Thus:

[tex] EI\left(-\frac{d^2z}{dy^2}\right) = Mgz[/tex]

[tex] \frac{d^2z}{dy^2} + \frac{Mg}{EI}z = 0[/tex]


I'm thinking that it's correct so far, and by differentiating the equation above I should get my answer. But I have no idea how to differentiate that thing above :(


Could someone help me out please


Thanks

Assuming what you have done is correct, the equation you have generated for z is a well known equation. If y were time, your equation would be saying acceleration is proportional to minus the displacement. This is the equation of a harmonic oscillator. Assume z as a function of y is harmonic (sine or cosine of ky) and take two derivatives. It will all fall into place.
 
  • #3
So I get this ?

z = a - x = a*sinw(L-y) / sinwL

dz/dy = -w*acosw(L-y) / sinwL

d^2z/dy^2 = -w^2*asinw(L-y) / sinwL

= -w^2 * 2


If this is what you were reffering to, what happens now ?
 
  • #4
[tex] \frac{d^2z}{dy^2} + Cz = 0[/tex] [tex]C=\frac{Mg}{EI}[/tex]
[tex] \frac{d^2z}{dy^2} =-Cz[/tex]
[tex] z(y)=?[/tex]
 
  • #5
I'm not following :(((
 
  • #6
It's hard to give any help without knowing what background you have, what kind of course, this is, etc. In particular, it seems very peculiar that you would be doing a problem like this without know how to solve a linear second order differential equation with constant coefficients.
 
  • #7
HallsofIvy said:
It's hard to give any help without knowing what background you have, what kind of course, this is, etc. In particular, it seems very peculiar that you would be doing a problem like this without know how to solve a linear second order differential equation with constant coefficients.

We have studied the things you just mentioned. linear, 2nd order, homogenous, inhomogenous, auxillary equations etc.


It's just we never did anything like that, we were always dive, simpler questions. The lecturer just isn't on the same level as we are :(


Edit: If someone did the working out, I'm sure I could figure it out and next time be able to do it myself :)
 
  • #8
OlderDan was trying to point out that the resulting equation was similar to that of a harmonic oscillator. da_willem then showed you the final differential equation you needed to solve to find z(y). The final result should be very similar to the equations of SHM.

I haven't taken Diff Eq so I can't be more specific, sorry.
 
  • #9
Logistics said:
So I get this ?

z = a - x = a*sinw(L-y) / sinwL

dz/dy = -w*acosw(L-y) / sinwL

d^2z/dy^2 = -w^2*asinw(L-y) / sinwL

= -w^2 * 2


If this is what you were reffering to, what happens now ?

Where did the *2 come from? You mean *z. So

d^2z/dy^2 + w^2*z = 0

Compare this to your DE and identify omega. Put your expression for z into the defining equation for z and solve it for x.
 

Related to How Does Bending Affect the Shape of a Clamped Rod with a Mass?

1. How do I know where to start when solving an equation?

When solving an equation, it is important to follow the order of operations, which is PEMDAS (Parentheses, Exponents, Multiplication/Division, Addition/Subtraction). Start by simplifying any expressions within parentheses, then evaluate any exponents, and continue in the order of operations until you reach the solution.

2. What should I do if I encounter fractions or decimals in the equation?

If the equation contains fractions or decimals, you can eliminate them by multiplying both sides of the equation by the least common multiple (LCM) of all the denominators. This will result in whole numbers, making it easier to solve the equation.

3. How can I check if my solution is correct?

To check if your solution is correct, you can substitute it back into the original equation. If both sides of the equation are equal, then your solution is correct. You can also use a calculator to verify your solution.

4. What should I do if I am stuck and cannot solve the equation?

If you are stuck and cannot solve the equation, try breaking it down into smaller parts. Start by simplifying one side of the equation, and then work on the other side. You can also ask for help from a friend, teacher, or tutor.

5. Are there any shortcuts or tricks to solving equations?

Yes, there are some common shortcuts or tricks that can help you solve equations faster. For example, when solving equations with variables on both sides, you can eliminate the variable on one side by using the addition or subtraction property of equality. You can also use the distributive property to simplify expressions and solve equations more efficiently.

Similar threads

  • Introductory Physics Homework Help
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
9
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
346
  • Introductory Physics Homework Help
Replies
12
Views
313
  • Introductory Physics Homework Help
Replies
13
Views
1K
  • Introductory Physics Homework Help
Replies
13
Views
2K
  • Introductory Physics Homework Help
Replies
27
Views
3K
  • Introductory Physics Homework Help
10
Replies
335
Views
9K
  • Introductory Physics Homework Help
Replies
5
Views
874
  • Introductory Physics Homework Help
Replies
15
Views
380
Back
Top