- #1

- 37

- 1

attempt at solution:

consider a small of length dx at dist x from axis.

for equilibrium of rod T = T-dT + dmω^2 x

dT = dm ω^2 x

integrating

T(x) =(d * a* ω^2 (x^2 - l^2))/2

this is negative of answer.where am i wrong?

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- Thread starter manasi bandhaokar
- Start date

- #1

- 37

- 1

attempt at solution:

consider a small of length dx at dist x from axis.

for equilibrium of rod T = T-dT + dmω^2 x

dT = dm ω^2 x

integrating

T(x) =(d * a* ω^2 (x^2 - l^2))/2

this is negative of answer.where am i wrong?

- #2

- 261

- 26

Your second equation, the one for [itex]dT[/itex], seems ok to me. I do not understand how can you obtain your last result from it. What are the limits of your integral?

Seems to me you're missing a sign in the integration, or, equivalently, you are swapping the integration limits.

- #3

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- #4

TSny

Homework Helper

Gold Member

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I think the problem is associated with the minus sign in writing T - dT for the tension at the outer end of dx. The change in the function T(x) as you go from x to x + dx should be written as dT not - dT. The change in the function will be a negative quantity. So, the value of T at x+dx is T + dT where dT is a negative quantity.for equilibrium of rod T = T-dT + dmω^2 x

dT = dm ω^2 x

integrating

T(x) =(d * a* ω^2 (x^2 - l^2))/2

this is negative of answer.where am i wrong?

By writing the tension at x + dx as T - dT, your symbol dT does not represent the change in the function T(x), it's the negative of the change in T(x).

Last edited:

- #5

- 37

- 1

oh!I think the problem is associated with the minus sign in writing T - dT for the tension at the outer end of dx. The change in the function T(x) as you go from x to x + dx should be written as dT not - dT. The change in the function will be a negative quantity. So, the value of T at x+dx is T + dT where dT is a negative quantity.

By writing the tension at x + dx as T - dT, your symbol dT does not represent the change in the function T(x), it's the negative of the change in T(x).

thanks!

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