• Support PF! Buy your school textbooks, materials and every day products Here!

Tension in rotating rod.

  • #1
a thin uniform rod of length l and density d is rotating with angular velocityω about an axis passing through one of its ends and perpendicular to it.find the tension int the rod as a function of x.(x= dist from axis of rotation).area of cross section = a

attempt at solution:

consider a small of length dx at dist x from axis.
upload_2016-4-24_21-45-30.png

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?
 

Answers and Replies

  • #2
251
20
I guess that [itex]a[/itex] and [itex]d[/itex] are the cross-sectional area and the density of the rod, respectively.

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
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
32,730
5,033
Further to @FranzDiCoccio 's comments, since it is a thin rod I would interpret "density" as meaning mass per unit length. You do not need a variable for the cross sectional area.
 
  • #4
TSny
Homework Helper
Gold Member
12,405
2,841
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?
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).
 
Last edited:
  • #5
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).
oh!
thanks!
 

Related Threads on Tension in rotating rod.

Replies
3
Views
187
Replies
2
Views
9K
Replies
9
Views
974
  • Last Post
Replies
4
Views
963
  • Last Post
Replies
21
Views
3K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
12
Views
11K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
2
Views
4K
Top