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Tension in a rotating rod at various places

Homework Statement
A rod of length ##L## is pivoted at one end and is rotated with a uniform angular velocity in a horizontal plane. Let ##T_1## and ##T_2## be the tensions at the points ##\frac{L}{4}## and ##\frac{3L}{4}## (respectively) away from the pivoted ends. Compare the magnitudes of ##T_1## and ##T_2##.
Homework Equations
Centripetal force needed to keep a particle of ##m## moving in a circle of radius ##r## with angular velocity ##\omega##: ##\;\;\mathbf{F_C = m\omega^2r}##.
(The answer given in the text says ##\boxed{T_1\; >\; T_2}## but, as I show below, I think it's just the opposite).

rotation.png


I begin by putting an image relevant to the problem above. Taking a small particle each of the same mass ##m## at the two positions, the centripetal forces are ##T_1 = \frac{m\omega^2 L}{4}## and ##T_2 = \frac{3m\omega^2 L}{4}##.

Clearly, from above, we have ##T_2 > T_1##, contrary to the answer given in the book.
 

kuruman

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You got your masses swapped. The mass to the right of points 1 and 2 is what counts. At 3L/4 there is less mass being accelerated by the tension.
 

PeroK

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Homework Statement: A rod of length ##L## is pivoted at one end and is rotated with a uniform angular velocity in a horizontal plane. Let ##T_1## and ##T_2## be the tensions at the points ##\frac{L}{4}## and ##\frac{3L}{4}## (respectively) away from the pivoted ends. Compare the magnitudes of ##T_1## and ##T_2##.
Homework Equations: Centripetal force needed to keep a particle of ##m## moving in a circle of radius ##r## with angular velocity ##\omega##: ##\;\;\mathbf{F_C = m\omega^2r}##.

(The answer given in the text says ##\boxed{T_1\; >\; T_2}## but, as I show below, I think it's just the opposite).

View attachment 250853

I begin by putting an image relevant to the problem above. Taking a small particle each of the same mass ##m## at the two positions, the centripetal forces are ##T_1 = \frac{m\omega^2 L}{4}## and ##T_2 = \frac{3m\omega^2 L}{4}##.

Clearly, from above, we have ##T_2 > T_1##, contrary to the answer given in the book.
The tension at the point ##L/4## is not simply rotating a point mass there. It's providing the force to rotate the rest of the bar, all the way out to ##L##.
 
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It is not a single particle at each location that is relevant. It is the body extending from the point in question to the outer free end that needs to be considered. Draw a FBD for the outer end of the rod from a point of interest to the outboard end. Then sum forces on that FBD and that will get the answer you were shown.
 

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