Tension in a rotating rod at various places

[brotherbobby]

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 $T_2$.

Relevant 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).

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]

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]

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 $L$.

 

[Dr.D]

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.