- #1

decentfellow

- 130

- 1

## Homework Statement

A uniform rope of mass ##M## and length ##L## is pivoted at one end and whirls with uniform angular velocity ##\omega##. What is the tension in the rope at distance ##r## from the pivot? Neglect gravity.

## Homework Equations

$$\vec{F}=m\vec{a}$$

## The Attempt at a Solution

Consider the differential element as shown in the figure.

I considered the setup of the tension acting on the differential element as such because I thought that net force that should be acting on the differential element should be in the direction of the centripetal acceleration. So, we get

$$T+dT-T=(dm)\omega^2x\implies dT=\left(\dfrac{M}{L}\omega^2x\right)dx$$

On, integrating the expression with proper limits we get

$$\int^{T}_{T_0}{dT}=\int^{r}_{0}\left({\dfrac{M}{L}\omega^2x}\right)dx \implies T-T_0=\dfrac{M}{2L}\omega^2r^2$$

Now, to find ##T_0##, we put the value of that ##T## which acts at the end of the rope. The value of ##T##, i.e. the value of the tension acting in the direction opposite to the direction of the acceleration of the element we see that the value of ##T## is ##0## as there is no tension acting in the outward direction of the rope.

So, we get $$-T_0=\dfrac{M}{2L}\omega^2L^2\implies T_0=-\dfrac{M}{2L}\omega^2L^2$$

$$\therefore T=T_0+\dfrac{M}{2L}\omega^2r^2=\dfrac{M}{2L}\omega^2\left(r^2-L^2\right)$$

The answer that I get has an opposite sign than that of the book's, which is ##\dfrac{M}{2L}\omega^2(L^2-r^2)##. Is it because the direction in which I considered ##T## to act is opposite to what I had assumed. If that is so, then doesn't that mean that ##T+dT(\gt T)## is in the direction opposite to that of the centripetal acceleration, so how would the centripetal acceleration be supplied to the differential element if the net force is in the direction opposite to that of the centripetal accleration.