Can Tension in a Whirling Rope be Modeled as a Centrifugal Force?

[Hamiltonian]

Homework Statement

A uniform rope of mass M and length L is pivoted at one end and whirls with uniform angular velocity w. what is the tension in the rope at a distance r from the pivot (neglect gravity)

Relevant Equations

f = mw^2r

the point on the string at a distance r from the pivot is rotating in a circle of radius r and hence a centrifugal force of magnitude mw^2r can be said to act on it where m = (M/L)r .

hence the T = centrifugal force

T = (M/L)(wr)^2

but my book says otherwise.
also can the string with mass be considered to be made up of two parts of length r and (L-r) connected by a massless string

 

[PeroK]

the point on the string at a distance r from the pivot is rotating in a circle of radius r and hence a centrifugal force of magnitude mw^2r can be said to act on it where m = (M/L)r .

hence the T = centrifugal force

The centrifugal force is not a real force. It's the centripetal force that is real.

 

[etotheipi]

As @PeroK points out, a centrifugal force will only arise in a rotating frame of reference, in which case it will point in the outward $\hat{r}$ direction. The inward pointing ($-\hat{r}$) resultant tension force is the real centripetal force in the problem, and will be there in both frames.

One way to do it is to consider a small element of rope from a radius $r$ to a radius $r+dr$, with mass $dm = \mu dr$, and it's probably easier conceptually to analyse it in an inertial frame first. It is acted upon by an inward force of magnitude $T(r)$ and an outward force of magnitude $T(r+dr)$. Since the element is small, you can treat it as a point particle and apply the circular motion condition.

 

[hutchphd]

Draw a free body diagram for a segment of the spinning rope. This will provide a (simple) differential equation for T which is in fact the centripetal force supplied by the rope.

 

[etotheipi]

N.B. I just looked at your figure, yes you can consider the section of rope of length $L-r$ in isolation, but since this is an extended body the net force will be proportional to the centre of mass acceleration (i.e. the effective radius is $r + \frac{L-r}{2} = \frac{L+r}{2}$). But in fact, this is probably a more straightforward solution since then you don't need to worry about the constant of integration you will obtain from the integral in the method in #3.

 

[Hamiltonian]

The centrifugal force is not a real force. It's the centripetal force that is real.

yea, you can consider the tension to be the centripetal force

 

[etotheipi]

the point on the string at a distance r from the pivot is rotating in a circle of radius r and hence a centrifugal force of magnitude mw^2r can be said to act on it where m = (M/L)r .

One further comment... you correctly identify the linear density as $M/L$, but a point-like section of rope at radius $r$ will not have a mass $(M/L)r$, but rather $(M/L)dr$, if $dr$ is the infinitesimal length of that section of rope under consideration. And the resultant tension force will also be an infinitesimal, $dT$

 

[Hamiltonian]

One further comment... you correctly identify the linear density as $M/L$, but a point-like section of rope at radius $r$ will not have a mass $(M/L)r$, but rather $(M/L)dr$, if $dr$ is the infinitesimal length of that section of rope under consideration. And the resultant tension force will also be an infinitesimal, $dT$

we essentially need to find the tension in the massless string connecting the two parts of the original string with mass(see my diagram)

 

[etotheipi]

we essentially need to find the tension in the massless string connecting the two parts of the original string with mass(see my diagram)

You can do it this way, but then you will need to consider the tension force as equal to the acceleration of the centre of mass of the element of length $L-r$.