Work done on a conical pendulum

[brotherbobby]

Homework Statement

The diagram below shows a conical pendulum of mass $m$, length $L$ and vertical height $h$ rotating with uniform speed in a plane carving out a circle of radius $r$. The tension in the (massless) wire is $T$. Calculate the work done by the tension on the pendulum bob after one complete revolution.

Relevant Equations

1. Horizontal and vertical components of the tension force : $T_V = T \cos \theta$ and $T_H = T \sin \theta$. (Please note the angle $\theta$ is the semi vertical angle - it is drawn or defined with respect the actual vector and the vertical, not the horizontal).

2. Newton's law for the acceleration of a body in an inertial frame : $\vec a = \frac{\vec F}{m}$.

3. Centripetal force on a body moving with a speed $\vec v$ in rotational motion : $\vec F = -\frac{mv^2}{r} \hat r$. (Please note that this value does not depend as to whether the rotational motion is uniform or non-uniform. In case of non-uniform motion, both $\vec v$ and $\vec F$ would be functions of time).

The diagram for the problem is shown alongside. In the vertical ($\hat z$) direction we have $T \cos \theta = mg$.

In the plane of the pendulum, if we take the pendulum bob at the left extreme end as shown in the diagram, we have $T \sin \theta = \frac{mv^2}{r}$ (the $\hat x$ axis of the pendulum is rotating along with it).

The first equation gives us the tension in the string in terms of the variables provided : $T = mg \sec \theta = mg \frac{L}{h}$.

How to calculate the work done by the tension force $-$ this is where am stuck.

In particular, what is the angle that the tension has with the velocity $\vec v$ at a given instant? If I could find that angle, say some $\alpha$, I can connect it to the semi-vertical angle $\theta$ as some $\alpha (\theta)$ and evaluate the work done by the tension by the line integral : $W_T = \int_0^{2\pi} \vec T(\theta) \cdot d\vec l$, where $d\vec l = \vec v dt$ and the (uniform) speed can be connected to the tension, the radius of the pendulum and its mass from dividing the two equations above : $v^2 = gr \tan \theta$.

$\color{blue}{Any\; help\; would\; be\; welcome}$.

 

[BvU]

what is the angle that the tension has with the velocity →vv→\vec v at a given instant

In the picture the $\vec T$ is clearly in the plane of the paper, whereas $\vec v$ is perpendicular to it, towards the viewer !

 

[brotherbobby]

In the picture the $\vec T$ is clearly in the plane of the paper, whereas $\vec v$ is perpendicular to it, towards the viewer !

Yes indeed, sorry I should have realized it. What you mean is that even though the tension $T(\theta)$ is not along the $\hat z$ direction, it is nonetheless perpendicular to $\vec v$ at this instant?

And since this instant is the same as the next, if we keep rotating our plane of paper so that $T(\theta)$ always lies on it, the velocity vector $\vec v$ will be always perpendicular to the tension for the whole motion?

If that is true, am I right in assuming that the work done by the tension force $T(\theta)$ is 0?

 

[BvU]

Yes. There is no motion in the direction of $\vec T$.