Virtual work = 0 implies net torque = 0

[Pushoam]

Homework Statement

Homework Equations

$\vec \tau = \Sigma \vec r_i \times \vec F_i$

The Attempt at a Solution

$\vec \tau = \Sigma \vec r_i \times \vec F_i$

$\delta W = \Sigma F_i \delta x_i$

Let's take our Cartesian coordinate system such that the axis of rotation becomes its z-axis.

Let's say that the body rotates through an angle $\theta$ so that the displacement $\vec dx_i = \vec \theta \times \vec r_i$.

Now, how to calculate virtual displacement?

 

[Charles Link]

This is one that is outside of my area of expertise, but I think you can use a generalized type of coordinate and then write it as $dW=\tau \, d \theta$ where $\tau$ is a torque. In more detail, as vectors, it may read $dW=\vec{\tau} \cdot d \vec{\theta}$ , where $\vec{\theta }$ is along the axis of rotation.

 

[TSny]

Let's say that the body rotates through an angle $\theta$ so that the displacement $\vec dx_i = \vec \theta \times \vec r_i$.
Now, how to calculate virtual displacement?

Your equation $\vec dx_i = \vec \theta \times \vec r_i$ is essentially what you need. Imagine the particles of the body undergo virtual displacements corresponding to a virtual rotation $\delta \theta \, \hat \theta$ of the body. What is the corresponding virtual displacement $\vec {\delta r_i}$ of the $i$^th particle?

Note that the virtual work can be written as $\delta W = \displaystyle \sum_{n=1}^N \vec {F_i} \cdot \vec {\delta r_i}$

 

[Dr.D]

In writing dxi = theta x ri you are equating a differential quantity with a finite quantity. This cannot be correct.

Instead, write dx = d(theta) x ri, where ri is constant for for each particle.

 

[Charles Link]

It is perhaps worth mentioning that the sum of the forces on a body or system can add to zero, and there can still be a net torque on the system. (For example, two forces of equal magnitude in opposite directions can both be applying a torque in the same direction). Thereby the system is then not at equilibrium, but can be undergoing a change in its angular momentum and rotational energy. The $\delta W$ from any torques needs to be zero as well to have equilibrium.

 

[Dr.D]

It is perhaps worth mentioning that the sum of the forces on a body or system can add to zero, and there can still be a net torque on the system.

This is most certainly true.