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The generalized force

Classical mechanics

The generalized force

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The generalized force

Using generalized coordinates, we introduce the concept of generalized force here, and use it to obtain the equation of motion of a classical object in terms of its kinetic energy.

Towards the end of the previous article, we had arrived at an important result called the D’Alembert principle. The principle describes the equilibrium condition for a mechanical system, without the need for the inclusion of the constraint forces. For every force that threatens to change the momentum of a particle, there exists an opposing force \mathbf{\dot{p}} that can keep the body in equilibrium. Therefore, any situation where mechanical equilibrium exists, we can view the set up as a balance between an applied force and an opposing \mathbf{\dot{p}}-force. This reasoning helps us figure out what are the conditions that keep a system in mechanical equilibrium. Our job now is to incorporate the generalized coordinates and obtain new concepts such as the generalized force, which ultimately help us describe the motion of objects.

Force to generalized force

Consider the equation for the D’Alembert principle that we have already derived:

(1)   \begin{equation*} \sum_i(\mathbf{F}_{i}-\mathbf{\dot{p}}_i).\delta \mathbf{r}_i =0  \end{equation*}

Now, if the coordinates {\mathbf{r}_i} being used here are responsible for this equation to appear fussy or complicated (in a given problem), we know that we can always switch to other coordinates with the help of the transformation equations between the coordinates. For e.g., if we chose to switch to some generalized coordinates {q_j}, we have at our rescue the transformation equations,

    \[\mathbf{r}_1=\mathbf{r}_1(q_1, q_2, \cdots q_n, t)\]

    \[\mathbf{r}_2=\mathbf{r}_2(q_1, q_2, \cdots q_n, t)\]

    \[\mathbf{r}_3=\mathbf{r}_3(q_1, q_2, \cdots q_n, t)\]

and so on.

Or, in short hand notation, we shall write them as,

    \[\mathbf{r}_i=\mathbf{r}_i(q_j,t)\]

Then the virtual displacement, which just is an infinitesimal variation of the coordinate \mathbf{r} at a frozen instant of time t, can be written as (using the chain rule),

    \[\delta \mathbf{r}_i=\sum_j {\partial \mathbf{r}_i \over \partial q_j} \delta q_j\]

Replacing \delta \mathbf{r}_i with this expansion, in equation (1), we can write the virtual work as,

    \[\sum_i\mathbf{F}_{i}.\delta \mathbf{r}_i=\sum_i\sum_j\mathbf{F}_{i}.{\partial \mathbf{r}_i \over \partial q_j} \delta q_j\]

We here define the expression,

(2)   \begin{equation*} \sum_i\mathbf{F}_{i}.{\partial \mathbf{r}_i \over \partial q_j}=Q_j  \end{equation*}

as the components of a “generalized force”.
(Observe that just like generalized coordinates needn’t be measured in meters, the generalized force needn’t be necessarily measured in newtons. But the product of the generalized force and the generalized coordinates must be in joules or any equivalent unit of work.)

With this, we can comfortably define virtual work as a product of the generalized forces and the generalized coordinates.

(3)   \begin{equation*} \sum_i\mathbf{F}_{i}.\delta \mathbf{r}_i=\sum_j Q_j \delta q_j  \end{equation*}

We’ve successfully expressed the first term of equation (1) in terms of the generalized coordinates. We have to do the same with the second term \sum_i\mathbf{\dot{p}}_i.\delta \mathbf{r}_i.

The other term

Firstly, since by definition of momentum \mathbf{p}_i=m_i\mathbf{\dot{r}}_i, we have, \mathbf{\dot{p}}_i=m_i\mathbf{\ddot{r}}_i (assuming the masses of the particles don’t change with time). Then, the second term of equation (1) shall be rewritten as,

    \[\sum_i \mathbf{\dot{p}}_i.\delta \mathbf{r}_i=\sum _i m_i\mathbf{\ddot{r}}_i.\delta \mathbf{r}_i\]

But we already know what to replace \delta \mathbf{r}_i with. So,

(4)   \begin{equation*} \sum_i\mathbf{\dot{p}}_i.\delta \mathbf{r}_i=\sum_i m_i\mathbf{\ddot{r}}_i.\sum_j {\partial \mathbf{r}_i \over \partial q_j} \delta q_j  \end{equation*}

Now, say we want to calculate the time derivative:

    \[\sum_i\frac{d}{dt}\left(m_i\mathbf{\dot{r}}_i.{\partial \mathbf{r}_i \over \partial q_j}\right)= \sum_i m_i\mathbf{\ddot{r}}_i.{\partial \mathbf{r}_i \over \partial q_j}+m_i\mathbf{\dot{r}}_i.\frac{d}{dt} \left({\partial \mathbf{r}_i \over \partial q_j}\right)\]

Oh, the first term on the right side here seems to have something to do with the right side of our equation (4).
Therefore, we can rewrite equation (4) as,

(5)   \begin{equation*} \sum_i \mathbf{\dot{p}}_i.\delta \mathbf{r}_i=\sum_{i,j}\left[\frac{d}{dt}\left(m_i\mathbf{\dot{r}}_i.{\partial \mathbf{r}_i \over \partial q_j}\right)-m_i\mathbf{\dot{r}}_i.\frac{d}{dt} \left({\partial \mathbf{r}_i \over \partial q_j}\right)\right]\delta q_j  \end{equation*}

The second term in the parenthesis on the right side here contains a derivative of \mathbf{r}_i first with respect to q_j and then with respect to t. We could as well interchange the order of these differentiations and write,

    \begin{align*} m_i\mathbf{\dot{r}}_i.\frac{d}{dt} \left({\partial \mathbf{r}_i \over \partial q_j}\right)&= m_i\mathbf{\dot{r}}_i.{\partial \over \partial q_j}\left(\frac{d\mathbf{r}_i}{dt} \right)\\ &= m_i\mathbf{\dot{r}}_i.{\partial \mathbf{v}_i \over \partial q_j} \end{align*}

Where we have used the definition of the Cartesian velocity components: \mathbf{v}_i=\dfrac{d \mathbf{r}_i}{dt}=\sum_k \dfrac{\partial \mathbf{r}_i}{\partial q_k}\dot{q}_k+\dfrac{\partial \mathbf{r}_i}{\partial t}; since the \mathbf{r}_i‘s are a function of the q_k‘s. This equation here also implies that \dfrac{\partial}{\partial \dot{q}_j}\dfrac{d \mathbf{r}_i}{dt}=\dfrac{\partial \mathbf{v}_i}{\partial \dot{q}_j}=\dfrac{\partial \mathbf{r}_i}{\partial q_j}. Now, putting all these results into equation (5), we have,

(6)   \begin{equation*} \sum_i \mathbf{\dot{p}}_i.\delta \mathbf{r}_i=\sum_{i,j}\left[\frac{d}{dt}\left(m_i\mathbf{\mathbf{v}}_i.{\partial \mathbf{v}_i \over \partial \dot{q}_j}\right)-m_i\mathbf{v}_i.\dfrac{\partial \mathbf{v}_i}{\partial q_j}\right]\delta q_j  \end{equation*}

Introducing the kinetic energy

We hope you still are aware of where we are headed to. We have been rewriting equation (1). So, putting in the results of equation (3) and equation (6) into equation (1), we have,

(7)   \begin{equation*} \sum_j Q_j \delta q_j-\sum_{i,j}\left[\frac{d}{dt}\left(m_i\mathbf{v}_i.{\partial \mathbf{v}_i \over \partial \dot{q}_j}\right)-m_i\mathbf{v}_i.\dfrac{\partial \mathbf{v}_i}{\partial q_j}\right]\delta q_j=0  \end{equation*}

Closely observe the first term in the parenthesis here, does it not look like the expansion,

    \begin{align*} \dfrac{\partial}{\partial \dot{q}_j}\left(\dfrac{1}{2} m_i \mathbf{v}_i.\mathbf{v}_i\right)&=\dfrac{1}{2} m_i \left(\mathbf{v}_i.\dfrac{\partial \mathbf{v}_i}{\partial \dot{q}_j} + \dfrac{\partial \mathbf{v}_i}{\partial \dot{q}_j}. \mathbf{v}_i\right)\\ &= m_i \mathbf{v}_i.\dfrac{\partial \mathbf{v}_i}{\partial \dot{q}_j} \end{align*}

Similarly,

    \[\dfrac{\partial}{\partial q_j}\left(\dfrac{1}{2} m_i \mathbf{v}_i.\mathbf{v}_i\right)=m_i \mathbf{v}_i.\dfrac{\partial \mathbf{v}_i}{\partial q_j}\]

So, we shall rewrite equation (7) as,

    \[\sum_j Q_j \delta q_j-\sum_j\left[\frac{d}{dt} \dfrac{\partial}{\partial \dot{q}_j}\left(\sum_i \dfrac{1}{2} m_i \mathbf{v}_i.\mathbf{v}_i \right)-\dfrac{\partial}{\partial q_j}\left(\sum_i \dfrac{1}{2} m_i \mathbf{v}_i.\mathbf{v}_i\right)\right]\delta q_j=0\]

Here, \sum_i \dfrac{1}{2} m_i \mathbf{v}_i.\mathbf{v}_i is just the famous classical kinetic energy T, of the i masses. Hence, we may write,

    \[\sum_j Q_j \delta q_j-\sum_j\left[\frac{d}{dt} \dfrac{\partial}{\partial \dot{q}_j}\left(T \right)-\dfrac{\partial}{\partial q_j}\left(T\right)\right]\delta q_j=0\]

Another way of writing this is,

    \[\sum_j \left[Q_j-\frac{d}{dt} \dfrac{\partial T}{\partial \dot{q}_j}+\dfrac{\partial T}{\partial q_j}\right]\delta q_j=0\]

The virtual displacement \delta q_j was chosen arbitrarily (nothing particular about it was assumed), and so for the above equation to hold for a non-zero displacement, the term in the parenthesis must vanish:

    \[Q_j-\frac{d}{dt} \dfrac{\partial T}{\partial \dot{q}_j}+\dfrac{\partial T}{\partial q_j}=0\]

Or,

(8)   \begin{equation*} Q_j=\frac{d}{dt} \dfrac{\partial T}{\partial \dot{q}_j}-\dfrac{\partial T}{\partial q_j}  \end{equation*}

We have here a recipe for obtaining the equation of motion of any object moving due to a force. Knowing the forces and using the relation (2), \sum_i\mathbf{F}_{i}.{\partial \mathbf{r}_i \over \partial q_j}=Q_j, we can know the generalized forces. Then all that remains is for us to calculate the kinetic energy of the object and put it all in the above equation (8), and we know how the object moves.

In the following article, we will see how the concept of the generalized force we understood here, can be used to deduce a set of marvelous equations called the Lagrange equations, which form an alternate explanation of object motions.

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Roshan Sawhil is a Physics postgraduate who rejoices both doing and explaining Physics. He also finds doing Philosophy as a leisure activity quite interesting. You can find and connect with him on Facebook and Twitter.

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