This principle is just an alternative way of deriving the Euler Lagrange equation with new concepts like virtual displacement and virtual work.
Virtual displacement is defined as the infinitesimal displacement in the coordinates of the system,with out any change in all the forces and constraints at same instant of time "t". Accordingly, virtual work is defined as, $$ \delta{W_{virtual}} = \vec{F_i}\cdot\vec{\delta{r_i}}$$
From the concept of system of particles, any force equation can be written as, $$ \frac{d^2}{dt^2}\sum_i m_i \vec{r_i} = \sum_i \vec{F_{i(e)}} + \sum_{i,j \,(i\neq j)\,} \vec{F_{ij}} \,\,\,\ldots...eq.(1)$$
Or simply, $$ \vec{F_{net}} = F_{external} + F_{internal} $$
Internal forces includes the constraints imposed on the system.
Net work done on the "i"th particle is,
$$ \delta{W_i} = \vec{F^{ext}_i}\cdot\delta{\vec{r_i}} + \vec{F^{int}_i}\cdot\delta{\vec{r_i}} $$
Further we assume that, the net virtual work done by the internal forces or the constraints is zero.
Now comes the principle of virtual work, If the system is in equilibrium, then the net force acting on the system is zero which will imply the net virtual work done on all the particles will be zero.
This gives us, $$ \sum_i \vec{F^{ext}_i} \cdot \delta{\vec{r_i}} = 0 $$
But this principle works only for statics where the net force is zero. And also we need to write the virtual displacement in terms of generalized coordinates. Only then we can independently work with each equations and equalize it to zero to get the condition.
Note: Now you may wonder, why do we need to define this new concept of virtual work?
The concept of virtual work is to clearly understand the properties of an equilibrium point. We define equilibrium at a point where the object has no tendency to do any work. It is just as same as the least action principle. Virtual work measures the tendency to do any work and we minimize it to zero in the principle of virtual work. Thus we are doing the same thing in two different ways.
To make a similar static situation in the moving objects, we put an additional effective reverse force in the principle of virtual work using Newton's second law,
$$ \vec{F^{net}_i} = \frac{d\vec{p_i}}{dt} = \dot{\vec{p_i}} $$ $$ \vec{F^{net}_i} - \dot{\vec{p_i}} = 0 $$
Again splitting the force into net external and internal, and assuming the work done by internal forces is zero, we finally arrive at D'Alembert's principle, $$ \sum_i \left(\vec{F^{ext}_i} - \dot{\vec{p_i}}\right)\cdot \delta{\vec{r_i}} = 0 \,\,\,...eq.(2)$$
Now, we will proceed in generalized coordinates to get Euler Lagrange equation.
Position coordinates in terms of generalized coordinates, $$ \vec{r_i} = \vec{r_i} (q_j,t) \,\,\, i= 1,2,..n \\~\\ j= 1,2,3..k \,\,\,...eq.(3)$$
Virtual displacement is given by,
$$\delta{\vec{r_i}} = \sum_j \frac{\partial\vec{r_i}}{\partial{q_j}} \delta{q_j} + \frac{\partial\vec{r_i}}{\partial{t}} dt $$
But "dt = 0" from the definition of virtual work because virtual work is measured on the same instance of time.
$$ \delta{\vec{r_i}} = \sum_j \frac{\partial\vec{r_i}}{\partial{q_j}} \delta{q_j} \,\,\,...eq.(3)$$
Now the velocity of the "i"th particle is given by, $$ \vec{v_i} = \frac{d\vec{r_i}}{dt} = \sum_j \frac{\partial\vec{r_i}}{\partial{q_j}} \frac{dq_j}{dt} + \frac{\partial{\vec{r_i}}}{\partial{t}} \,\,\,..eq.(4)$$
Lets just find out the expression for the first term in eq.(2) i.e. D'Alembert's principle,
$$ \sum_i \vec{F_i}\cdot\delta{\vec{r_i}} = \sum_i \vec{F_i} \cdot \left(\sum_j \frac{\partial{\vec{r_i}}}{\partial{q_j}} \delta{q_j}\right) = \sum_i \sum_j\, \vec{F_i^{ext}} \frac{\partial\vec{r_i}} {\partial{q_j}} \delta{q_j} $$
We give it a name for the expression, $$\sum_i \vec{F^{ext}_i} \frac{\partial\vec{r_i}}{\partial{q_j}} = Q_j $$ where $Q_j$ is called the generalized force.
Thus we end up with the simplified form, $$ \sum_i \vec{F_i^{ext}} \cdot \delta{\vec{r_i}} = \sum_j Q_j \delta{q_j} \,\,\,...eq.(5) $$
Now, taking the second part of the D'Alembert's principle,
$$ \sum_i \dot{\vec{p_i}} \cdot \delta{\vec{r_i}} = \sum_i \sum_j\, \dot{\vec{p_i}} \frac{\partial\vec{r_i}}{\partial{q_j}} \delta{q_j} = \sum_i\sum_j \,m_i \ddot{\vec{r_i}} \frac{\partial\vec{r_i}}{\partial{q_j}} \delta{q_j} $$ where masses are assumed to constant over time.
Making some alterations with the product rule gives,
$$ m_i \ddot{\vec{r_i}} \frac{\partial{r_i}}{\partial{q_j}}\delta{q_j} = \left[\frac{d}{dt}(m_i\dot{\vec{r_i}} \frac{\partial\vec{r_i}}{\partial{q_j}}) - m_i \dot{\vec{r_i}} \frac{d}{dt}(\frac{\partial\vec{r_i}}{\partial{q_j}}) \right] \delta{q_j} \,\,\,...eq.(6)$$
Making use of eq.(4) and commutation between differentiation we get $$ \frac{\partial\vec{v_i}}{\partial{\dot{q_j}}} = \frac{\partial{\vec{r_i}}}{\partial{q_j}} $$ and $$ \frac{d}{dt} (\frac{\partial\vec{r_i}}{\partial{q_j}}) = \frac{\partial}{\partial{q_j}}(\frac{d\vec{r_i}}{dt}) $$
Eq.(6) becomes, $$ \sum_i \sum_j\,\left[ \frac{d}{dt}\left( m_i \vec{v_i} \frac{\partial{\vec{v_i}}}{\partial\dot{q_j}}\right) - m_i \vec{v_i} \frac{\partial{\vec{v_i}}}{\partial{q_j}}\right] \delta{q_j} $$
again using the product rule,
$$ = \,\,\, \sum_i\sum_j \left\{ \frac{d}{dt}\left[\frac{\partial}{\partial{q_j}}(\frac{1}{2}m_i v_i^2)\right] - \frac{\partial}{\partial{q_j}} \left[\frac{1}{2}m_i v_i^2\right]\right\} \,\,\,...eq.(7)$$
Making use of the relation, $$ \sum_i \frac{1}{2}m_iv_i^2 = \sum_i T_i = T $$ where T is total Kinetic Energy of the system of particles.
eq.(7) can be rewritten as, $$ \sum_i \dot{\vec{p_i}} \cdot \delta{\vec{r_i}} = \sum_j \left[\frac{d}{dt}\left(\frac{\partial{T}}{\partial\dot{q_j}}\right) - \frac{\partial{T}}{\partial{q_j}}\right] \delta{q_j}\,\,\,...eq.(8) $$
Substituting eq.(5) and eq.(8) in D'Alembert's principle, we get,
$$ \sum_i \left[ \vec{F^{ext}_i} - \dot{\vec{p_i}}\right]\delta{\vec{r_i}} = \sum_j \left\{ Q_j - \left[\frac{d}{dt}\left(\frac{\partial{T}}{\partial{\dot{q_j}}}\right) - \frac{\partial{T}}{\partial{q_j}}\right]\right\} \delta{q_j} = 0 \,\,\,...eq.(9) $$
If we consider only conservative forces, then forces are derivable from the potential and it leads to,
$$ \vec{F^{ext}_i} = -\nabla{V_i} $$
Eq.(5) becomes, $$ Q_j = \sum_i \vec{F^{ext}_i} \cdot \frac{\partial{\vec{r_i}}}{\partial{q_j}} = \sum_j -\nabla{V} \frac{\partial{\vec{r_i}}}{\partial{q_j}} = -\frac{\partial{V}}{\partial{q_j}} $$
Substituting in eq.(9) gives, $$ \sum_j \left[\frac{d}{dt} \left(\frac{\partial{T}}{\partial\dot{q_j}}\right) - \frac{\partial{T}}{\partial{q_j}}\right] \delta{q_j} = 0 $$
Further, if we consider only simple potentials that depends only on the coordinates then, $$ \frac{\partial{V}}{\partial{q_j}} = 0 $$
Now we arrived at our final form,
$$ \sum_j \left[ \frac{d}{dt} \left(\frac{\partial (T-V)}{\partial{\dot{q_j}}}\right) - \frac{\partial(T-V)}{\partial{q_j}} \right] \delta{q_j} = 0 \,\,\,...eq.(10)$$
Since the equation is expressed in term of the displacements in generalized coordinates, they are all independent of each other. To make it zero, each of the coefficients should be zero.
That gives us our desired Euler Lagrange equation, $$ \frac{d}{dt} \frac{\partial{L}}{\partial{\dot{q_j}}} - \frac{\partial{L}}{\partial{q_j}} = 0 \,\,\,,\,\,\, j = 1,2,...k \\~\\ L = T - V $$
where k - is the number of generalized coordinates, L - Lagrangian, T- Kinetic Energy of the system of particles and V - Potential Energy of the system of particles.
Virtual displacement is defined as the infinitesimal displacement in the coordinates of the system,with out any change in all the forces and constraints at same instant of time "t". Accordingly, virtual work is defined as, $$ \delta{W_{virtual}} = \vec{F_i}\cdot\vec{\delta{r_i}}$$
From the concept of system of particles, any force equation can be written as, $$ \frac{d^2}{dt^2}\sum_i m_i \vec{r_i} = \sum_i \vec{F_{i(e)}} + \sum_{i,j \,(i\neq j)\,} \vec{F_{ij}} \,\,\,\ldots...eq.(1)$$
Or simply, $$ \vec{F_{net}} = F_{external} + F_{internal} $$
Internal forces includes the constraints imposed on the system.
Net work done on the "i"th particle is,
$$ \delta{W_i} = \vec{F^{ext}_i}\cdot\delta{\vec{r_i}} + \vec{F^{int}_i}\cdot\delta{\vec{r_i}} $$
Further we assume that, the net virtual work done by the internal forces or the constraints is zero.
Now comes the principle of virtual work, If the system is in equilibrium, then the net force acting on the system is zero which will imply the net virtual work done on all the particles will be zero.
This gives us, $$ \sum_i \vec{F^{ext}_i} \cdot \delta{\vec{r_i}} = 0 $$
But this principle works only for statics where the net force is zero. And also we need to write the virtual displacement in terms of generalized coordinates. Only then we can independently work with each equations and equalize it to zero to get the condition.
Note: Now you may wonder, why do we need to define this new concept of virtual work?
The concept of virtual work is to clearly understand the properties of an equilibrium point. We define equilibrium at a point where the object has no tendency to do any work. It is just as same as the least action principle. Virtual work measures the tendency to do any work and we minimize it to zero in the principle of virtual work. Thus we are doing the same thing in two different ways.
To make a similar static situation in the moving objects, we put an additional effective reverse force in the principle of virtual work using Newton's second law,
$$ \vec{F^{net}_i} = \frac{d\vec{p_i}}{dt} = \dot{\vec{p_i}} $$ $$ \vec{F^{net}_i} - \dot{\vec{p_i}} = 0 $$
Again splitting the force into net external and internal, and assuming the work done by internal forces is zero, we finally arrive at D'Alembert's principle, $$ \sum_i \left(\vec{F^{ext}_i} - \dot{\vec{p_i}}\right)\cdot \delta{\vec{r_i}} = 0 \,\,\,...eq.(2)$$
Now, we will proceed in generalized coordinates to get Euler Lagrange equation.
Position coordinates in terms of generalized coordinates, $$ \vec{r_i} = \vec{r_i} (q_j,t) \,\,\, i= 1,2,..n \\~\\ j= 1,2,3..k \,\,\,...eq.(3)$$
Virtual displacement is given by,
$$\delta{\vec{r_i}} = \sum_j \frac{\partial\vec{r_i}}{\partial{q_j}} \delta{q_j} + \frac{\partial\vec{r_i}}{\partial{t}} dt $$
But "dt = 0" from the definition of virtual work because virtual work is measured on the same instance of time.
$$ \delta{\vec{r_i}} = \sum_j \frac{\partial\vec{r_i}}{\partial{q_j}} \delta{q_j} \,\,\,...eq.(3)$$
Now the velocity of the "i"th particle is given by, $$ \vec{v_i} = \frac{d\vec{r_i}}{dt} = \sum_j \frac{\partial\vec{r_i}}{\partial{q_j}} \frac{dq_j}{dt} + \frac{\partial{\vec{r_i}}}{\partial{t}} \,\,\,..eq.(4)$$
Lets just find out the expression for the first term in eq.(2) i.e. D'Alembert's principle,
$$ \sum_i \vec{F_i}\cdot\delta{\vec{r_i}} = \sum_i \vec{F_i} \cdot \left(\sum_j \frac{\partial{\vec{r_i}}}{\partial{q_j}} \delta{q_j}\right) = \sum_i \sum_j\, \vec{F_i^{ext}} \frac{\partial\vec{r_i}} {\partial{q_j}} \delta{q_j} $$
We give it a name for the expression, $$\sum_i \vec{F^{ext}_i} \frac{\partial\vec{r_i}}{\partial{q_j}} = Q_j $$ where $Q_j$ is called the generalized force.
Thus we end up with the simplified form, $$ \sum_i \vec{F_i^{ext}} \cdot \delta{\vec{r_i}} = \sum_j Q_j \delta{q_j} \,\,\,...eq.(5) $$
Now, taking the second part of the D'Alembert's principle,
$$ \sum_i \dot{\vec{p_i}} \cdot \delta{\vec{r_i}} = \sum_i \sum_j\, \dot{\vec{p_i}} \frac{\partial\vec{r_i}}{\partial{q_j}} \delta{q_j} = \sum_i\sum_j \,m_i \ddot{\vec{r_i}} \frac{\partial\vec{r_i}}{\partial{q_j}} \delta{q_j} $$ where masses are assumed to constant over time.
Making some alterations with the product rule gives,
$$ m_i \ddot{\vec{r_i}} \frac{\partial{r_i}}{\partial{q_j}}\delta{q_j} = \left[\frac{d}{dt}(m_i\dot{\vec{r_i}} \frac{\partial\vec{r_i}}{\partial{q_j}}) - m_i \dot{\vec{r_i}} \frac{d}{dt}(\frac{\partial\vec{r_i}}{\partial{q_j}}) \right] \delta{q_j} \,\,\,...eq.(6)$$
Making use of eq.(4) and commutation between differentiation we get $$ \frac{\partial\vec{v_i}}{\partial{\dot{q_j}}} = \frac{\partial{\vec{r_i}}}{\partial{q_j}} $$ and $$ \frac{d}{dt} (\frac{\partial\vec{r_i}}{\partial{q_j}}) = \frac{\partial}{\partial{q_j}}(\frac{d\vec{r_i}}{dt}) $$
Eq.(6) becomes, $$ \sum_i \sum_j\,\left[ \frac{d}{dt}\left( m_i \vec{v_i} \frac{\partial{\vec{v_i}}}{\partial\dot{q_j}}\right) - m_i \vec{v_i} \frac{\partial{\vec{v_i}}}{\partial{q_j}}\right] \delta{q_j} $$
again using the product rule,
$$ = \,\,\, \sum_i\sum_j \left\{ \frac{d}{dt}\left[\frac{\partial}{\partial{q_j}}(\frac{1}{2}m_i v_i^2)\right] - \frac{\partial}{\partial{q_j}} \left[\frac{1}{2}m_i v_i^2\right]\right\} \,\,\,...eq.(7)$$
Making use of the relation, $$ \sum_i \frac{1}{2}m_iv_i^2 = \sum_i T_i = T $$ where T is total Kinetic Energy of the system of particles.
eq.(7) can be rewritten as, $$ \sum_i \dot{\vec{p_i}} \cdot \delta{\vec{r_i}} = \sum_j \left[\frac{d}{dt}\left(\frac{\partial{T}}{\partial\dot{q_j}}\right) - \frac{\partial{T}}{\partial{q_j}}\right] \delta{q_j}\,\,\,...eq.(8) $$
Substituting eq.(5) and eq.(8) in D'Alembert's principle, we get,
$$ \sum_i \left[ \vec{F^{ext}_i} - \dot{\vec{p_i}}\right]\delta{\vec{r_i}} = \sum_j \left\{ Q_j - \left[\frac{d}{dt}\left(\frac{\partial{T}}{\partial{\dot{q_j}}}\right) - \frac{\partial{T}}{\partial{q_j}}\right]\right\} \delta{q_j} = 0 \,\,\,...eq.(9) $$
If we consider only conservative forces, then forces are derivable from the potential and it leads to,
$$ \vec{F^{ext}_i} = -\nabla{V_i} $$
Eq.(5) becomes, $$ Q_j = \sum_i \vec{F^{ext}_i} \cdot \frac{\partial{\vec{r_i}}}{\partial{q_j}} = \sum_j -\nabla{V} \frac{\partial{\vec{r_i}}}{\partial{q_j}} = -\frac{\partial{V}}{\partial{q_j}} $$
Substituting in eq.(9) gives, $$ \sum_j \left[\frac{d}{dt} \left(\frac{\partial{T}}{\partial\dot{q_j}}\right) - \frac{\partial{T}}{\partial{q_j}}\right] \delta{q_j} = 0 $$
Further, if we consider only simple potentials that depends only on the coordinates then, $$ \frac{\partial{V}}{\partial{q_j}} = 0 $$
Now we arrived at our final form,
$$ \sum_j \left[ \frac{d}{dt} \left(\frac{\partial (T-V)}{\partial{\dot{q_j}}}\right) - \frac{\partial(T-V)}{\partial{q_j}} \right] \delta{q_j} = 0 \,\,\,...eq.(10)$$
Since the equation is expressed in term of the displacements in generalized coordinates, they are all independent of each other. To make it zero, each of the coefficients should be zero.
That gives us our desired Euler Lagrange equation, $$ \frac{d}{dt} \frac{\partial{L}}{\partial{\dot{q_j}}} - \frac{\partial{L}}{\partial{q_j}} = 0 \,\,\,,\,\,\, j = 1,2,...k \\~\\ L = T - V $$
where k - is the number of generalized coordinates, L - Lagrangian, T- Kinetic Energy of the system of particles and V - Potential Energy of the system of particles.
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