Physics Asked on February 17, 2021
The translation form $r_i$ to $q_j$ language start forms the transformation equation:
$r_i=r_i (q_1,q_2,…,q_n,t)$ (assuming $n$ independent coordinates)
Since it is carried out by means of the usual “chain rules” of the calculus of partial differentiation.
$$
v_iequiv frac{d r_i}{dt}= sum_k frac{partial r_i}{partial q_k} dot{q_k} + frac{partial r_i}{partial t}.
$$
Similarly, the arbitrary virtual displacement $δr_i$ can be connected with the virtual displacement $δq_j$ by:
$$
delta r_i= sum_j frac{partial r_i}{partial q_j} delta q_j.
$$
I am having trouble understanding how the first equation is derived and where the second equations is coming form.
I am having trouble applying the chain rule in this context i was wondering if a more detailed derivation can be given.
I am having trouble understanding where the second equation comes form (arbitrary virtual displacement)?
These equation are in terms of d’Alembert principle and Lagrange’s equations
A total derivative of a function $f(x_1,x_2,cdots ,x_n)$ is defined as $$dfequiv sum_{i=0}^nleft(frac{partial f}{partial x_i}right)dx_i.$$ In your situation, the total differential of a function $r_i(q_1,cdots ,q_n,t)$ is given by $$dr_i= sum_{j=0}^nleft(frac{partial r_i}{partial q_j}right)dq_j+frac{partial r_i}{partial t}dt,$$ or $$boxed{v_iequivfrac{dr_i}{dt}= sum_{j=0}^nleft(frac{partial r_i}{partial q_j}right)dot{q_j}+frac{partial r_i}{partial t}.}$$ Considering virtual displacement ($dt=0$) we get: $$boxed{delta r_i= sum_{j=0}^nleft(frac{partial r_i}{partial q_j}right)delta q_j.}$$
Correct answer by Young Kindaichi on February 17, 2021
The definition of the total differential is $$ d r_i= sum_j frac{partial r_i}{partial q_j} d q_j + frac{partial r_i}{partial t} d t. $$ Dividing both sides by the differential with respect to time we get your first equation (the chain rule): $$ frac{d r_i}{dt}= sum_j frac{partial r_i}{partial q_j} frac{d q_j}{dt} + frac{partial r_i}{partial t}. $$
Answered by Roger on February 17, 2021
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