Machine Dynamics Problems

72 B. Dyniewicz, c.i. Bajer
Finally the potential energy of the string, with respect to (25) can be described by
the equation
2
Ep = l_NfI i ~2 q/ (I) - P'I qj(I) sin inix
4 i=1 l i=1 /
Now, when we have kinetic and potential energy described in generalized
coordinates and the derivative of generalized coordinates with respect to time, we
can formulate the Lagrange equation, which general form is given by the equation.
(30)
(31)
In order to obtain the Lagrange equation describing our problem, we must compute
respective required terms. From (13) and (22) we have derivative of kinetic energy
of travelling mass Ekm with respect to qi and ~i .
(32)
(33)
We compute the derivative of the kinetic energy for hole system (26) with respect
to qi and ~i , taking into account (32) and (33).
8£ k 8£km [2 ~ ij Jr2 irax jmn j: ( ) ~ in iJrut. imn j: ( )]
--=--=m u £...,.--cos--cos--.". t +u£...,.-cos--sm--.". t
8;i O;i ;,}=1 [2 I 1 J i,}=1[ [ [J
es, 1 00. 8£ 1 00. [00 jr: imx imx
-. =-pA[L;;(t)+~=-pAIL;;(t)+m u L -sin-cos-;;(t)+
O;i 2 ;=1 8;; 2 ;=1 ;,}=1 I I I
(34)
~ jtt , invt . invt j; ()~
+ c: -Sln--Sln--." .,t
i,}=1 / I I J
The derivative of the potential energy (30) with respect to generalized
qj.
8E 1 00.2 2 00'
p _ 7\TI" 1 Jr t: () p'" isto:
----lV,£...,.--." t - L,.sm--
8qi 2 i=1 /2 1 i=1 I
(35)
coordinates
(36)
The derivative of (35) with respect to t is as follows.