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Note that the only contributor to the moment is the friction force. This is because the spring force Fk2 passes through the center of mass body 2, so it produces no moment about the center of mass. Recall the constraint equation developed earlier, x2 = Rθ. Substituting this into Eq. 9↑, the following expression for the friction force is determined:
(10) Ff = (Ix2̈)/(R2)
Equation 10↑ can now be substituted back into Eq. 8↑ and rearranged to give the equation of motion for body 2:
(11) m + (I)/(R2)x2̈ + k2x2 − k2x1 = 0
Equations 6↑ and 11↑ are the equations of motion governing the system. These are coupled second-order equations. Not surprisingly, the coupling between the equations of motion is due to k2, which also couples the two bodies together physically. These equations need to be solved simultaneously to give the response, which can be done analytically using various mathematical methods or numerically. The solution of the equations is left for another topic.

2 Lagrange’s Method

The equations of motion for the same system are now derived using Lagrange’s method. Lagrange’s method does not require a free-body diagram; instead, expressions for the kinetic energy, potential energy, and generalized force are developed and used with Lagrange’s equation to derive the equations of motion.

2.1 Kinetic energy

The first step is to get the system kinetic energy. The equation is
(12) KE = (1)/(2)mvv + (1)/(2) ωIω
where the summation is evaluated over the number of bodies in the system which is two for the present system. The linear velocity of body 1 is x1̇, and body 1 does not rotate, so the kinetic energy of body 1 is
(13) KE1 = (1)/(2)Mx1̇2
Similarly, the kinetic energy of body 2 is
(14) KE2 = (1)/(2)mx2̇2 + (1)/(2)Iω2
where ω = θ̇. Substituting the relation from the constraints, the system kinetic energy is
(15) KE = (1)/(2)Mx1̇2 + (1)/(2)x2̇2m + (I)/(R2)

2.2 Potential energy

In this problem, the kinetic energy comes from the springs, as there is no gravitational potential energy or other sources of potential energy. The potential energy stored in spring 1 is
(16) PE1 = (1)/(2)k1x21
and the potential energy stored in spring 2 is
(17) PE2 = (1)/(2)k2(x2 − x1)2 = (1)/(2)k2(x22 − 2x1x2 + x21)
The system potential energy is just the sum of these potential energies, or
(18) PE = (1)/(2)x21(k1 + k2) + (1)/(2)k2x22 − k2x1x2
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