bananahorsLv1

6 Apr 2022

3. Two material points, m1, and m2 are attached to the ends of a thin rod. The rod can rotate on a vertical plane around it's point 0, which is fixated. Material points m1, and m2 are located at a distance I1 and I2 accordingly, from the point O. The reaction forces are ideal, and in addition to the gravitational force, there's also a force of resistance of the medium, proportional to the speed and mass with the coefficient of proportionality (F * = mv).

(a) Taking as the generalized coordinate the angle that the rod coincides with the vertical, such as shown in the figure, find the Lagrangian L, the generalized force Q: corresponding to the resistance force of the medium and compose the Lagrange equation.

(b) Calculate the momentum of this system and the moment of the forces acting on it, in relation to a fixed point 0 , and show that the momentum equation follows directly from the equation which has the same form as Lagrange's equation, found in the previous part of the problem.

 

Now let p=0, m1 = 2m, m2 = m, 1=2 and 2=1 and let at the initial moment t = 0 rod was in the vertical position =0, where the material point m 1 was in the lowest position.

(c) If the material point m 1 at the initial moment had a rate of intensity find (t)

(d) If the material point m 1 at the initial moment had a velocity of intensity v0 =4 calculate the time \tau required for the rod to take a horizontal position = / 2.