Question: Data: m 1 = 1 2 k g , m 2 = 6 k g m 3 = 6 k g r 1 = 0

Data:
m1=12kg,m2=6kg
m3=6kg
r1=0.3m,r2=0.1m,
L1=0.2m,L2=1m,
M0=50Nm,
k1=2000Nm,
k2=1000Nm,
c=40Nsm,
e=0.8,
=0.2,0=0.25
g=9.81ms2
The rigid homogeneous disk 1 and the rigid homogeneous circular disk 2 are connected by an id Disk 1 can move on the horizontal rough ground as shown in the Figure. (The coefficient of s kinetic friction between ground and disk 1 are 0 and , respectively). The connecting rope is woun 2, and moment M0 is acting on disk 1.
The system starts from rest at the instant t=0. Disk 1 is slipping, because the coefficient of static not enough large to keep rolling. Disk 2 will rotate about the fixed axis passing through B.
The homogeneous slender bar 3 is in equilibrium in the vertical position. Disk 1 will hit the bar 3t=t1. Friction in the bearings B and C is to be neglected. The impact factor is e , the spring stiffnes and k2, respectively. The damping coefficient of viscous damping is c .
Data: m 1 = 1 2 k g , m 2 = 6 k g m 3 = 6 k g r 1

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