3. Refer to the diagram, which shows an overhead view of four point masses connected by...

The moment of inertia of the system about the pin is given as follows: 1 md? +m,d?+m,d? +m,d? (1.56 kg) (2.30 m) +(3.87 kg)(1.75 m) +(2.94 kg)(0.912 m) +(4.68 kg)(3.40 m) = 76.65 kg m2 Torques is given as follows do dt Therefore, the angular velocity achieved by the structure over 5.30 s is given as follows: 5.30 s I 5.30 B)dt At213 + Jo I m)0 30 s (6.87 N.m/s2/3 +(0.654 N .m 76.65 kg-m = (0.912 rad/s) This angular velocity is maintained at 0.912 rad/s until 9.78 s. Since, the surface is frictionless Calculate the translational velocity of each mass as follows Velocity of mass m (directed along negative x-direction) is given as follows: =(2.30 m)(0.911 6) -2.097 m/s Velocity of mass m2 (directed along negative y-direction) is given as follows: = (1.75 m) (0.9116) 1.595 m/s

Velocity of mass m (directed along positive x-direction) is given as follows: (0.912 m) (0.91 16) 0.813 m/s Velocity of mass m4 (directed along positive y-direction) is given as follows: (3.40 m) (0.9116) -3.099 m/s Total velocity along x- direction is given as follows: V = (0.813-2.097) m/s -1.284 m/s Total velocity along y- direction is given as follows: v, = (3.099-1.595) m/s 1.504 m/s After 12.0 s time elapsed since the pin is removed is given as follows: t (12.0-9.78) s -2.22 s Therefore, The change in x-coordinate of the pin is given as follows: (1.284 m/s)(2.22 s) =-2.85 m The change in y-coordinate of the pin is given as follows: y=v,¢ = (1.504 m/s) (2.22 s) = 3.34 m Therefore, the final coordinate of the pin is (-2.58 m, 3.34 m).

The total moment of inertia of all masses about the pin point is 1-т,d? + m,a; + m, а; +m,d? 1.56kg)(2.30 m +(3.87kg )(1.75m) (5.2524kg m2)+(11.85187kg (2.94kg)(0.912 m (4.68kg)(3.40 m) +(2.4453kg m2) +(54.1008kg m2) 2. 76.65kg m Write the mathematical expression for torque is do dt r is torque, I is total moment of inertia of all masses about the pin point, o is system and dt is interval of time Here, angular speed of the masses Simplify and substitute the given values in the above expression. rdt da I (4fA + В) dt

Integrating the above expression on both sides 5.30s (4i+B)di 2/3 + В ) dt 0 do= I 0 5.30s ((6.87N-m/3/) 4N-m))dt (0.654N 0 76.65kg m 3 -5.30s + (6.87N m/s2/5) 5.30s (0.654 N m JC 5 76.65kg m (630s)- (6.87N m/s5) 5/3 2/3 m)[(5.30s)-0 -0(0.654N -m 76.65kg m2 =0.9116 rad/s

(m,) in x-direction is The linear velocity of the mass = (2.30m)(0.9116rad/s) =2.097m/s The linear velocity of the mass (m2) in x-direction is =(0.912m) (0.9116 rad/s) -0.831m/s The linear velocity of the mass (m2) in y-direction is -(1.75m) (0.9116 rad/s) 1.595 m/s The linear velocity of the mass (m,) in y-direction is (3.40m) (0.9116rad/s) = 3.099m/s

The linear velocity of masses m, and mg are in opposite direction so the total velocity along x-direction can be determine as 1 =(0.831m/s)-(2.097 m/s) =-1.266 m/s The linear velocity of masses m, and m, are in opposite direction so the total velocity along y-direction can be determine as V. ニ1 (3.099m/s)-(1595m/s) -1.504 m/s The surface is friction less so the angular speed (=0.9116rad/s) will remains constant up till 9.78s The pin is removed at 9.78s so time after the pin removed will be t-(12s)-(9.78s) =2.22s

The distance travelled by the pin in x -direction after pin removed is =(-1.266m/s)(2.22s) -2.81m x is x-coordinate of the pin after the pin removed Here The distance travelled by the pin in y-direction after pin removed is -(1.504m/s)(2.22s) = 3.3388m Here, y is y-coordinate of the pin after the pin removed Thus, the final coordinate of the pin is|(-2.81m ,3.3388m