## Answers

in Counter clockwise (CCW) direction 0 to 5.30 s

A = 6.87 N-m/s^(2/3) B = 0.654 N-m , torque is removed after 5.30 secs

at t = 9.78 s , the pivot or pin is removed , so the point of intersection is free to move

The moment of Inertia of system = 1.56 x 2.30^3 Kgm^2 + 3.87 x 1.75m^2 + 2.94 x 0.912^2 + 4.68 x 3.4^2 kgm^2

= 76.6 Kg-m^2

from t=0 to t= 5.3 secs this non-uniform angular acceleration takes place.

From t = 5.3 sec to t= 9.78 secs ,the system continues with uniform angular velocity

=

= 0.91 Radians/s ( ang velocity at 5.3 secs)

=

=

= 1.84 Radians (angle covered upto 5.3 secs)

from 5.3 to 9.78 secs

=5.9 R =338 degrees CCW or 22 deg clock wis

Let us assume that the masses system has not turned at all and calculate velocities of each of them

v1= 0.91 x 2.30 = 2.1 m/s in negative Y-direction ;corresponding momentum= 2.1 x1.56 =3.28 Kgp/s

v2= 0.91 x 1.75 = 1.6 m/s in negative X-direction ;corresponding momentum=3.87 x 1.6 = 6.19 Kgm/s

v3= 0.91 x 0.912 = 0.83 m/s in positive Y-dirction ; corresponding momentum =2.94 x 0.83=2.44kgm/s

v4= 0.91 x 3.40 = 3.1 m/s in positive x direction; corresponding momentum= 4.68 x 3.1 =14.51

net momentum in +X direction=8.32 Kgm/s

net momentum in negative Y-direction =0.84 Kgm/s

net velocity with Resp to X-direction=8.32/(1.56+3.87+2.94+4.68)=0.64m/s

net velocity in negative Y direction=0.84/(1.56+3.87+2.94+4.68)=0.06 m/s

from 9.78 secs to 12.0 secs

distance moved by pin in X axis = (12.0-9.78) x0.64m/s = 1.42 m

distance moved by pin along Negative Y-axis = -0.08m

.