Question

3. Refer to the diagram, which shows an overhead view of four point masses connected by two massless rods that intersect each other at right angles (Overhead View) The masses and distances are as follows: m,-1.56 kg m2-3.87kg m, -2.94 kg m, 4.68 kg dı = 2.30 m d2 = 1.75 m d,-0.9 12 m 4-3.40 m This rigid structure is lying flat, initially at rest in the position shown, ona frictionless surface. It is pinned to that surface at the rods' intersection point The structure can rotate about the (frictionless) pin, but the pin doesn't move. Let an x-y coordinate system be defined with its origin at the pin (pin) Throughout the time interval 0sts5.30 s, a counter-clockwise torque is applied to the structure. This torque varies with time: t(t) - A B, where 1 is measured in seconds. A 6.87 N·m/s2/3, and B 0.654 N·m The torque is removed (becomes zero) at t-5.30 s. Im Att- 9.78 s, the pin is removed (frictionlessly-pulled straight up), so that the entire rigid structure is then free to move on the frictionless surface. Calculate the location (x- and y- coordinates) of the rods' intersection point at t 12.0s.

Answer 1

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).

Answer 2

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