8 files changed, 88 insertions, 0 deletions

diff --git a/3131/CH19/EX19.1/Ex19_1.sce b/3131/CH19/EX19.1/Ex19_1.sce
new file mode 100644
index 000000000..95fd683c5
--- /dev/null
+++ b/3131/CH19/EX19.1/Ex19_1.sce
@@ -0,0 +1,28 @@
+clear all; clc;
+disp("Ex 19_1")
+a=8//rad/sec
+b=0.25//m
+v_G_disk=a*b
+printf('\n\nSince the disk is rotating about a fixed axis through point B then v_g = %0.0f m/s',v_G_disk)
+m=10//kg
+H_G=0.5*m*b^2*a
+printf('\n\n(clockwise)+H_G = %0.2f kgm^2/s',H_G)
+H_B=H_G+m*v_G*b
+printf('\n\n(clockwise)+H_B = %0.2f kgm^2/s',H_B)
+disp(" ")
+disp("I_B=(3/2)*m*r^2")
+H_B_new=(3/2)*m*b^2*a
+printf('\n\n(clockwise)+H_B = %0.2f kgm^2/s',H_B_new)
+disp(" ")
+disp("The bar undergoes General Plane Motion")
+//Refer figure 19-c
+c=3.464//m
+w=v_G_disk/c
+printf('\n\nw = %0.4f rad/s',w)
+v_G_bar=w*v_G
+printf('\n\nFor the bar :-  v_G = %0.3f m/s',v_G_bar)
+H_G_new=(1/12)*5*4^2*w
+printf('\n\nFor the bar :-  (clockwise)+H_G = %0.2f kgm^2/s',H_G_new)
+//Moments of I_Gw and m.v_G about the IC yield
+H_IC=H_G_new+2*5*v_G_bar
+printf('\n\n(clockwise)+H_IC = %0.1f kgm^2/s',H_IC)
diff --git a/3131/CH19/EX19.2/Ex19_2.sce b/3131/CH19/EX19.2/Ex19_2.sce
new file mode 100644
index 000000000..dbd6179fe
--- /dev/null
+++ b/3131/CH19/EX19.2/Ex19_2.sce
@@ -0,0 +1,15 @@
+clear all; clc;
+disp("Ex 19_2")
+a=100 //weight of disk in N
+m=100/9.81 // mass of disk in kg
+r=0.3 //radius of the disk in m
+I_A=0.5*m*r^2
+printf('\n\nMoment of inertia of the disk about its fixed axis of rotation is = %0.3f kgm^2',I_A)
+//Principle of Impulse Momentumyields the following 3 equations
+//0+A_x*(2)=0 ... equation 1
+//0+A_y*(2)-100*(2)-40*(2)=0 .. equation 2
+//0+4*(2)+[40*(2)]*(0.3)=0.459*w_2
+//Solving these equations yields :
+disp("A_x is = 0")
+disp("A_y is = 140 N")
+disp("w_2 is = 69.7 rad/sec in clockwise direction")
diff --git a/3131/CH19/EX19.3/Ex19_3.sce b/3131/CH19/EX19.3/Ex19_3.sce
new file mode 100644
index 000000000..bc5b01b75
--- /dev/null
+++ b/3131/CH19/EX19.3/Ex19_3.sce
@@ -0,0 +1,8 @@
+clear all; clc;
+disp("Ex 19_3")
+m=100 //kg
+k_G=0.35//m
+I_G=m*k_G^2
+printf('\n\nMoment of inertia of the spool about its mass centre is = %0.2f kgm^2',I_G)
+//Applying Impulse Momentum Principle, we get 2 equtions with integral terms in them and on solving them, we get :
+disp("w_2=1.05 rad/sec in clockwise direction")
diff --git a/3131/CH19/EX19.4/Ex19_4.sce b/3131/CH19/EX19.4/Ex19_4.sce
new file mode 100644
index 000000000..e379cb706
--- /dev/null
+++ b/3131/CH19/EX19.4/Ex19_4.sce
@@ -0,0 +1,7 @@
+clear all; clc;
+disp("Ex 19_4")
+//Applying Impulse Momentum Principle, we get the following two equations:
+//0.4*w_1+T*(3)*0.2=0.4*w_2 .. equation 1
+//-6*2+T*3-58.86*3=-6*(v_B)2 .. equation 2
+//Solving the above two equations simultaneosly
+disp("(v_B)2=13.0 m/s in the downward direction")
diff --git a/3131/CH19/EX19.5/Ex19_5.sce b/3131/CH19/EX19.5/Ex19_5.sce
new file mode 100644
index 000000000..815c307cf
--- /dev/null
+++ b/3131/CH19/EX19.5/Ex19_5.sce
@@ -0,0 +1,4 @@
+clear all; clc;
+disp("Ex 19_5")
+//Referring figure 19-8
+disp("r_p=r(bar)+ ((k_G)^2)/r(bar)")
diff --git a/3131/CH19/EX19.6/Ex19_6.sce b/3131/CH19/EX19.6/Ex19_6.sce
new file mode 100644
index 000000000..b1ca58357
--- /dev/null
+++ b/3131/CH19/EX19.6/Ex19_6.sce
@@ -0,0 +1,8 @@
+clear all; clc;
+disp("Ex 19_6")
+// Using Principle of Conservation of Angular Momentum,
+//Eqn 1 : (0.2-0.03)*10*(v_G)1+0.156*w_1=0.2*10*(v_G)2+0.156*w_2
+//Since no slipping occurs,equation 1 reduces to (v_G)2=0.892*(v_G)1
+//Using Conservation of Energy Principle, Equation 2 is: 0.5*10*(v_G)2^2 +  0.5*0.156*w_2^2+0=0+{98.1*0.03}
+//Put w_2=5*(v_G)2 and equation 1 in the above equation,
+disp("(v_G)1=0.729 m/s")
diff --git a/3131/CH19/EX19.7/Ex19_7.sce b/3131/CH19/EX19.7/Ex19_7.sce
new file mode 100644
index 000000000..8ae0c1a83
--- /dev/null
+++ b/3131/CH19/EX19.7/Ex19_7.sce
@@ -0,0 +1,9 @@
+clear all; clc;
+disp("Ex 19_7")
+//Using Principle of Conservation of Angular Momentum:
+//Equation 1 becomes : 1.039=0.003*(v_B)2+2.50*(v_G)2+0.417*w_2
+//Since the rod is pinned at O, form figure 19-10c
+//(v_G)2=0.5*w_2
+//(v_B)2=0.75*w_2
+//Puttin the above two values in equation 1, we get the value of w_2
+disp("w_2 = 0.623 rad/sec in the anticlockwise direction")
diff --git a/3131/CH19/EX19.8/Ex19_8.sce b/3131/CH19/EX19.8/Ex19_8.sce
new file mode 100644
index 000000000..cce238320
--- /dev/null
+++ b/3131/CH19/EX19.8/Ex19_8.sce
@@ -0,0 +1,9 @@
+clear all; clc;
+disp("Ex 19_8")
+//Referring figure 19-12b and 19-12c
+//Using the principle of Conservation of Angular Momenutum and since (v_G)2=0.5*w_2
+//Equation 1 becomes : 5.0=0.5*(v_B)2+1.67*w_2
+//Using the coefficient of restitution theory, we have 4.0=0.5*w_2-(v_B)2
+//Solving the above two equatons,
+disp("(v_B)2= 2.177 m/sec towards the left")
+disp("w_2= 3.64 rad/sec in the anticlockwise direction")