10 files changed, 442 insertions, 0 deletions
diff --git a/1862/CH9/EX9.1/C9P1.sce b/1862/CH9/EX9.1/C9P1.sce
new file mode 100755
index 000000000..0177c958b
--- /dev/null
+++ b/1862/CH9/EX9.1/C9P1.sce
@@ -0,0 +1,22 @@
+clear
+clc
+//To find magnitude of torque due to gravity about the pivot point o
+
+// GIVEN::
+
+//refer to figure 9-5 from page no. 178
+//mass of body 
+m = 0.17//in kg
+//length of rod
+L = 1.25//in meters
+//angle of pendulum with vertical
+theta = 10//in degrees
+//acceleration due to gravity
+g = 9.8//in m/s^2
+
+// SOLUTION:
+
+//magnitude of torque
+tow = L*m*g*sind(theta)//in N.m
+
+printf ("\n\n Magnitude of torque tow = \n\n %.2f N.m" ,tow);
diff --git a/1862/CH9/EX9.10/C9P10.sce b/1862/CH9/EX9.10/C9P10.sce
new file mode 100755
index 000000000..1becf5d74
--- /dev/null
+++ b/1862/CH9/EX9.10/C9P10.sce
@@ -0,0 +1,42 @@
+
+clear
+ clc
+//to find acceleration of the falling block
+//to find tension in the chord
+//to find angular acceleration of the disk
+
+
+// GIVEN::
+
+//refer to figure 9-26(a) from page no. 192
+//mass of disk
+M = 2.5//in kg
+//radius of disk
+R = 20//in cm
+//mass of block
+m = 1.2//in kg
+//acceleration due to gravity
+g = 9.8//in m/s^2
+
+// SOLUTION:
+
+//refer to figure 9-26(b) from page no. 192
+//applying newton's second law in y direction for block
+//and applying rotational form of newton's second law for disk
+//we get 2 equations and 2 unknowns
+A = [m 1;(1/2*M) -1]
+B = [(m*g);0]
+c = A\B
+//acceleration of block
+a = c(1)//in m/s^2
+//tension in the string
+T = c(2)//in N
+//angular acceleration of disk
+az = a/(R*10^-2)//in rad/s^2
+a_z = az/(2*%pi)//in rev/s^2
+
+printf ("\n\n Acceleration of block a = \n\n %.1f m/s^2",a);
+printf ("\n\n Tension in the string T = \n\n %.1f N",T);
+printf ("\n\n Angular acceleration of disk az in rad/s^2 = \n\n %.1f rad/s^2",az);
+printf ("\n\n Angular acceleration of disk a_z in rev/s^2 = \n\n %.1f rev/s^2",a_z);
+
diff --git a/1862/CH9/EX9.12/C9P12.sce b/1862/CH9/EX9.12/C9P12.sce
new file mode 100755
index 000000000..8728efd0a
--- /dev/null
+++ b/1862/CH9/EX9.12/C9P12.sce
@@ -0,0 +1,34 @@
+
+clear
+ clc
+//to find velocity of center of mass at time t
+//to find value of t
+
+// GIVEN::
+
+//refer to figure 9-33(a) from page no. 192
+//radius of solid cylinder
+R = 12//in cm
+//mass of solid cylinder
+M = 3.2//in kg
+//initial angular velocity of solid cylinder
+w0 = 15//in rev/s
+//coefficient of kinetic friction between surface and cylinder
+mew_k = 0.21
+//acceleration due to gravity
+g = 9.8//in m/s^2
+
+// SOLUTION:
+
+//refer to figure 9-33(b) from page no. 192
+w_0 = w0*2*%pi//in rad/rev
+//applying newton's second law in x direction
+//and applying rotational form of newton's second law 
+//velocity of center of mass
+vcm = (1/3*w_0*(R*10^-2))//in m/s
+//value of t
+t = vcm/(mew_k*g)//in seconds
+
+printf ("\n\n Velocity of center of mass vcm = \n\n %.1f m/s",vcm);
+printf ("\n\n Value of t = \n\n %.1f seconds",t);
+
diff --git a/1862/CH9/EX9.13/C9P13.sce b/1862/CH9/EX9.13/C9P13.sce
new file mode 100755
index 000000000..a48176b78
--- /dev/null
+++ b/1862/CH9/EX9.13/C9P13.sce
@@ -0,0 +1,46 @@
+
+clear
+ clc
+//to find rotational velocity when it reaches end of the string
+
+// GIVEN::
+
+//refer to figure 9-34(a) from page no. 196
+//total mass of yo-yo
+M = 0.24//in kg
+//radius of disk
+R = 2.8//in cm
+//radius of shaft
+R0 = 0.25//in cm
+//length of the string
+L = 1.2//in meters
+//initial velocity of yo-yo
+v0 = 1.4//in m/s
+//acceleration due to gravity
+g = 9.8//in m/s^2
+
+// SOLUTION:
+
+//refer to figure 9-34(b) from page no. 196
+//momemt of inertia
+I = (1/2*(M*R^2))
+//applying newton's second law
+//and applying rotational form of newton's second law 
+//angular acceleration
+az = (g*100/R0)*(1/(1+R^2/(2*R0^2)))//in rad/s^2
+//angle through which yo-yo rotates
+fi = L/(R0*10^-2)//in rad
+//initial angular velocity
+w0z = v0/(R0*10^-2)//in rad/s
+//solving using equation to find out time
+y = poly([-fi w0z (1/2*az)],'t','coeff')
+c = roots(y)
+//taking only positive value as it is time
+t2 = c(2)//in seconds
+//rotational velocity when it reaches end of the string
+wz = w0z+(az*t2)//in rad/s^2
+
+printf ("\n\n Angular acceleration az = \n\n %.1f rad/s^2",az);
+printf ("\n\n Time for calculating rotational velocity  t2 = \n\n %.2f seconds",t2);
+printf ("\n\n initial angular velocity w0z = \n\n %3i rad/s",w0z);
+printf ("\n\n Rotational velocity when it reaches end of the string wz = \n\n %3i rad/s^2",wz);
diff --git a/1862/CH9/EX9.2/C9P2.sce b/1862/CH9/EX9.2/C9P2.sce
new file mode 100755
index 000000000..bc92c89d9
--- /dev/null
+++ b/1862/CH9/EX9.2/C9P2.sce
@@ -0,0 +1,62 @@
+clear
+clc
+//To find rotational inertia
+//to find angular acceleration
+
+// GIVEN::
+
+//refer to figure 9-9 from page no. 181
+//mass of first partical
+m1 = 2.3//in kg
+//mass of second partical
+m2 = 3.2//in kg
+//mass of third partical
+m3 = 1.5//in kg
+//force applied to m2
+F = 4.5//in N
+//angle made by force with horizontal
+theta = 30//in degrees
+
+// SOLUTION:
+
+//consider firstly the axis passes through m1
+r1f = 0.0//in m
+r2f = 3.0//in m
+r3f = 4.0//in m
+//rotational inertia about the axis
+I1 = (m1*r1f^2)+(m2*r2f^2)+(m3*r3f^2)//in Kg.m^2
+
+//consider secondly the axis passes through m2
+r1s = 3.0//in m
+r2s = 0.0//in m
+r3s = 5.0//in m
+//rotational inertia about the axis
+I2 = (m1*r1s^2)+(m2*r2s^2)+(m3*r3s^2)//in Kg.m^2
+
+//consider thirdly the axis passes through m3
+r1t = 4.0//in m
+r2t = 5.0//in m
+r3t = 0.0//in m
+//rotational inertia about the axis
+I3 = (m1*r1t^2)+(m2*r2t^2)+(m3*r3t^2)//in Kg.m^2
+I1 = round(I1)
+I2 = round(I2)
+I3 = round(I3)
+
+//from figure fi
+fi = asind(3/5)//in degrees
+//angle between F and line connecting m3 and m2
+fi1 = theta + fi//in degrees 
+//value of moment arm
+r_perpendicular = r3s*sind(fi1)//in m
+//magnitude of torque about m3
+tow_z = r_perpendicular*F//in N.m
+//using rotational inertia about axis through m3
+//angular acceleration
+az = -(tow_z)/I3//in rad/s^2
+
+printf ("\n\n Rotational inertia about the axis when the axis passes through m1 is I1 = \n\n %2i Kg.m^2",I1);
+printf ("\n\n Rotational inertia about the axis when the axis passes through m2 is I2 = \n\n %2i Kg.m^2",I2);
+printf ("\n\n Rotational inertia about the axis when the axis passes through m3 is I3 = \n\n %3i Kg.m^2",I3);
+printf ("\n\n Magnitude of torque about m3 tow_z = \n\n %.1f N.m",tow_z);
+printf ("\n\n Angular acceleration az = \n\n %.2f rad/s^2",az);
diff --git a/1862/CH9/EX9.3/C9P3.sce b/1862/CH9/EX9.3/C9P3.sce
new file mode 100755
index 000000000..45e356548
--- /dev/null
+++ b/1862/CH9/EX9.3/C9P3.sce
@@ -0,0 +1,49 @@
+
+clear
+ clc
+//To find rotational inertia
+
+// GIVEN::
+
+//refer to figure 9-9 from page no. 181
+//mass of first partical
+m1 = 2.3//in kg
+//mass of second partical
+m2 = 3.2//in kg
+//mass of third partical
+m3 = 1.5//in kg
+
+// SOLUTION:
+//locating center of mass
+
+x1 = 0//in m
+x2 = 0//in m
+x3 = 4.0//in m
+//x coordinate of center of mass
+x_cm = (m1*x1+m2*x2+m3*x3)/(m1+m2+m3)//in m
+
+y1 = 0//in m
+y2 = 3.0//in m
+y3 = 0//in m
+//y coordinate of center of mass
+y_cm = (m1*y1+m2*y2+m3*y3)/(m1+m2+m3)//in m
+//squqred distance from center of mass to each of particals
+//for first partical
+r1_square = x_cm^2 + y_cm^2//in m^2
+//for second partical
+r2_square = x_cm^2 + (y2-y_cm)^2//in m^2
+//for third partical
+r3_square = (x3-x_cm)^2 + y_cm^2//in m^2
+//rotational inertia
+I_cm = (m1*r1_square+m2*r2_square+m3*r3_square)//in Kg.m^2
+
+r2_square = nearfloat("succ",3.40)
+r3_square = nearfloat("pred",11.74)
+I_cm = ceil(I_cm)
+
+printf ("\n\n x coordinate of center of mass x_cm = \n\n %.2f m",x_cm);
+printf ("\n\n y coordinate of center of mass y_cm = \n\n %.2f m",y_cm);
+printf ("\n\n Squqred distance from center of mass for first partical r1_square = \n\n %.2f m^2",r1_square);
+printf ("\n\n Squqred distance from center of mass for second partical r2_square = \n\n %.2f m^2",r2_square);
+printf ("\n\n Squqred distance from center of mass for third partical r3_square = \n\n %2i m^2",r3_square);
+printf ("\n\n Rotational inertia I_cm = \n\n %.1f Kg.m^2",I_cm);
diff --git a/1862/CH9/EX9.6/C9P6.sce b/1862/CH9/EX9.6/C9P6.sce
new file mode 100755
index 000000000..10192a5e5
--- /dev/null
+++ b/1862/CH9/EX9.6/C9P6.sce
@@ -0,0 +1,30 @@
+
+clear
+ clc
+//to find forces that is scale reading
+
+
+// GIVEN::
+
+//refer to figure 9-22(a) from page no. 189
+//mass od beam
+m = 1.8//in kg
+//massof block
+M = 2.7//in kg
+//acceleration due to gravity
+g = 9.8//in m/s^2
+
+// SOLUTION:
+
+//refer to figure 9-22(b) from page no. 189
+//consider our system as beam and block together
+//equating net torque to zero
+//force Fr
+Fr = (g/4)*(M+2*m)//in N
+//equating forces iny direction as 0 for equillibrium condition
+//force F1
+F1 = (M+m)*g - Fr//in N
+F1 = round(F1)
+
+printf ("\n\n Force Fr = \n\n %2i N",Fr);
+printf ("\n\n Force F1 = \n\n %2i N",F1);
diff --git a/1862/CH9/EX9.7/C9P7.sce b/1862/CH9/EX9.7/C9P7.sce
new file mode 100755
index 000000000..d0179a9b4
--- /dev/null
+++ b/1862/CH9/EX9.7/C9P7.sce
@@ -0,0 +1,37 @@
+
+clear
+ clc
+//to find forces exerted on the ladder by the ground and by the wall
+
+
+// GIVEN::
+
+//refer to figure 9-23(a) from page no. 189
+//length of ladder
+L = 12//in meters
+//mass of ladder
+m = 45//in kg
+//distance of upper end of ladder above the ground
+h = 9.3//in meters
+//mass of firefighter
+M = 72//in kg
+//acceleration due to gravity
+g = 9.8//in m/s^2
+
+// SOLUTION:
+
+//refer to figure 9-23(b) from page no. 189
+//distance from the wall to the foot of ladder
+a = sqrt(L^2 - h^2)//in meters
+//considering equillibrium conditions
+//finding normal reaction by ground
+N = (M+m)*g//in N
+//force exerted on ladder by the wall
+Fw = (g*a*(M/2 + m/3))/h//in N
+N = round(N)
+Fw = round(Fw)
+printf ("\n\n Distance from the wall to the foot of ladder a = \n\n %.1f m",a);
+//answer is slightly different than book.But answer of scilab program is same as that of calculator
+printf ("\n\n Forces exerted on the ladder by the ground N = \n\n %3i N",N);
+//answer is slightly different than book.But answer of scilab program is same as that of calculator
+printf ("\n\n Forces exerted on the ladder by the wall Fw = \n\n %3i N",Fw);
diff --git a/1862/CH9/EX9.8/C9P8.sce b/1862/CH9/EX9.8/C9P8.sce
new file mode 100755
index 000000000..ed6a936a3
--- /dev/null
+++ b/1862/CH9/EX9.8/C9P8.sce
@@ -0,0 +1,58 @@
+
+clear
+ clc
+//to find tension in the wire
+//to find force exerted by the hinge on the beam
+
+
+// GIVEN::
+
+//refer to figure 9-24(a) from page no. 190
+//length of the beam
+L = 3.3//in meters
+//mass of beam
+m = 8.5//in kg
+//distance at which wire is connected
+d = 2.1//in meters
+//angle made by beam with horizontal
+theta = 30//in degrees
+//mass of body
+M = 56//in kg
+//acceleration due to gravity
+g = 9.8//in m/s^2
+
+// SOLUTION:
+
+//refer to figure 9-24(b) from page no. 190
+//angle alpha from geometry
+alpha = atand((d-(L*sind(theta)))/(L*cosd(theta)))//in degrees
+k = M*g+m*g;
+j = m*g/2;
+//applying equllibrium conditions to get 4 equations
+A = [0 1 0 -1 ; 1 0 1 0 ; 1 -tand(theta) 0 0 ; 0 0 1 -tand(alpha)];
+b = [0 ; k ; j ; 0];
+c = A\b
+Fv = c(1)
+Fh = c(2)
+Tv = c(3)
+Th = c(4)
+
+Fv = round(Fv)
+Fh = round(Fh)
+Th = round(Th)
+//resultant tension in the wire
+T = sqrt(Th^2 + Tv^2)//in N
+//resultant force exerted by the hinge on the beam
+F = sqrt(Fh^2+ Fv^2)//in N
+T = round(T)
+F = round(F)
+//angle made by vector F with horizontal
+fi = atand(Fv/Fh)//in degrees
+
+printf ("\n\n Vertical force Fv = \n\n %3i N",Fv);
+printf ("\n\n Horizontal force Fh = \n\n %3i N",Fh);
+printf ("\n\n vertical tension in in wire Tv = \n\n %3i N",Tv);
+printf ("\n\n Horizontal tension in in wire Th = \n\n %3i N",Th);
+printf ("\n\n Resultant tension in the wire T = \n\n %3i N",T);
+printf ("\n\n Resultant force exerted by the hinge on the beam F = \n\n %3i N",F);
+printf ("\n\n angle made by vector F with horizontal fi = \n\n %.1f degrees",fi);
diff --git a/1862/CH9/EX9.9/C9P9.sce b/1862/CH9/EX9.9/C9P9.sce
new file mode 100755
index 000000000..a895df04f
--- /dev/null
+++ b/1862/CH9/EX9.9/C9P9.sce
@@ -0,0 +1,62 @@
+
+clear
+ clc
+//to find magnitude of torque
+//to find resultant angular acceleration of the system
+
+
+// GIVEN::
+
+//refer to figure 9-25 from page no. 191
+//force exerted
+F = 115//in N
+//distance from axis of rotation at which force is exerted
+r = 1.50//in meters
+//angle of apllication of force
+theta1 = 32//in degrees
+//direction of horizontal component
+theta2 = 15//in degrees
+//acceleration due to gravity
+g = 9.8//in m/s^2
+//radius od disk
+R = 1.5//in meters
+//thicknes of disk
+d = 0.40//in cm
+//mass of child
+m = 25//in kg
+//radius of position of child
+r1 = 1.0//in meters
+
+
+// SOLUTION:
+
+//refer to figure 9-25 from page no. 191
+//horizontal component of force
+Fh = F*cosd(theta1)//in N
+//component of force perpendicular to r
+F_perpendicular = Fh*cosd(theta2)//in N
+//vertical torque along the axis of rotation
+tow = r*F_perpendicular//in N.m
+
+//volume of disk
+volume = %pi*(R*100)^2*d//in m^3
+//consider density of steel
+density = 7.9//in g/cm^3
+//mass of merry-go-round
+M = (volume*density)*10^-3//in kg
+//rotational inertia of disk
+Im = ((1/2)*M*R^2)///in kg.m^2
+//rotational inertia of child
+Ic = m*r1^2///in kg.m^2
+//total rotational inertia
+It = Im + Ic//in kg.m^2
+//angular acceleration of the system
+alpha_z = tow/It//in rad/s^2
+
+printf ("\n\n Horizontal component of force Fh = \n\n %.1f N",Fh);
+printf ("\n\n Component of force perpendicular to r F_perpendicular  = \n\n %.1f N",F_perpendicular);
+printf ("\n\n Vertical torque along the axis of rotation tow = \n\n %3i N.m",tow);
+printf ("\n\n Rotational inertia of disk Im = \n\n %3i kg.m^2",Im);
+printf ("\n\n Rotational inertia of child Ic = \n\n %3i kg.m^2",Ic);
+printf ("\n\n Total rotational inertia It = \n\n %3i kg.m^2",It);
+printf ("\n\n Angular acceleration of the system alpha_z = \n\n %.2f rad/s^2",alpha_z);