diff options
Diffstat (limited to '3886/CH20')
-rw-r--r-- | 3886/CH20/EX20.1/20_1.sce | 14 | ||||
-rw-r--r-- | 3886/CH20/EX20.1/20_1.txt | 6 | ||||
-rw-r--r-- | 3886/CH20/EX20.10/20_10.sce | 19 | ||||
-rw-r--r-- | 3886/CH20/EX20.10/20_10.txt | 5 | ||||
-rw-r--r-- | 3886/CH20/EX20.11/20_11.sce | 16 | ||||
-rw-r--r-- | 3886/CH20/EX20.11/20_11.txt | 5 | ||||
-rw-r--r-- | 3886/CH20/EX20.2/20_2.sce | 14 | ||||
-rw-r--r-- | 3886/CH20/EX20.2/20_2.txt | 5 | ||||
-rw-r--r-- | 3886/CH20/EX20.3/20_3.sce | 14 | ||||
-rw-r--r-- | 3886/CH20/EX20.3/20_3.txt | 5 | ||||
-rw-r--r-- | 3886/CH20/EX20.4/20_4.sce | 12 | ||||
-rw-r--r-- | 3886/CH20/EX20.4/20_4.txt | 5 | ||||
-rw-r--r-- | 3886/CH20/EX20.5/20_5.sce | 9 | ||||
-rw-r--r-- | 3886/CH20/EX20.5/20_5.txt | 5 | ||||
-rw-r--r-- | 3886/CH20/EX20.6/20_6.sce | 11 | ||||
-rw-r--r-- | 3886/CH20/EX20.6/20_6.txt | 6 | ||||
-rw-r--r-- | 3886/CH20/EX20.7/20_7.sce | 20 | ||||
-rw-r--r-- | 3886/CH20/EX20.7/20_7.txt | 6 | ||||
-rw-r--r-- | 3886/CH20/EX20.8/20_8.sce | 11 | ||||
-rw-r--r-- | 3886/CH20/EX20.8/20_8.txt | 5 | ||||
-rw-r--r-- | 3886/CH20/EX20.9/20_9.sce | 10 | ||||
-rw-r--r-- | 3886/CH20/EX20.9/20_9.txt | 6 |
22 files changed, 209 insertions, 0 deletions
diff --git a/3886/CH20/EX20.1/20_1.sce b/3886/CH20/EX20.1/20_1.sce new file mode 100644 index 000000000..f2ee73e0f --- /dev/null +++ b/3886/CH20/EX20.1/20_1.sce @@ -0,0 +1,14 @@ +//Fly wheel
+//omega=3*t^3-2*t+2
+//theta=t^3-t^2+2*t+C
+//When t=1 theta=4
+C=2
+//theta=t^3-t^2+2*t+2
+//When t=3
+theta=3*3*3-3*3+2*3+2 //radian
+omega=3*3*3-2*3+2 //rad/sec
+//angular acceleration alpha
+//alpha=6*t-2
+//when t=3
+alpha=6*3-2 //rad/sec^2
+printf("\ntheta=%.3f radian\nomega=%.3f rad/sec\nalpha=%.3f rad/sec^2",theta,omega,alpha)
diff --git a/3886/CH20/EX20.1/20_1.txt b/3886/CH20/EX20.1/20_1.txt new file mode 100644 index 000000000..879cb21be --- /dev/null +++ b/3886/CH20/EX20.1/20_1.txt @@ -0,0 +1,6 @@ +
+--> exec('E:\My program EM\20. Rotation of rigid bodies\20.1.sce', -1)
+
+theta=26.000 radian
+omega=23.000 rad/sec
+alpha=16.000 rad/sec^2
\ No newline at end of file diff --git a/3886/CH20/EX20.10/20_10.sce b/3886/CH20/EX20.10/20_10.sce new file mode 100644 index 000000000..54161666c --- /dev/null +++ b/3886/CH20/EX20.10/20_10.sce @@ -0,0 +1,19 @@ +//welded cylinder
+//refer fig. 20.8 (a) and (b)
+//Mass moment of inertia of the bar about A
+IBA=((200)/(2*9.81))+((200*0.5^2)/(9.81))
+//Moment of inertia of the cylinder about A
+ICA=((500*0.2*0.2)/(2*9.81))+((500*1.2*1.2)/(9.81))
+//mass moment of inertia of the system about A
+I=6.7958+74.41
+//Rotational moment about A
+Mt=200*0.5+500*1.2 //N-m
+//Equating it to I*alpha
+alpha=((700)/(81.2097)) //rad/sec
+//Instantaneous acceleration of rod AB is vertical with magnitude
+Iaccnrod=0.5*8.6197 //m/sec
+//Instantaneous acceleration of cylinder is vertical with magnitude
+Iaccncylinder=1.2*8.6197 //m/sec
+//Applying D'Alembert's dynamic equilibrium equation to the system of forces
+RA=200+500-((200*4.3100)/(9.81))-((500*10.344)/(9.81)) //N
+printf("\nalpha=%.3f rad/sec\nRA=%.3f N",alpha,RA)
diff --git a/3886/CH20/EX20.10/20_10.txt b/3886/CH20/EX20.10/20_10.txt new file mode 100644 index 000000000..99b2ee1db --- /dev/null +++ b/3886/CH20/EX20.10/20_10.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\20. Rotation of rigid bodies\20.10.sce', -1)
+
+alpha=8.620 rad/sec
+RA=84.913 N
\ No newline at end of file diff --git a/3886/CH20/EX20.11/20_11.sce b/3886/CH20/EX20.11/20_11.sce new file mode 100644 index 000000000..455b3d26d --- /dev/null +++ b/3886/CH20/EX20.11/20_11.sce @@ -0,0 +1,16 @@ +//Rods welded
+//refer fig. 20.9 (a) and (b)
+//Mass moment of inertia of AB about axis of rotation
+AB=((200*1.2*1.2)/(12*9.81))+((200*0.6*0.6)/(9.81))
+//Mass moment of inertia of rod CD about A
+CD=((100*0.6*0.6)/(12*9.81))+((100*1.2*1.2)/(9.81))
+//Total mass moment of the system about A
+I=9.786+147.0
+//Let alpha be the instantaneous angular acceleration
+//Kinetic equation for motion gives
+alpha=(300*0.75)/(156.786) //rad/sec
+//Writing the dynamic equilibrium condition
+VA=200+100 //N
+HA=300-((200*0.6*alpha)/(9.81))-((100*1.2*alpha)/(9.81)) //N
+printf("\nVA=%.3f N\nHA=%.3f N",VA,HA)
+
diff --git a/3886/CH20/EX20.11/20_11.txt b/3886/CH20/EX20.11/20_11.txt new file mode 100644 index 000000000..c53337f4a --- /dev/null +++ b/3886/CH20/EX20.11/20_11.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\20. Rotation of rigid bodies\20.11.sce', -1)
+
+VA=300.000 N
+HA=264.891 N
\ No newline at end of file diff --git a/3886/CH20/EX20.2/20_2.sce b/3886/CH20/EX20.2/20_2.sce new file mode 100644 index 000000000..49b9e8bf2 --- /dev/null +++ b/3886/CH20/EX20.2/20_2.sce @@ -0,0 +1,14 @@ +//Flywheel
+//alpha=12-t
+//omega=12*t-(t^2)/2+C
+//When t=4 sec omega=60 rad/sec
+C1=20
+//When t=6 sec
+omega=12*6-((6*6)/(2))+20 //rad/sec
+//theta=6*t^2-(t^3)/6+20*t+C2
+//When t=0 theta0=C2
+//When t=6 sec theta6=180+C2
+//Angular displacement during 6 seconds=180 rad
+//Number of revolution
+N=180/(2*%pi)
+printf("\nomega=%.3f rad/sec\nNumber of revolution=%.3f ",omega,N)
diff --git a/3886/CH20/EX20.2/20_2.txt b/3886/CH20/EX20.2/20_2.txt new file mode 100644 index 000000000..613895fa4 --- /dev/null +++ b/3886/CH20/EX20.2/20_2.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\20. Rotation of rigid bodies\20.2.sce', -1)
+
+omega=74.000 rad/sec
+Number of revolution=28.648
\ No newline at end of file diff --git a/3886/CH20/EX20.3/20_3.sce b/3886/CH20/EX20.3/20_3.sce new file mode 100644 index 000000000..dd394d594 --- /dev/null +++ b/3886/CH20/EX20.3/20_3.sce @@ -0,0 +1,14 @@ +//Wheel rotating about fixed axis
+//Initial velocity
+omega0=2.0944 //rad/sec
+t=70 //sec
+//Angular displacement
+theta=100*%pi //radian
+//Using kinematic equation
+alpha=0.06839 //rad/sec^2
+//Angular velocity at the end of 70 seconds interval
+omega=2.0944+0.06839*70 //rad/sec
+//Let the time required for the velocity to reach 100 rpm be t
+t=(((200*%pi)/(60))-(2.0944))*((1)/(0.06839)) //sec
+printf("\nomega=%.3f rad/sec\nt=%.3f sec",omega,t)
+
diff --git a/3886/CH20/EX20.3/20_3.txt b/3886/CH20/EX20.3/20_3.txt new file mode 100644 index 000000000..4bd4b6eb1 --- /dev/null +++ b/3886/CH20/EX20.3/20_3.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\20. Rotation of rigid bodies\20.3.sce', -1)
+
+omega=6.882 rad/sec
+t=122.497 sec
\ No newline at end of file diff --git a/3886/CH20/EX20.4/20_4.sce b/3886/CH20/EX20.4/20_4.sce new file mode 100644 index 000000000..0c24701ba --- /dev/null +++ b/3886/CH20/EX20.4/20_4.sce @@ -0,0 +1,12 @@ +//Fly-wheel
+//theta=200*%pi //radian
+omega0=(120*2*%pi)/(60) //rad/sec
+omega=(160*2*%pi)/(60) //rad/sec
+//Using kinematic relation
+alpha=0.0977 //rad/sec^2
+//Also
+t=(16.755-4*%pi)/0.0977 //sec
+//theta' be the total angular displacement in reaching the velocity of 160 rpm
+theta=(1436.1)/(2*%pi) //revolution
+printf("\nt=%.3f sec\ntheta=%.3f revolution",t,theta)
+
diff --git a/3886/CH20/EX20.4/20_4.txt b/3886/CH20/EX20.4/20_4.txt new file mode 100644 index 000000000..067f65a7e --- /dev/null +++ b/3886/CH20/EX20.4/20_4.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\20. Rotation of rigid bodies\20.4.sce', -1)
+
+t=42.872 sec
+theta=228.562 revolution
\ No newline at end of file diff --git a/3886/CH20/EX20.5/20_5.sce b/3886/CH20/EX20.5/20_5.sce new file mode 100644 index 000000000..65728eafe --- /dev/null +++ b/3886/CH20/EX20.5/20_5.sce @@ -0,0 +1,9 @@ +//Power driven wheel
+omega0=30*%pi
+omega=0
+theta=720*%pi //rad
+//thus using kinematic equations
+alpha=-1.9635 //rad/sec^2
+//Also
+t=(30*%pi)/(1.9635) //sec
+printf("\nRetardation is %.3f rad/sec^2\nt=%.3f sec",-alpha,t)
diff --git a/3886/CH20/EX20.5/20_5.txt b/3886/CH20/EX20.5/20_5.txt new file mode 100644 index 000000000..5c5eb3091 --- /dev/null +++ b/3886/CH20/EX20.5/20_5.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\20. Rotation of rigid bodies\20.5.sce', -1)
+
+Retardation is 1.964 rad/sec^2
+t=48.000 sec
\ No newline at end of file diff --git a/3886/CH20/EX20.6/20_6.sce b/3886/CH20/EX20.6/20_6.sce new file mode 100644 index 000000000..135af1d5a --- /dev/null +++ b/3886/CH20/EX20.6/20_6.sce @@ -0,0 +1,11 @@ +//The step pulley
+//refer fig. 20.3
+theta=20 //radian
+alpha=2 //rad/sec^2
+omega0=0
+//Using kinematic relation
+t=sqrt(20) //sec
+//Velocity of A
+vA=8.944 //m/sec
+vB=0.6*8.944 //m/sec
+printf("\nt=%.3f sec\nvA=%.3f m/sec\nvB=%.3f m/sec",t,vA,vB)
diff --git a/3886/CH20/EX20.6/20_6.txt b/3886/CH20/EX20.6/20_6.txt new file mode 100644 index 000000000..d451d6ba3 --- /dev/null +++ b/3886/CH20/EX20.6/20_6.txt @@ -0,0 +1,6 @@ +
+--> exec('E:\My program EM\20. Rotation of rigid bodies\20.6.sce', -1)
+
+t=4.472 sec
+vA=8.944 m/sec
+vB=5.366 m/sec
\ No newline at end of file diff --git a/3886/CH20/EX20.7/20_7.sce b/3886/CH20/EX20.7/20_7.sce new file mode 100644 index 000000000..9797210e8 --- /dev/null +++ b/3886/CH20/EX20.7/20_7.sce @@ -0,0 +1,20 @@ +//A flywheel
+omega0=41.888 //rad/sec
+omega=29.322 //rad/sec
+t=120 //sec
+//Kinematic equation gives
+alpha=(29.3224-41.888)/(120) //rad/sec^2
+//Weight of flywheel
+Wf=50000 //N
+//Radius of gyration
+k=1 //m
+I=(50000/9.81) //kg-m^2
+//(1) Retarding torque acting on the flywheel Tr
+Tr=5096.84*0.1047 //N-m
+//(2) Change in K.E.
+C.K.E=(5096.84*((41.888^2)-(27.322^2)))/(2) //N-m The answer provided in the textbook is wrong
+//(3) Change in its angular momentum
+C.A.M=5096.84*(41.888-29.322) //N-sec
+printf("\nRetarding torque acting on the flywheel Tr=%.3f N-m\nChange in K.E.=%.3f N-m\nChange in its angular momentum=%.3f N-sec",Tr,C.K.E,C.A.M)
+
+
diff --git a/3886/CH20/EX20.7/20_7.txt b/3886/CH20/EX20.7/20_7.txt new file mode 100644 index 000000000..f475b5e6a --- /dev/null +++ b/3886/CH20/EX20.7/20_7.txt @@ -0,0 +1,6 @@ +
+--> exec('E:\My program EM\20. Rotation of rigid bodies\20.7.sce', -1)
+
+Retarding torque acting on the flywheel Tr=533.639 N-m
+Change in K.E.=2569094.975 N-m
+Change in its angular momentum=64046.891 N-sec
\ No newline at end of file diff --git a/3886/CH20/EX20.8/20_8.sce b/3886/CH20/EX20.8/20_8.sce new file mode 100644 index 000000000..43c40ae5d --- /dev/null +++ b/3886/CH20/EX20.8/20_8.sce @@ -0,0 +1,11 @@ +//Pulley
+//refer fig. 20.6
+//Let a be the resulting acceleration and T be the tension in the rope
+//Angular acceleration of pulley
+//alpha=1.667*a rad/sec^2
+//Dynamic equilibrium condition for the block gives
+//T=(600-(600*a)/(9.81))
+//From kinetic equation for pulley
+T=(200*7.358)/(9.81) //N
+a=(600*9.81)/(800) //m/sec^2
+printf("\nT=%.3f N\na=%.3f m/sec^2",T,a)
diff --git a/3886/CH20/EX20.8/20_8.txt b/3886/CH20/EX20.8/20_8.txt new file mode 100644 index 000000000..8f083a423 --- /dev/null +++ b/3886/CH20/EX20.8/20_8.txt @@ -0,0 +1,5 @@ +
+--> exec('E:\My program EM\20. Rotation of rigid bodies\20.8.sce', -1)
+
+T=150.010 N
+a=7.357 m/sec^2
\ No newline at end of file diff --git a/3886/CH20/EX20.9/20_9.sce b/3886/CH20/EX20.9/20_9.sce new file mode 100644 index 000000000..a075f95f6 --- /dev/null +++ b/3886/CH20/EX20.9/20_9.sce @@ -0,0 +1,10 @@ +//Composite pulley
+//refer fig. 20.7 (a) and (b)
+//Let aA be acceleration of 4000 N block and aB that of 2000 N block,and alpha be the angular velocity of pulley, then
+//aA=0.5*alpha
+//aB=0.75*alpha
+//Writing dynamic equilibrium equation for the two blocks and from kinetic equation of pulley
+alpha=500/245.97 //rad/sec^2
+TA=4000*(1-(0.5*2.033)/(9.81)) //N
+TB=2000*(1+(0.75*2.033)/(9.81)) //N
+printf("\nalpha=%.3f rad/sec^2\nTA=%.3f N\nTB=%.3f N",alpha,TA,TB)
diff --git a/3886/CH20/EX20.9/20_9.txt b/3886/CH20/EX20.9/20_9.txt new file mode 100644 index 000000000..807d4fa4c --- /dev/null +++ b/3886/CH20/EX20.9/20_9.txt @@ -0,0 +1,6 @@ +
+--> exec('E:\My program EM\20. Rotation of rigid bodies\20.9.sce', -1)
+
+alpha=2.033 rad/sec^2
+TA=3585.525 N
+TB=2310.856 N
\ No newline at end of file |