Indian Institute of Technology Guwahati Tutorial-1 (Vector and Tensor)

Course: CL 202 Fluid Mechanics Date: 08/08/2015

Instructors: Dr. Partho S.G. Pattader

Dr. Dipankar Bandyopadhyay

Prob 1.

Solution:

Prob 2.

Find the velocity of the arbitrary point of a rigid body rotating about a fixed point O ?

Fig: Velocity v and centripetal acceleration of rigid body rotating about a fixed point O

Solution:

Assuming during the time interval t , the body undergo infinitesimal rotation , causing an

arbitrary point M of the body to experience a displacement r . Then

r= r (1)

Where r is the radius vector of M . Dividing Eq. (1) by t and taking the limit 0t , we get

0

rv lim

t t

Is the velocity of the point M, and

0limt t

Is the instantaneous angular velocity of the body about the point O .

Prob 3.

Prove that:

Solution:

Prob. 4

Solution:

Prob. 5

Solution:

Prob 6.

Using the Divergence theorem find out the value of ∫

over the surface of a

rectangular parallelepiped 0 ,0 ,0x a y b z c . Where ( ) ̂

( ) ̂ ( ) ̂.

Solution:

Gauss Divergence Theorem:

If F is continuously differentiable vector function in the region E bounded by the closed

surfaceS , thenS V

F.N Fds div dv , Therefore divergence

2 2 2F ( ) ( ) ( )div x yz y zx z xyx y z

2( )x y z

V 0 0 0

F 2 ( )

c b a

div dv x y z dxdydz

( )abc a b c

Prob. 7

Solution:

Prob. 8

Verify that 3 3

1 10ijk jk

j k

if

jk kj .

Solution:

3 3

11 11 12 12 13 13 21 21 22 22 23 23 31 31 32 32 33 331 1

ijk jk i i i i i i i i ij k

with same indices is zero, therefore above equation reduces to

12 12 13 13 21 21 23 23 31 31 32 32i i i i i i

When with indices reverses, sign changes

Therefore above equation reduces to following expression

12 12 21 13 13 31 23 23 32( ) ( ) ( )i i i

Here jk kj

Then the equation reduces to

12 12 21 13 13 31 23 23 32( ) ( ) ( ) 0i i i

Prob. 9

The components of symmetric tensor are

3xx 2xy 1xz

2yx 2yy 1yz

1zx 1zy 4zz

The components of vector v are

5xv 3yv 2zv

Evaluate (a)[ .v] (b)[v. ] (c)[ . ] (d) (v.[ .v])

Solution (a):

[ .v] (3)(5) (2)(3) ( 1)( 2) 23x

[ .v] (2)(5) (2)(3) (1)( 2) 14y

[ .v] ( 1)(5) (1)(3) (4)( 2) 10z

(b)

[v. ] (5)(3) (3)(2) ( 2)( 1) 23x

[v. ] (5)(2) (3)(2) ( 2)(1) 14y

[v. ] (5)( 1) (3)(1) ( 2)(4) 10z

Generally [ .v] [v. ] , Since is symmetric [ .v] [v. ] .

(c)

2( : ) ij ij ij

i j i j

2 2 2 2 2 2 2 2 2( : ) (3) (2) ( 1) (2) (2) (1) ( 1) (1) (4) 41

(d)

(v [ v]) [ v]i i

i

v

(5)(23) (3)(14) ( 2)( 10) 117

Prob 10

Show that 2 2 2([ ].[ ]) ( . )v w v w v w v w

Solution: Expanding the above equation,

v w v w ( v w )( v w ) ( v v )( w w )ijk j k imn m n i i j j i i j ji j k m n i j i j

( )v v w w ( v v w w ) ( v v w w )jm kn jn km j m k n i j i j i i j jj k m n i j i j

(v v w w v v w w ) ( v v w w ) ( v v w w )j j k k j k k j i j i j i i j ji k i j i j

When ,i j j k , then

v v w wj j k kj k

Prob 11

Solution

ith Direction:

jth Direction:

kth Direction:

Prob 12

Solution