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Example 1

The cross product of the position vector (P) and translational momentum (mV) of a point defines the angular momentum, H_p, of the point: H_p = (P) × mV. If point A of mass 1 kg is located at 2 i m from the origin and is moving with a velocity of 3.0 j, what is the angular momentum about the origin, O of point A?

ANS: Using the cross product definition of the angular momentum:

(1) H_O = (P_A) × m_AV_A = 2 i × (1kg)(3.0 j) = 6 i × jkgm ⁄ s = 6 kN.

In computing the cross products of unit vectors, it is important to understand the direction of the operation so that the appropriate sign can be determined. The easiest way to remember this is to write out the units in the order of the right hand rule system. For the Cartesian system here, that is ijkijk. If the cross product goes from left to right (i.e., i × j, j × k, k × i), then the sign is positive. If the cross product goes in the opposite direction (i.e., j × i, i × k, k × j), then the direction and sign are negative. A cross product on itself (i.e., i × i, j × j, k × k) is always zero.

Example 2

The curl of the velocity of a fluid element (V) defines the rotationally of the fluid at that location (of the fluid element). Compute the rotation of the flow given the velocity of the fluid elements:

a) V = 2xi + xy² j + 4 k

ANS:

(2) ∇ × V = i⎛⎝(∂4)/(∂y) − (∂xy²)/(∂z)⎞⎠ − j⎛⎝(∂4)/(∂x) − (∂2x)/(∂z)⎞⎠ + k⎛⎝(∂xy²)/(∂x) − (∂2x)/(∂y)⎞⎠ = i(0 − 0) − j(0 − 0) + k(y² − 0) = y² k

Here, the fluid is rotating in the positive z or k direction with a changing speed of y².

b) V = 2xi + y² j + 4zk

ANS:

(3) ∇ × V = i⎛⎝(∂4z)/(∂y) − (∂y²)/(∂z)⎞⎠ − j⎛⎝(∂4z)/(∂x) − (∂2x)/(∂z)⎞⎠ + k⎛⎝(∂y²)/(∂x) − (∂2x)/(∂y)⎞⎠ = i(0 − 0) − j(0 − 0) + k(0 − 0) = 0

Here, the fluid has no rotation and is classified as an irrotational flow. Thus, the equations of fluid dynamics governing this flow can be simplified to the potential equations.

CROSS PRODUCTS AND CURL EXAMPLES