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Elliptical cylinder.

Short Answer

Expert verified

The required values are mentioned below.

s¨u=u¨+sinhucoshusin2u+sin2vu˙2+2sinucosusin2u+sin2vu˙v˙-sinucosusin2u+sin2vv˙2hus¨v=u¨+sinvcosusin2u+sin2vu˙2+2sinhvcoshusin2u+sin2vu˙v˙-sinvcosusin2u+sin2vv˙2hus¨z=z¨

Step by step solution

01

Given Information

An elliptical cylinder.

02

Definition of cylindrical coordinates.

The coordinate system primarily utilized in three-dimensional systems is the cylindrical coordinates of the system. The cylindrical coordinate system is used to find the surface area in three-dimensional space.

03

Find all the values.

Velocity in elliptical cylinder iss˙=u˙asinh2u+sin2ve^u+v˙asinh2u+sin2ve^v+ze^z˙.

Find the acceleration.

localid="1659333210449" s¨=u¨+Γijuq˙iq˙jhue^u+v¨+Γijvq˙iq˙jhve^u+z¨e^z2sqiqj=Γijksqk

Solve further.

2su2=acoshucosve^x+asinhusinve^y2suv=-asinhusinve^x+acoshucosve^y2sv2=acoshucosve^x-asinhusinve^y

Find the value of e^xand e^y.

e^x=sinhucosvasinh2u+sin2vsu-coshusinvasinh2u+sin2vsve^y=coshusinvasinh2u+sin2vsu-sinhucosvasinh2u+sin2vsv

Find the coefficients.

2su2=sinhucoshusinh2u+sin2vsu-sinvcosvsinh2u+sin2vsvΓuuu=sinhucoshusinh2u+sin2vΓuuu=-sinvcosvsinh2u+sin2v2su2=sinvcosvsinh2u+sin2vsu-sinhucoshusinh2u+sin2vsv

Find the further coefficients.

Γuvu=sinucosusinh2u+sin2v;Γuvu=sinhvcoshvsinh2u+sin2v2sv2=-sinhucoshusinh2u+sin2vsu+sinvcosvsinh2u+sin2vsvΓvvu=-sinhucoshusinh2u+sin2vΓvvv=-sinvcosvsinh2u+sin2v

Find the components of acceleration.

s¨u=u¨+sinhucoshusin2u+sin2vu˙2+2sinucosusin2u+sin2vu˙v˙-sinucosusin2u+sin2vv˙2hus¨v=u¨+sinvcosusin2u+sin2vu˙2+2sinhvcoshusin2u+sin2vu˙v˙-sinvcosusin2u+sin2vv˙2hus¨z=z¨.

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Most popular questions from this chapter

For Example 1, write out the components of U,V, and U×Vin the original right-handed coordinate system and in the left-handed coordinate system S' with the axis reflected. Show that each component ofU×VinS'has the “wrong” sign to obey the vector transformation laws.

If P and S are 2nd-rank tensors, show that 92=81coefficients are needed to write each component of P as a linear combination of the components of S. Show that 81=34is the number of components in a 4th-rank tensor. If the components of the 4th -rank tensor are Cijkm , then equation gives the components of P in terms of the components of S. If P and S are both symmetric, show that we need only 36different non-zero components inCijkm . Hint: Consider the number of different components in P and S when they are symmetric. Comment: The stress and strain tensors can both be shown to be symmetric. Further symmetry reduces the 36components of C in (7.5)to 21or less.

Write out the sumsPijej for each value of and compare the discussion of (1.1).Hint: For example, ifi=2 [or y in(1.1) ], then the pressure across the face perpendicular to thex2axis is P21e1+P22e22+P23e3, or, in the notation of (1.1), Pyxi+Pyyj+Pyzk.

What are the physical components of the gradient in polar coordinates? [See (9.1)].The partial derivatives in (10.5) are the covariant components ofu. What relationdo you deduce between physical and covariant components? Answer the samequestions for spherical coordinates, and for an orthogonal coordinate system withscale factorsh1,h2,h3.

Show that the sum of the squares of the direction cosines of a line through the origin is equal to 1 Hint: Let (a,b,c)be a point on the line at distance 1 from the origin. Write the direction cosines in terms of (a,b,c).

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