## Friday, June 21, 2019

### Solutions To Theoretical Mechanics Problems (2)

The Problems Again:

1)  Prove the relations below:

i) dA/dt =  h/ 2

ii) P  =   2 p  ab h -1

And show:   P  =   2 p Ö (a 3 ) /  Öm

Where:   m  =  G (m1 + m2)

2) Because of the Earth’s rotation a plumb bob does not hang precisely in the direction of Earth’s gravitational attraction, i.e. in the direction of its weight mg.  Given the radius of the Earth = 6.4  x 10 6 m  and your latitude is 45 degrees, find the angle by which the bob deviates from the direction of g. (Hint: Sketch a diagram with forces, angle shown to assist in solution.)

3) An asteroid moves in an elliptic orbit around the Sun. The lengths of the major and minor axes are 2a and 2b, respectively. If the asteroid’s velocity at the point of closest approach (where it crosses the major axis) is   v o   then how much time is needed for the object to make one complete orbit?  (Hint:  Take the area of an ellipse to be:   p ab  )

Solutions:

1)   (i) We know:   h  =  L/ m  =  r 2   dq  /dt

Working from 1st principles using the diagram below:

We have:

D =    ½  [r   x (r +  r )] =   r^  x   Dr

Now, the areal velocity at P  is by definition:

lim Dt ®0  (D A /D t)

Then :  dA/ dt =  ½   r   x  r'  =     ½  (  2   dq  /dt   )

Or   dA/dt =  h/ 2

(ii)  A  =   ½  h (t   -  t 0)

But if  (t   -  t 0)  = P   then, for an ellipse A =  p  ab

And:  p  ab   =   ½  h (P)   Or:   P =  2 p  ab  / h

Now Rem:  h =    [m a(1 - e2)]½

And write:

P =  2 p  ab  / h  =    2 p  ab/ [m a(1 - e2)]½

Subst. for  e:  = (a2 - b2)½ / a

Cranking out the algebra (details left to readers), we find:

P  =   2 p Ö (a ) /  Öm

(2)  For clarity we use the following three diagrams to show the situation, first from a 'global' view then focusing on the forces, angle to be found.

First the global view with latitude circle and latitude angle (phi) shown: Second, the deviation of the bob owing to Earth's rotation with g forces shown: Last, the angle we need to obtain, i.e. between true vertical and the deviated one from rotation, This angle which I label s   and can be found using the law of cosines.  We obtain:

sin ( s ) /  (  w 2  R cos ϕ)  =  sin  ( ϕ  ) /  g'

(But   g'     g   given   sin s      s )

Then:     s    = ( w 2 R )  [sin ϕ cos ϕ  ]  /  g

= ( w 2 R )  [sin (p / 4)  cos (p / 4 ) /  9.8  N/ kg

And substitute in:

w  = 7.3 x  x 10 -5 rad/ sec ;   R   =  6.4  x 10 6 m

sin (p / 4)  =   cos (p / 4 ) =  Ö2/  2

(3)   From:  dA/dt =  h/ 2 ;    We get:

d A/ dt     =   ½    dq /  dt

ò  d A    =   ½  2  ò    dq  (dt) / dt

ò  d A       = ½  2  ò  2p  o    dq

A  =   p ab   =   ½  2  ò  2p  o    dq

But:    ò  2p  o    dq   =   p   radians  so that:

p ab   =   ½   (p )   2

Subst.  r =  v o / w     so:

ab   =     v o / w  2

But:  w    = p  /  T

So:  ab =  v o 2  T 2   /  (p) 2

T 2    =   (p) 2  ab /   v o 2

T     =     p  Ö (ab)  /   v o