46)
Given we effectively have a flat rotating cylinder than it behooves us to make
use of cylindrical coordinates. The
divergence is:

**Ñ**=

**^h**¶ / ¶r +

**^m**/ r (¶ / ¶f) +

**^k**¶ / ¶z

And
we have:

**r**=**r^h + z****^k**
(z
= constant)

The
velocity is:

**v**= d

**r**/ dt = r’

**^h**+ r f’

**^m**= r w

**^m**

**Ñ**x

**v**= (

**^h**x

**^m**) w +

**^m**/ r (¶ / ¶f) [r w ^m]

+

**^k**x

**^m**¶ (r w)/ ¶z

But: (¶ / ¶f) ^m = -

**^h**
So
that:

**Ñ**x

**v**= 2 (

**^h**x

**^m**) w = 2 w (

**)**

__B__
47)
Based on the presentation in (46) it must be in the

**+z direction**(**)**__D__
48) curl

**ix**^{2}+ jy^{2}+ k (x^{2}– y^{2})
=

**Ñ**x**i**x^{2}+**j**y^{2}+**k**(x^{2}– y^{2})
=
(

**i x j**) (¶ / ¶x) y^{2}+ (i x k) ¶ / ¶x (x^{2 }+^{ }y^{2 })
+
(

**j x i**) ¶ / ¶y (x^{2}) + (**j x k**) ¶ / ¶y (x^{2 }-^{ }y^{2 })
+
(

**k x i**) ¶ / ¶z (x^{2}) + (**k x j**) ¶ / ¶z (x^{2 }-^{ }y^{2 })
= -2 (

**i**y +**j**x)
49)
First,
construct a sketch of the string, e.g. in u, x
coordinates as shown in order to do a brief analysis as shown:

We have in terms of the string tension T:

T

_{u}= T sin q
As
usual per these approaches, assume q is

*very small*, in which limit,*sin**q » tan q*
T
sin q = T tan q = T (¶ u/ ¶x)

Take
the force difference:

[T

**]**_{u}**- [T**_{x + dx}**]**_{u}**= ¶ / ¶x (T (¶ u/ ¶x) dx**_{x}
To
ensure

*no net horizontal forces*due to tension, we limit the situation to small slopes. Also, neglect the possibility of a vertical force per unit length, so:
r dx (¶

^{2 }u/ ¶ t^{2}) = ¶ / ¶x (T (¶ u/ ¶x) dx
Which
simplifies to:

r (¶

^{2 }u/ ¶ t^{2}) = T (¶^{2}u/ ¶x^{2})
Now,
transform to standard wave equation in 1-dimension:

¶

^{2}u/ ¶x^{2}- 1/ c^{2}(¶^{2 }u/ ¶ t^{2}) = 0
From
this we can solve for the velocity c:

c= (T/ r )

^{½}^{ }

The

*boundary conditions*can be written:
u(x,
t) = u(L, t) = 0

and:

u(x,t)
= X(x) Y(t)

Then,
on forming the particular 2

^{nd}order differential equation:
c

^{2}/ X (d^{2 }X/ dt^{2}) = 1/ Y (d^{2 }Y/ dt^{2})
From
which:

1/
Y (d

^{2 }Y/ dt^{2}) = - w^{2}
And: c

^{2}/ X (d^{2 }X/ dt^{2}) = - w^{2}^{}

The

*solutions*are of the form:
X = C
cos wx/ c
+ D sin wx/ c

Check
boundary conditions: X(0) = C = 0

And
X(L) = D sin wL/ c
= 0

Then
w will assume values:

w = n p c/ L = n p / L [T/ r ]

^{½}
The
frequency is f = w/ 2p
= n p / L [T/ r ]

^{½ }w/ 2p^{}
Or
f = n/ 2L [T/ r ]

^{½}^{ }

Note:
Though in the actual GRE there will be

*multiple choice options*, this is generally the type of problem you want to avoid because – in the absence of memory (say already knowing the*wave equation for a string*) and the useful DEs, the derivation is excessively long. Bear in mind this is a timed test. So you have 3 hours (180 mins.) to get through__100 problems__. That works out to roughly 1 m 48s per problem. That means if you need to send something like 8 minutes deriving the answer for one problem, you will lose valuable time for 4-5 others (even given 2 may be easy) . If, however, you can make an__educated guess__, say based on a knowledge of dimensional analysis, then that might be worth a try!
50)
(

**)**__E__
A
rigid body may move in any of three directions. It may also rotate about any of
three axes. Therefore, it must have six degrees of freedom.

51)
(

**)**__A__
If
a rigid body is constrained to rotate about a given axis it has only one
rotational degree of freedom. It cannot have any translational freedom since
this would change the axis of rotation.

52)
(

**)**__B__
It
has one degree of rotational freedom since the axis of rotation must be
parallel to the plane. It can only have one degree of translational freedom to
keep the disk perpendicular to the plane. Hence the total is 1 R + 1T = 2 DF.

(To
Be Continued)

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