Solutions to Final Exam for GP I,
SNME, Fall 2011.
1. 10 %
A small block with mass 0.250 kg is attached
to a string passing through a hole in a
frictionless, horizontal surface (see figure).
The block is originally revolving in a circle
with a radius of 0.800 m about the hole with
a tangential speed of 4.00 m/s. The string
is then pulled slowly from below, shortening
the radius of the circle in which the block
revolves. The breaking strength of the string is 30.0 N.
What is the radius of the circle when the string breaks ?
Solution
Tension T of the string is radial so that it gives no torque on the block. Angular
momemtum is therefore conserved, i.e., L  mvr  const .
Hence
2
v2 m  L 
L2
T m
 


r
r  mr 
m r3
L  0.250  4.00  0.800  0.800
 0.800
30 
2
0.250 r 3

r
3
 0.800
2
30  0.250
 0.440 m
2. 10 %
You are trying to raise a bicycle wheel
of mass m and radius R up over a curb
of height h. To do this, you apply a
horizontal force F at the center of the
wheel (see figure).
What is the smallest magnitude of the
force F that will succeed in raising the wheel over the curb ?
Solution
Perpendicular distance of F to the corner of the curb is R – h.
Perpendicular distance of mg to the corner of the curb is
R 2   R  h   2 Rh  h 2 .
2
Torques about the corner of the curb:
  F  R  h  m g
2Rh  h 2
Setting   0 gives the minimum force required to lift the wheel:
F
mg
2 Rh  h 2
Rh
3. 20 %
You measure the period of a physical pendulum about one pivot point to be T.
Then you find another pivot point on the opposite side of the center of mass that
gives the same period. The two points are separated by a distance L.
 2 
g L
 .
 T 
2
Show that
Solution
For a physical pendulum
I
.
mgd
T  2
Let the distances of the pivot points to the CM be d1 and d2 . We have
m g d 2 T 2  4 2 I 2
m g d1 T 2  4 2 I1

m g  d1  d 2  T 2  4 2  I1  I 2 
From the parallel axis theorem, we have
I1  I cm  m d12

m g 1d
d2
I 2  I cm  m d 22
2
T4
2
m
2
1
d
g T 2  4 2  d1  d 2   4 2 L

2
2
d
 2 
g L

 T 
4.
2
20%
A cylindrical container of an incompressible
liquid of density  rotates with constant
angular speed  about its axis of symmetry,
which we take to be the y axis (see figure).
Find the height h of the liquid surface as a
function of r.
Solution
Pressure p is a function of both y and r. Let the liquid surface at r  0 be the
origin of the y –axis. Then for the pressure p  r  on the plane y  0 ,
d p r
v2

  2 r
dr
r
p  r   p  0 

1
 2 r2
2
Using
p  r   pA   g h  r 
with
p  0  p A
where pA is the atmospheric pressure, we have
1
2
 g h r   2 r2
or
h r 
1 2 2
 r
2g
5. 20%
A 5.00 m, 0.732 kg wire is used to
support two uniform 235 N posts of
equal length (see figure).
A strong wind is blowing, causing the
wire to vibrate in its 7th overtone (8th
harmonic).
Assume the wire is essentially horizontal and that the speed of sound in air is 344
m/s.
What are the frequency and wavelength of the sound this wire produces?
Hint: The frequency of the sound wave (in air) is equal to the frequency of the
standing wave in the vibrating wire.
Solution
Torques about left pivot point:
T L sin 57  w
T
L
cos57  0
2
235 cos57
 76.31 N
2 sin 57
Velocity of wave on wire is given by v 
T

.
Hence,
76.31
 22.83 m / s
0.732 / 5.00
For the 8th harmonic, wavelength of standing wave is
v

2 Lwire 2  5.00

 1.25 m
n
8
Its frequency is
f 
v

22.83
 18.26 Hz
1.25

The frequency of the sound wave is therfore 18.3 Hz. Its wavelength is
v
344
  sound 
 18.8 m
f
18.26
6. 10% + 10%
A spherical shell has inner and outer radii a and b, respectively, and the temperatures
at the inner and outer surfaces are T1 and T2. The thermal conductivity of the
material of which the shell is made is k.
(a) Find the total heat current through the shell.
(b) Find T within the shell as a function of r, the distance from the center of the
shell.
n
 x dx 
Hint:
x n 1
for all n  1 ,
 n  1!
and
x
1
dx  ln x .
Solution
(a)
H   k  4 r 2 

dT  
dT
dr
H dr
4 k r 2
T2  T1  
H
4 k

b
a
H b  a 
dr
H 1 1

  
2
r
4 k  b a 
4 kab
T T 
H  4 k ab  1 2 
 ba 
(b)
T  T1  
T  T1 
H
4 k

r
a
H r  a
dr
ra b


   T1  T2 
2
r
ba r 
4 ka r
ra b
   T1  T2 
ba r