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(PHYS111)[2009](f)final~=tkk933^_94273.pdf
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Physics 111 (Fall 2009)
Final Examination Solution

12 2
1. (a) Rotational inertia of the disk . m(2r) . 2mr
2 1 212
Rotational inertia of the rod about the disk center . (4m)r . mr
12 3 2127 2
Total rotational inertia . Ii . 2mr . mr . mr
33
After the disturbance,
Rotational inertia of the disk . 2mr2
Using the parallel axis theorem,

1 2 2132
Rotational inertia of the rod about the disk center . (4m)r . (4m)r . mr
12 3 213 219 2Total rotational inertia . If . 2mr . mr . mr
33
Using the conservation of angular momentum,

I .. I .
ii ff
Hence the final angular velocity is
Ii . 7mr2 /3 . 7 ..1
.. .... ..5 ... .5 .1.84 rads (answer)
fi 2
If .19mr /3 ..19 .
1 L2
(b) Kinetic energy of rotation K . I.2 .
22I
Since the angular momentum is conserved,
KfIi 7mr2/3 7

.. .. 0.368 (answer)
Ki If 19mr2 /3 19
2. (a) Period
m
0.5
T. 2.

. 2.
. 0.1814 s . 0.181s (answer)
k 600
(b) Using the conservation of energy, 1 2121 2
kA . kx0 . mv0
2 22 121 212
(600) A . (600)(0.03) . (0.5)(1.2) . 0.63 J
22 2
. 0.04583 m . 0.0458 m (answer)

(c) x(t) . xm cos(.t ..) v(t) . ..xm sin(.t ..)
Here, ..

At t = 0,
x0 . xm cos.

v0 . ..xmsin.
Dividing,

v0 . .. tan.
x
0
v0 1.2
tan... .. ..1.1547
.x0 (34.64)(0.03)
...49.1o (answer)
3. (a) Since y(x, t) = (0.21 m)sin(20.x . 360.t), we have k = 20. and . = 360.. 2. 2.
Wavelength: ... . 0.1m (answer)
k20.
. 360.

Frequency: f.. .180 Hz (answer)
2. 2.
Wave velocity: v. f.. (180)(0.1) . 18 ms.1 (answer)

.
(b) Since v .
.

...v2 . (0.16)(18)2 . 51.84 N . 51.8 N (answer)
(c) Reflected wave: yr (x,t) . (0.21m)sin(20.x . 360.t ..). y(x,t) . yi (x,t) . yr (x,t) . (0.21m)sin(20.x . 360.t) . (0.21m)sin(20.x . 360.t ..). At x = 0.12 m, y(x,t) . (0.21m)sin(2.4.. 360.t) . (0.21m)sin(2.4.. 360.t ..). Using the sum formula,
. ... ..
y(x,t) . (0.42 m)sin.2.4.. .cos.360.t ..
. 2 .. 2 .

At x = 0.12 m, the amplitude is
... ...
A(x) . (0.42 m)sin.2.4.. .. (0.42 m)sin.0.4.. .
. 2 .. 2 .

At the fixed end, A(x) =0. Therefore,
.
0.4.. . 0
2
...0.8. or 1.2. (answer)

4. (a) The two resonance positions differ by half wavelength. Hence the wavelength is .. 2(0.586 . 0.419) . 0.334 m
Frequency:
v 343

f.. .1027 Hz .1030 Hz (answer)
. 0.334
0.419
When h = 41.9 cm, the number of wavelengths ..1.25. There are 2.5 loops.
0.334
0.586
When h = 58.6 cm, the number of wavelengths .. 1.75. There are 3.5 loops.
0.334

v 1
5. Mach cone angle: sin... . 0.625 . .. 38.68o
vS 1.6

h 5000
BC .. . 6245 m
tan. tan 38.68o h 5000
AC .. . 8660 m
tan. tan 30o
BC . AC 6245 . 8660
. t.. . 27.2 s (answer)
vs (1.6)(343)
6. (a) The total gravitational force acting on one of the revolving stars
22 2
G(2M )(M ) G(M )(M )2GM GM 9GM
Sun Sun Sun Sun Sun Sun Sun
.. ... (answer)
2 2 222
r (2r ) r 4r 4r
E E EEE
(b) Using Newtons second law,
v29GM 2
9GM
Sun Sun
M .. v .
Sun 2
r 4r 4r
EE E
Orbital perio