Transcript Lecture 1: Rotation of Rigid Body

PHY 126-07 Final Exam
Problem 1 (20 points)
A small ball of mass 2.00 kg is suspended by two wires, wire 1 and wire 2. If two wires
make 90o and wire 2 makes q=30o with respect to the vertical, find the tension T1 and T2.
Q
P
wire 2
q = 30o
wire 1
T2 90o
T1
mg
Solution
Let’s denote the length of wire 2 by L. Then the condition that the ball does not rotate
about the pivot point of wire 2, Q, is: -L mg sinq + LT1=0 (5 points), namely,
T1=mg sinq = (2.00 kg)(9.80 m/s2)sin30o=9.80 N (5 points). The condition that the net
force is zero in the horizontal direction gives: T1cosq - T2sinq = 0 (5 points), namely,
T2=T1/tan30o =17.0 N (5 points).
(The condition that the net force is zero in the vertical direction gives:
T1cosq + T2sinq = mg can be used too.)
Problem 2 (10+10 points)
An innovative musician is making a new music instrument using cylindrical glasses of height
h=1 m with some water in them where the top (bottom) of each glass is open (closed). Right
now he is adjusting the height of the water in one of the glasses so that sound of middle C
note at frequency 256 Hz creates resonance. Assume that the speed of sound is 344 m/s.
(a) What is the maximum height of water for the instrument to resonate with middle C note?
(b) What is the next lowest frequency of sound that this glass instrument can resonate with
if the water level is the same as in Part (a).
Solution
(a) This instrument works as a stopped pipe. Therefore the
frequencies of possible n-th harmonics are expressed by
fn=nv/(4Ln) where n=1,3,5,…, v is the speed of sound, and
L
Ln is the height of air column. Since Ln =nv/(4f), for f=256 Hz,
L1 becomes shortest at (344 m/s)/(4 x 256 1/s)=0.336 m
when n=1. So the maximum water height for middle C note is
1.000 m - 0.336 m=0.664 m. For a higher harmonics, the height
of the air column is larger.
(b) When the water level is adjusted as in Part (a), the fundamental
frequency f1 is 256 Hz. The next lowest frequency is when n=3,
namely the third harmonics f3=3f1=3 x 256 Hz = 768 Hz.
open
h
water
closed
Problem 3 (20 points)
Jane and John will host a big party. They are preparing iced tea for the party. Their plan is
to have iced tea at 3.0oC in an insulated container without ice left in it. They first made 15
liters of tea at 20oC. Then they plan to put some ice at -5.0oC into the container to cool the
tea. How much of ice is needed to make this iced tea according to their plan? Treat the tea
as water. If needed, use the following constants: the density of water rwater=1000 kg/m3, the
latent heat of fusion of water Lf=3.33 x 105 J/kg, specific heat of ice cice=2090 J/(kg K) and
the specific heat of water cwater=4190 J/(kg K).
Solution
The mass of tea is mwater=(15 L) x (1.0 kg/L)=15 kg (3 points).
The amount of heat needed to melt the ice is Qmelt=mice x (3.33 x 105 J/kg) (3 points).
To heat the ice from -5.0oC to 0.0oC the amount of heat needed is Qice=
mice x cice x [0.0oC –(-5.0oC)]=micex(10.5x103 J/kg) (3 points).
The heat needed to raise the temperature of tea from the ice from 0.0oC to 3.0oC is
Qice-water=mice x cwater x [3.0-(0.0oC)]=micex(12.6x103 J/kg) (3 points).
The heat the tea acquires is Qwater=mwater x cwater x [3.0oC -(20.0oC)]=
(15 kg)x cwaterx(-17.0oC)=1068 kJ (3 points).
From Qmelt+Qice+Qice-water+Qwater=0 (2 points), micex(333x 103 J/kg) + micex(10.5x103 J/kg) +
micex (12.6x103 J/kg) + (1068x103 J) = 0. So micex(356x103 J)=1068x103 J and therefore
the amount of ice needed is mice=3.0 kg (5 points).
Problem 4 (5+5+5+5 points)
You are experimenting with a converging lens with a focal length of 10 cm. Find the image
position (whether on the left or right of the lens and how far from the lens), magnification,
and whether the image is upright or inverted, and real or virtual when an object is placed at
(a) 25 cm, (b) 10 cm, (c) 7.5 cm and (d) -20 cm. If the image is formed at infinity, just stating
that gets you 5 points. Assume that the incoming light rays come from the left of the lens
toward the lens.
Solution
1/s + 1/s’ = 1/f where s is the object distance, s’ is the image distance and f is the focal length.
(a) f=10 cm and s = 25 cm, so s’=17 cm. The magnification m=-s’/s=16.7 cm/25 cm=-2/3.
So the image is formed at 17 cm to the right of the lens, real and inverted.
(b) f=10 cm and s=10 cm, so s’=+infinity. So the image is formed at infinity.
(c) f=10 cm and s=7.5 cm, so s’=-30 cm. The magnification m=-s’/s=+4. So the image is
formed at 30 cm to the left of the lens; virtual and upright.
(d) f=10 cm and s=-20 cm, so s’=6.7 cm. The magnification m=-s’/s=+1/3. So the image is
formed at 6.7 cm to the right of the lens; real and upright. (Note the object is a virtual
image.)
Problem 5 (5+5+5+5+5 points)
An experimental power plant at the Natural Energy Lab generates electricity from the
temperature gradient of the ocean. The surface and deep-water temperatures are 27oC
and 6oC, respectively.
(a) What is the maximum theoretical efficiency of this power plant?
(b) If the power plant is to produce 210 kW of power, at what rate must heat be extracted from
the warm water?
(c) If the power plant is to produce 210 kW of power, at what rate must heat be absorbed by the
cold water?
(d) To raise the temperature of water whose mass is dm by DT, how much of heat dQ is needed?
(For this question you may use the symbol c for specific heat of water).
(e) The cold water that enters the plant leaves it at a temperature of 10oC.
What must be the flow rate of cold water through the system in kg/h?
[Use, if needed, the fact that specific heat of water is c=4190 J/(kg K)].
Solution
(a) The maximum efficiency is realized by a Carnot engine. Therefore
(b) Pout /   210 kW / 0.070  3.0 MW ,
(c) 3.0 MW  210 kW  (1 /   1)(210 kW )  2.8 MW .
(d) dQ  c  dm  DT
6
(e) dm d Q / dt (2.8 10 W)(3600s/h )


 6 105 kg/h.
dt
cDT
[4190J/(kg K) ](4K )
  1
279 .15 K
 7.0%
300 .15 K
Problem 6 (3+3+3+3+3+3+2 points)
Three moles of an ideal gas are taken around the cycle acb shown in Figure. For this gas
Cp = 29.1 J/(mol K). Process ac is at constant pressure, process ba is at constant volume,
and process cb is adiabatic. The temperature of the gas in states a,c, and b is Ta=300 K,
Tc=492 K, and Tb=600 K. The universal gas constant R is 8.31 J/(mol K).
(a) For a process at constant pressure, when the temperature changes by DT, how much
does the volume change DV?
(b) For process ac, how much work is done (plug-in numbers)?
(c) For an ideal gas, find the molar heat capacity at constant volume CV in terms of Cp and
the universal gas constant R (no numbers).
(d) For process cb, how much is the change of internal energy, DU in terms of the number of
moles n, CV, and the change of temperature DT (no numbers)?
(e) For process cb, how much work is done (plug-in numbers)?
p
(f) For process ba, how much work is done (plug-in numbers)?
b
(g) Find the total work done for the cycle acba?
Solution
(a) From the ideal gas equation, pV=nRT. So pDV=nRDT.
(b) Wac=pDV=nRDT=(3.00 mol)[8.31 J/(mol K)](492 K – 300 K)
=4.79x103 J.
a
(c) CV=Cp-R.
O
(d) DU=nCVDT.
(e) As Qcb=0 for an adiabatic process, Wcb=Qcb-DU=-nCVDT=-n(Cp-R)DT
=-(3.00 mol)[29.1 J/(mol K)-8.31 J/(mol K)](600 K – 492 K)=-6.74x103 J.
(f) For a process at constant volume, the work done is zero. So Wba=0.00 J.
(g) The total work Wtot=Wac+Wcb+Wba=4.78x103 J – 6.74x103 J=-1.95x103 J.
c
V