Name________________________________________
Physics 110 Quiz #3, October 11, 2013
Please show all work, thoughts and/or reasoning in order to receive partial credit. The
quiz is worth 10 points total.
I affirm that I have carried out my academic endeavors with full academic honesty.
____________________________
A remote controlled airplane has a mass of 1.5kg . It is
attached to a 2.0m long string (you’re holding the other
end) and the plane is flown in a horizontal circle at a
constant speed. The string makes a 20 o angle with
€ takes 1.5s to
respect to the vertical and the plane
€
complete one circular orbit. The airflow over the wings
of the plane generates a lifting force FL that is always
€
perpendicular to the wings of the plane.
€
y
FL
x
θ
FT
FW
1. What is the lifting force€FL generated by the wings?
 2πr  2


2
v2
4 π 2 rm 4 π ( L sin θ ) m
4 π 2 Lm
 T 
€
F
:
F
sin
θ
=
ma
=
m
=
m
=
=
→
F
=
∑ x T
x
T
r
r
T2
T2
T2
∑ Fy : FL − FW − FT cosθ = may = 0 → FL =FW + FT cosθ = 1.5kg × 9.8 sm2 + 52.6N = 64.2N
(
)
directed vertically upwards from the wings.
€
2. What is the tension force FT in the string?
2
4 π 2 mL 4 π (1.5kg)(2m)
=
= 52.6N directed along the string away from the
2
T2
1.5s
(
)
€
plane.
FT =
€
3. Suppose that the angle that the string makes with the vertical is constant and that the
plane is able to fly at a speed that is greater than that of the initial problem. In this
case
€
€
€
€
a.
b.
c.
d.
FL ↑ and
FL ↓ and
FL ↑ and
F ↓ and
€L
€
€
€
FT ↑.
FT ↑.
FT ↓.
FT ↓.
Useful formulas:
Motion in the r = x, y or z-directions
1
2
rf = r0 + v 0r t + ar t
Uniform Circular Motion
v2
ar =
r
v2
Fr = mar = m
r
2πr
v=
T
mm
FG = G 1 2 2
r
2
v fr = v 0r + ar t
2
2
v fr = v 0r + 2arΔr
Vectors
Circles
Triangles
Spheres
C = 2πr
A = 12 bh
A = 4 πr 2
A = πr 2
V = 43 πr 3
Quadratic equation : ax 2 + bx + c = 0,
whose solutions are given by : x =
−b ± b 2 − 4ac
2a
Useful Constants
€
€
2
magnitude of avector = v x + v 2 y
g = 9.8 m s2
v 
direction of avector → φ = tan−1 y 
 vx 
N A = 6.02×1023 atoms mole
Linear Momentum/Forces
→
€
Geometry /Algebra
→
→
→
F = ma
→
K r = 12 Iω
Heat
2
TC = 59 [TF − 32]
2
Fs = −k x
U S = 12 kx 2
F f = µFN
WT = FdCosθ = ΔE T
TF = 95 TC + 32
Lnew = Lold (1+ αΔT )
Anew = Aold (1+ 2αΔT )
Vnew = Vold (1+ βΔT ) : β = 3α
PV = Nk B T
W R = τθ = ΔE R
W net = W R + WT = ΔE R + ΔE T
ΔE R + ΔE T + ΔU g + ΔU S = 0
3
2
€
Fluids
kB T = 12 mv 2
ΔQ = mcΔT
ΔQ kA
PC =
= ΔT
Δt L
ΔQ
PR =
= εσAΔT 4
ΔT
ΔU = ΔQ − ΔW
ΔE R + ΔE T + ΔU g + ΔU S = −ΔE diss
Rotational Motion
kg 2
vsound = 343 m s
U g = mgh
→
2
kB = 1.38×10−23 J K
Work/Energy
K t = 2 mv
→
p f = p i + F Δt
→
σ = 5.67×10−8 W m 2 K 4
€
1
p = mv
→
G = 6.67×10−11 Nm
Simple Harmonic Motion/Waves
ω=
€
k
2π
= 2πf =
m
T
TS = 2π
m
k
TP = 2π
l
g
1
v =±
k  x 2 2
A1− 
m  A2 
x ( t ) = x max sin(ωt ) or x max cos(ωt )
v ( t ) = v max cos(ωt ) or − v max sin(ωt )
a( t ) = −amax sin(ωt ) or − amax cos(ωt )
v max = ωx max ; amax = ω 2 x max
Sound
v = fλ =
v = fλ = (331+ 0.6T) ms
v
2L
I = 2π 2 f 2 ρvA 2
I
; Io = 1×10−12 mW2
I0
v
v
f n = nf1 = n ; f n = nf1 = n
2L
4L
f n = nf1 = n
β = 10log
€
€
FT
µ