Name________________________________________
Physics 110 Quiz #4, October 16, 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.
____________________________
Suppose that you are a roller coaster designer tasked with a new thrill ride. The cars and
their riders (of mass mtotal = 1000kg ) start on the ground an immediately a mechanism
pulls them up a large hill of height h . The cart and riders crest the hill and starting
essentially from rest at the top of the hill, travel down to the hill to the first loop. The top
of the loop is 15m above the ground and the cart and riders travel around the loop and
€ moving to the right. The cart and riders then travel towards a ramp that is
exit the loop
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inclined at an angle of 45 0 with respect to the ground. The entire track is frictionless
with the exception of the inclined portion with the ramp which as a coefficient of friction
€
µ.
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1. What is the minimum height h that is needed so that the cart and riders (the blue box)
will make it around the loop?
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h
To calculate the minimum height, we’ll use the fact that energy is conserved and that
at the top of loop, the minimum speed will lead to the minimum height. We find the
minimum speed by examining the forces on the cart and riders assuming that
vertically up is the positive y-direction. When the normal force vanishes we have the
condition for the minimum speed.
v2
v2
2
→ v min
= rg
∑ Fy : − FN − FW = mac = −m r → FW = mg = m min
r
Applying conservation of energy we have for the minimum height
2
ΔE = ΔKE + ΔU g + ΔU s = ( 12 mv 2f − 12 mv 2f ) + ( mgy f − mgy i ) = 12 mv min
+ ( mg(2r) − mgh) = 0
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∴h =
2mgr + 12 mrg
= 2.5r = 2.5 × 7.5m = 18.8m
mg
2. Suppose that to ensure rider safety you make the height of the hill 20% greater than
the value you found in part 1. What will the speed at the bottom of the loop?
€ than the value in part 1,
Assuming that the height of the starting hill is 20% greater
and taking the ground as the zero of our potential energy, we have by conservation of
energy
2
ΔE = ΔKE + ΔU g + ΔU s = ( 12 mv 2f − 12 mv 2f ) + ( mgy f − mgy i ) = 12 mv bottom
− mg(1.2h ) = 0
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∴v bottom = 2(1.2gh ) = 2.4 × 9.8 sm2 ×18.8m = 21.0 ms
€
3. When the cart and riders enter the inclined portion of the ride they come to rest at a
distance x measured along the incline. Considering just the cart and riders, the net
work done by all external forces on the cart and rider as the cart and riders travel
along the incline is given by
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€
€
€
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a.
b.
c.
d.
W net
W net
W net
W net
= 0J .
= −1.2mgh .
= mgx sin θ .
= −µmgx cos θ .
The net work done on the cart and riders by either calculating the change in kinetic
energy or by determining the net force that acts on the cart and riders over the
distance given.
From the change in kinetic energy we have:
2
W net = ΔKE = 12 mv 2f − 12 mv i2 = − 12 mv bottom
= −1.2mgh
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From the
we have:
 net force

W net = Fnet ⋅ Δx = Fnet Δx cos φ = ( mgsin θ + µmgcos θ ) x cos(180) = −( mgsin θ + µmgcos θ ) x
which one could work out to show gives choice B.
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
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2
magnitude of avector = v x + v 2 y
Linear Momentum/Forces
→
→
23 atoms
N A = 6.02×10
→
→
F = ma
→
σ = 5.67×10
kg 2
vsound = 343 m s
m 2K 4
Heat
2
TC = 59 [TF − 32]
K r = 12 Iω 2
TF = 95 TC + 32
U g = mgh
Lnew = Lold (1+ αΔT )
→
Fs = −k x
U S = 12 kx 2
F f = µFN
WT = FdCosθ = ΔE T
Anew = Aold (1+ 2αΔT )
Vnew = Vold (1+ βΔT ) : β = 3α
PV = Nk B T
W R = τθ = ΔE R
3
2
kB T = 12 mv 2
W net = W R + WT = ΔE R + ΔE T
ΔE R + ΔE T + ΔU g + ΔU S = 0
ΔQ = mcΔT
ΔE R + ΔE T + ΔU g + ΔU S = −ΔE diss
PC =
ΔQ kA
= ΔT
Δt L
ΔQ
PR =
= εσAΔT 4
ΔT
ΔU = ΔQ − ΔW
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Rotational Motion
2
kB = 1.38×10−23 J K
mole
Work/Energy
K t = 2 mv
→
p f = p i + F Δt
→
−8 W
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1
p = mv
→
G = 6.67×10−11 Nm
g = 9.8 m s2
v 
direction of avector → φ = tan−1 y 
 vx 
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Geometry /Algebra
Fluids
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 = (331+ 0.6T)
v = fλ =
m
s
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
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FT
µ