Download Seat: PHYS 1500 (Fall 2012) Exam #2, V1 Name: 1. From book Mult

Seat:
PHYS 1500 (Fall 2012) Exam #2, V1
5 pts
1. From book Mult Choice 8.16 A student lies on a very light, rigid board with a
scale under each end. Her feet are directly over one scale and her body is positioned
as shown. The left scale reads 60 lb and the right scale reads 100 lb. What is the
student’s weight?
(a) 60 lb
5 pts
Name:
(b) 80 lb
(c) 100 lb
(d) 160 lb
2. From book Concep Ques 9.3 A 0.2 kg plastic cart and a 20 kg lead cart can roll
without friction on a horizontal surface. Equal forces are used to push both carts
forward for a time of 1 s, starting from rest. After the force is removed at t = 1 s, is
the momentum of the plastic cart greater than, less than, or equal to the momentum
of the lead cart? Explain.
Because they have the same force acting on them for the same amount of
time, they must have the same impulse. The impulse also equals the change
in momentum. They both start with 0 momentum. Therefore, their final
momenta are the same.
5 pts
3. From book HWK Prob 3.37 You are on the edge of a merry-go-round (take your
mass to be 70 kg and that of the merry-go-round to be 500 kg) that spins at 18 rpm. The
diameter of the merry-go-round is 4.6 m. a) What are the merry-go-round’s period (in
s) and frequency in (rev/s)? b) What is your speed? c) What is your acceleration? (a)
The conversion from rpm (rotations per minute) to frequency in (rev/s)
is divide by 60. The period is T = 1/f . (b) The objects speed is the
circumference divided by the period: v = 2πr/T = πdiameter/T . (c) The
acceleration is the centripetal acceleration a = v 2 /r.
18 rev
rev
1
f=
= 0.30
T=
s = 3.33 s
60 s
s
0.3
π 4.6 m
m
(4.34 m/s)2
m
v=
= 4.34
a=
= 8.17 2
(V 1)
3.33 s
s
2.3 m
s
16 rev
rev
1
= 0.27
s = 3.75 s
T=
60 s
s
0.27
π 4.4 m
m
(3.69 m/s)2
m
v=
= 3.69
a=
= 6.18 2
3.75 s
s
2.2 m
s
f=
22 rev
rev
1
= 0.37
T=
s = 2.73 s
60 s
s
0.37
π 4.8 m
m
(5.53 m/s)2
m
v=
= 5.53
a=
= 12.7 2
2.73 s
s
2.4 m
s
(V 2)
f=
(V 3)
10 pts
4. From book HWK Prob 6.49 A 55.0 g ball is on a 1.0 m long string so that the
ball moves at constant speed in a horizontal circle of radius 49.0 cm. a) What is the
tension in the string? b) What is the ball’s angular velocity?
To see how to do this problem, you should make a free body diagram like
the one above. You then need to use Newton’s 2nd law. The x-component
of the net force is T cos θ + 0 = max and the y-component of the net force is
T sin θ − mg = may . The x-component of the acceleration must be v 2 /r for
the ball to go in a circle. The y-component of the acceleration is 0 since
the ball moves in a horizontal circle. For part b), you can find the angular
velocity from ω = v/r. Lastly, you can find the angle from the triangle made
from the string and the radius of the circle: θ = cos−1 (r/1 m).
θ = cos−1 (0.49) = 60.7◦
v2 =
0.618 N cos(60.7◦ ) 0.49 m
m2
= 2.70 2
0.055 kg
s
θ = cos−1 (0.54) = 57.3◦
v2 =
T=
0.768 N cos(57.3◦ ) 0.54 m
m2
= 3.40 2
0.066 kg
s
θ = cos−1 (0.59) = 53.8◦
v2 =
T=
T=
0.935 N cos(53.8◦ ) 0.59 m
m2
= 4.23 2
0.077 kg
s
0.055 kg 9.8 m/s2
= 0.618 N
sin(60.7◦ )
v = 1.64
m
s
ω=
1.64 m/s
= 3.35 rad/s
0.49 m
(V 1)
0.066 kg 9.8 m/s2
= 0.768 N
sin(57.3◦ )
v = 1.84
m
s
ω=
1.84 m/s
= 3.41 rad/s
0.54 m
(V 2)
0.077 kg 9.8 m/s2
= 0.935 N
sin(53.8◦ )
v = 2.06
m
s
ω=
2.06 m/s
= 3.48 rad/s
0.59 m
(V 3)
5 pts
5. From book Mult Choice 6.29 Two planets orbit a star. Planet 1 has orbital radius
r1 and planet 2 has r2 = 4r1 . Planet 1 orbits with period T1 . Planet 2 orbits with
period
(a) T2 = 12 T1
5 pts
(b) T2 = 2T1
(c) T2 = 4T1
(d) T2 = 8T1
6. From book Concep Ques 8.1 An object is acted upon by two (and only two) forces
that are of equal magnitude and oppositely directed. Is the object necessarily in static
equilibrium? Explain and give an example.
No, it isn’t necessarily in static equilibrium. The reason is that the object
can have a size. Generalized example: if the two forces don’t act on a line,
then the object will have an angular acceleration. Specific example: push
up on the edge of your book with a force equal to the weight (the other
force is the weight from gravity), the book will spin.
5 pts
7. Based on book Prob 7.10 A thin rod has a length of 2.20 m, a pivot-point 50.0 cm
from the left end, and almost no mass. The rod is at an angle 30.0◦ above the horizontal.
A force of 3.80 N acts on the right end in the +y-direction. A force of 15.3 N acts on
the left end in the −x-direction. What is the torque on the rod?
You need to add the torques from the two forces using τ = rF sin(ϕ) for each
force. Define θ to be the angle given in the problem and the two torques
as in the figure below. ϕ1 = θ and ϕ2 = 90◦ − θ. Note τ1 is negative and τ2 is
positive.
τ = τ1 + τ2 = −0.50 m 15.3 N sin(30◦ ) + 1.70 m 3.80 N sin(60◦ ) = 1.77 N · m (V 1)
τ = τ1 + τ2 = −0.50 m 15.3 N sin(40◦ ) + 1.40 m 2.80 N sin(50◦ ) = −1.91 N · m (V 2)
τ = τ1 + τ2 = −0.50 m 15.3 N sin(30◦ ) + 2.10 m 1.80 N sin(60◦ ) = −0.55 N · m (V 3)
10 pts
8. From book HWK Prob 9.26 In a daring rescue in space, Jessica (mass of 77.0 kg
with suit) is traveling at 2.60 m/s in a direction 33.0◦ above the +x-axis. Meanwhile,
Michael (mass 98.0 kg with suit) is traveling in the −y-direction with a speed of
1.80 m/s. They collide and hold onto each other. Give the velocity of the resulting
Jessica-Michael unit.
This is a conservation of momentum problem. You need to find the momentum before the collision and set it equal to the momentum after the
collision. You must use components to get the answer. This gives two
equations:
Mj vx,j + Mm vx,m = (Mj + Mm )vx,jm
Mj vy,j + Mm vy,m = (Mj + Mm )vy,jm
All of Michael’s velocity is in the y-direction. The components of Jessica’s
velocity are vx,j = vj cos(θ) and vy,j = vj sin(θ).
vx,j = 2.6
vy,j = 2.6
m
m
sin(33◦ ) = 1.42
s
s
vx,j = 2.6
vy,j = 2.6
vy,jm =
m
m
cos(33◦ ) = 2.18
s
s
m
m
sin(33◦ ) = 1.42
s
s
vx,j = 2.6
vy,j = 2.6
m
m
cos(33◦ ) = 2.18
s
s
vy,jm =
m
m
cos(36◦ ) = 2.10
s
s
m
m
sin(36◦ ) = 1.53
s
s
vy,jm =
vx,jm =
77 kg 2.18 m/s
= 0.96 m/s
175 kg
77 kg1.42 m/s + 98 kg(−1.8 m/s)
= −0.38 m/s
175 kg
(V 1)
vx,jm =
83 kg 2.18 m/s
= 0.99 m/s
182 kg
83 kg1.42 m/s + 99 kg(−1.7 m/s)
= −0.28 m/s
182 kg
(V 2)
vx,jm =
77 kg 2.10 m/s
= 0.92 m/s
176 kg
77 kg1.53 m/s + 99 kg(−1.9 m/s)
= −0.40 m/s
176 kg
(V 3)
5 pts
9. From book Prob 7.16 Three coins are centered on three of the corners of a square
as shown. The length of a side is L = 12 cm. The horizontal position of the center of
mass is
(a) 0 cm
(d) 6 cm
5 pts
(b) 3 cm
(e) 9 cm
(c) 4 cm
(f) 12 cm
10. From book Concep Ques 3.14 You are driving your car in a circular path on level
ground at a constant speed of 20 mph. At the instant you are driving north, and
turning left, are you accelerating? If so, toward what point of the compass (N, W, S,
E, NW, SW, ...) does your acceleration vector point? If not, why not? Explain your
answer.
The car is accelerating towards the center of the circle. Since the car
is moving north but turning left, the center is to the west which is the
direction of the acceleration.
5 pts
11. From book HWK Prob 9.35 In class, Prof Robicheaux sat on a chair that was free
to spin. He started with the weights held in outstretched arms and finished with the
weights pulled into his body. With weights in outstretched arms the moment of inertia
was 9.00 kg·m2 while it was 4.00 kg·m2 with the weights pulled in. If he had an angular
speed of 6.60 rad/s with his arms outstretched, what was the angular speed with his
arms pulled in?
This is a conservation of angular momentum problem because there is no
external torque on this system. This means Lf = Li . You can figure out the
final angular velocity using If ωf = Ii ωi .
ωf =
Ii ωi
9 kg · m2 6.6 rad/s
=
= 14.9 rad/s
If
4 kg · m2
(V 1)
ωf =
Ii ωi
11 kg · m2 4.4 rad/s
=
= 9.68 rad/s
If
5 kg · m2
(V 2)
Ii ωi
8 kg · m2 4.9 rad/s
ωf =
=
= 13.1 rad/s
If
3 kg · m2
(V 3)
10 pts 12. From book Prob 7.64 The 3.0 kg, 40.0-cm-diameter disk is spinning at 300.0 rpm.
How much friction force must the brake apply to the rim to bring the disk to a halt in
3.0 s?
The brake gives a torque to the disk opposite the angular velocity. Using
the moment of ineria of the disk you can find the angular acceleration.
Strategy: from the angular velocity and the time to halt the disk, find the
angular acceleration you need: ωf = ωi +α∆t. Using the angular acceleration
and the moment of inertia you can find the torque needed: τ = Iα. From
the torque and the radius of the disk you can find the force the applied by
the brake: τ = −rF sin ϕ with ϕ = 90◦ and the − because the torque is in the
clockwise direction.
rot 2π rad 1 min
= 31.4 rad/s
min rot 60 s
0 = 31.4 rad/s + α3 s
α = −10.5 rad/s2
ω = 300
1
1
I = MR2 = 3 kg (0.20 m)2 = 0.06 kg·m2
τ = 0.06 kg·m2 (−10.5 rad/s2 ) = −0.63 N m
2
2
−0.63 N m
F=−
= 3.15 N
(V 1)
0.2 m
rot 2π rad 1 min
= 31.4 rad/s
min rot 60 s
0 = 31.4 rad/s + α4 s
α = −7.85 rad/s2
ω = 300
1
1
I = MR2 = 4 kg (0.25 m)2 = 0.125 kg·m2
τ = 0.125 kg·m2 (−7.85 rad/s2 ) = −0.98 N m
2
2
−0.98 N m
F=−
= 3.93 N
(V 2)
0.25 m
rot 2π rad 1 min
= 31.4 rad/s
min rot 60 s
0 = 31.4 rad/s + α5 s
α = −6.28 rad/s2
ω = 300
1
1
I = MR2 = 5 kg (0.30 m)2 = 0.225 kg·m2
τ = 0.225 kg·m2 (−6.28 rad/s2 ) = −1.41 N m
2
2
−1.41 N m
F=−
= 4.71 N
(V 3)
0.3 m
Equations
Basic Mathematic Formulas
cos θ = A/H tan θ = O/A H 2 = O2 + A2
sin θ = O/H
Ccirc = 2πr
Vsph =
4π 3
r
3
Acirc = πr2
√
−b ± b2 − 4ac
x=
2a
Asur of sph = 4πr2
Chapter 2
∆x = xf − xi
(vx )f = (vx )i +ax ∆t
āx = [(vx )f − (vx )i ]/(tf − ti ) = ∆vx /∆t
1
1
∆x = [(vx )i +(vx )f ]∆t
xf = xi +(vx )i ∆t+ ax ∆t2
(vx )2f = (vx )2i +2ax ∆x
2
2
v̄x = ∆x/∆t
Chapter 3
Ax = A cos θ
(vy )f = (vy )i +ay ∆t
Ay = A sin θ
A=
√
A2x + A2y
tan θ = Ay /Ax
⃗vav = ∆⃗d/∆t
⃗aav = (∆⃗v)/∆t
∆⃗d = ⃗df − ⃗di
1
1
∆y = [(vy )i +(vy )f ]∆t
yf = yi +(vy )i ∆t+ ay ∆t2
2
2
2
f = 1/T
v = 2πR/T
a = v /R
⃗ net
F
Chapter 4
⃗1 +F
⃗2 +F
⃗ 3 + ... = m⃗a
=F
(vy )2f = (vy )2i +2ay ∆y
⃗ 12 = −F
⃗ 21
F
Chapter 5
w = mg
fs max = µs n
fk = µk n
fr = µ r n
Chapter 6
s
θ=
r
ωav
θf − θi
∆θ
=
=
tf − ti
∆t
Fg = G
m1 m2
r2
ω = (2π rad)f
G = 6.67259 × 10−11
N · m2
kg2
v = ωr
T2 =
v2
a=
= ω2r
r
4π 2 3
r
GM
Chapter 7
ωf − ωi
∆ω
x1 m1 + x2 m2 + x3 m3 + ...
αav =
=
at = αr
xcg =
tf − ti
∆t
m1 + m2 + m3 + ...
1
1
∆θ = ωi ∆t + α∆t2
ω 2 = ωi2 + 2α∆θ
ω = ωi + α∆t
∆θ = (ωi + ω)∆t
2
2
∑
∑
τ = rF⊥ = r⊥ F = rF sin ϕ
I≡
mr2
τ = Iα
vobj = ωR
aobj = αR
∑
Fx = 0
∑
Chapter 8
Fy = 0
∑
τ =0
Fsp = −k∆x
Chapter 9
⃗J = F
⃗ avg ∆t
⃗p = m⃗v
⃗ avg ∆t = ∆⃗p = m⃗vf −m⃗vi
F
∑
L ≡ Iω
τ=
m1⃗v1i +m2⃗v2i = m1⃗v1f +m2⃗v2f
∆L
∆t
Chapter 10
W = ∆E = ∆K + ∆Ug + ∆Us + ∆Eth + ∆Echem + ...
1
Ktrans = mv 2
2
1
Krot = Iω 2
2
W = F∥ d = (F cos θ)d
1
Us = kx2
2
Ug = mgy
P =
∆E
∆t
P = Fv
Chapter 11
what you get
QH − QC
TC
∆Eth = W + Q
e=
emax = 1 −
what you pay
QH
TH
TC
QH
TH
Qc
COP =
COPmax =
COP =
COPmax =
Win
TH − TC
Win
TH − TC
e=
Chapter 12
n = N/NA = M (in grams)/Mmol
9
TC = T − 273
TF = TC + 32
5
Q = M c∆T
∑
Qk = 0
Q = ±M L
p = F/A
pV = N kB T = nRT
∆L = αLi ∆T
∆V = βVi ∆T
Q
Th − Tc
= kA
∆t
L
Q
= σAe(T 4 − T04 )
∆t
Chapter 13
ρ=
m
V
p=
F
A
p = p0 +ρgd
FB = ρf Vf g
A1 v1 = A2 v2
1
1
p1 + ρv12 +ρgy1 = p2 + ρv22 +ρgy2
2
2
Chapter 14
Fs = −kx
1
Us = kx2
2
√
T = 2π
v = −Aω sin(ωt)
m
k
f=
1
T
a = −Aω cos(ωt)
2
ω = 2πf
x = A cos(ωt)
√
T = 2π
L
g
Some Properties of Materials
Material
Lead
Iron/Steel
Copper
Aluminum
Water
density
11000 kg/m3
7900 kg/m3
8900 kg/m3
2700 kg/m3
1000 kg/m3
Linear Expansion Coef
2.9 × 10−5 ◦ C−1
1.2 × 10−5 ◦ C−1
1.7 × 10−5 ◦ C−1
2.3 × 10−5 ◦ C−1
N.A.
Specific Heat
128 J/(kg ◦ C)
448 J/(kg ◦ C)
387 J/(kg ◦ C)
900 J/(kg ◦ C)
4190 J/(kg ◦ C)