oldmidterm 02 – JYOTHINDRAN, VISHNU – Due: Mar 5 2007, 4:00 am
Question 1, chap 6, sect 3.
part 1 of 1
10 points
An amusement park ride consists of a large
vertical cylinder that spins about its axis fast
enough that any person inside is held up
against the wall when the floor drops away
(see figure). The coefficient of static friction
between the person and the wall is µ and the
radius of the cylinder is R.
R
1
m v2
r
Frictional force: fs ≤ µ N = fsmax
Solution: The maximum frictional force
due to friction is fmax = µ N , where N is the
inward directed normal force of the wall of
the cylinder on the person. To support the
person vertically, this maximal friction force
fsmax must be larger than the force of gravity
m g so that the actual force, which is less
than µN , can take on the value m g in the
positive vertical direction. Now, the normal
v2
force supplies the centripetal acceleration
R
on the person, so from Newton’s second law,
Centripetal force: F =
N =
m v2
.
R
Since
ω
fsmax = µ N =
What is the minimum tangential velocity
needed to keep the person from slipping downward?
s
gR
correct
1. v =
µ
p
2. v = µ g R
1p
gR
2
p
= gR
p
= 2µ gR
p
= 2 gR
p
= µ 2πgR
p
= 2µgR
p
= µ 2gR
3. v =
4. v
5. v
6. v
7. v
8. v
9. v
10. v =
1p
gR
µ
Explanation:
Basic Concepts:
µ m v2
≥ mg,
R
the minimum speed required to keep the person supported is at the limit of this inequality,
which is
2
µ m vmin
= m g, or
R
1
gR 2
.
vmin =
µ
Question 2, chap 6, sect 99.
part 1 of 1
10 points
The following figure shows a Ferris wheel
that rotates 2 times each minute and has a
diameter of 24 m.
The acceleration of gravity is 9.8 m/s2 .
What force does the seat exert on a 48 kg
rider at the lowest point of the ride?
Correct answer: 495.666 N (tolerance ± 1
%).
oldmidterm 02 – JYOTHINDRAN, VISHNU – Due: Mar 5 2007, 4:00 am
2
Explanation:
v2
=M g−
k̂
The period of the Ferris wheel is
r
= (22 kg)
T = 60 s/2 = 30 s .
(7.8 m/s)2
2
× 9.8 m/s −
k̂
The speed of the wheel is
15 m
= 126.368 N
2πr
v=
T
2 π (12 m)
=
30 s
= 2.51327 m/s ,
The force with which he presses against the
so the centripetal acceleration is
seat is equal but opposite to the normal force.
2
v
a=
Question 4, chap 6, sect 99.
r
2
part 1 of 1
10 points
(2.51327 m/s)
=
12 m
An object attached to the end of a string
= 0.526379 m/s2 .
swings in a vertical circle of radius 3.4 m as
The force exerted by the seat balances the
gravity and provides the centripetal force, so
Fl = m [g + a]
= (48 kg) (9.8 m/s2 + 0.526379 m/s2 )
shown. At an instant when θ = 21 ◦ , the speed
of the object is 15 m/s and the tension in the
string is 11 N.
The acceleration of gravity is 9.8 m/s2 .
= 495.666 N .
3.4 m
Explanation:
The boy is executing circular motion at the
top of the crest, so the net force is centripetal.
Two forces act on him: (1) his weight acting
down and (2) the supporting normal force
from the seat which acts up
~ net = M~a
F
2
~ g − N = M v k̂
F
r
M v2
~
k̂
⇒ N = Fg −
r
9.8 m/s
21 ◦
/s
A(n) 22 kg boy rides a roller coaster.
The acceleration of gravity is 9.8 m/s2 .
With what force does he press against the
seat when the car moving at 7.8 m/s goes over
a crest whose radius of curvature is 15 m?
Correct answer: 126.368 N (tolerance ± 1
%).
2
15 m
Question 3, chap 6, sect 99.
part 1 of 1
10 points
11 N
What is the mass m of the object?
Correct answer: 0.175538 kg (tolerance ± 1
%).
Explanation:
Let :
θ
v
r
g
= 21 ◦ ,
= 15 m/s ,
= 3.4 m , and
= 9.8 m/s2 .
oldmidterm 02 – JYOTHINDRAN, VISHNU – Due: Mar 5 2007, 4:00 am
R
a frictionless hole in the center of the table.
A second 93 kg mass is attached to the other
end of the string.
The acceleration of gravity is 9.8 m/s2 .
v 2 kg
T
θ
g
mg
v
5m
θ
93 kg
Basic Concepts:
Fr = m ar =
m v2
r
Solution: The centripetal acceleration is
v2
ar = . According to Newton’s second law,
r
the total force in the radial direction should
be equal to Fr = m ar .
The forces acting on the mass are the tension T on the string, directed radially inward,
and the weight force, whose component in the
radial direction is m g sin θ. The radial component of the resultant force is then
Fr = m ar = T + m g sin θ ,
2
v
m
− g sin θ = T .
R
T
v2
R
=
Determine the period (the time for one revolution).
Correct answer: 0.658151 s (tolerance ± 1
%).
Explanation:
Let :
or
Finally,
m=
3
m1
m2
r
g
= 2 kg ,
= 93 kg ,
= 5 m , and
= 9.8 m/s2 .
The linear velocity v can be expressed in
terms of the distance it travels each revolution:
2πr
v=
.
T
Consider the forces acting on each mass:
− g sin θ
(11 N)
m/s)2
(15
− 9.8 m/s2 sin(21 ◦ )
(3.4 m)
~
T
~
N
m1 ~g
~
T
m2 ~g
= 0.175538 kg .
Question 5, chap 6, sect 3.
part 1 of 1
10 points
A 2 kg mass moves in a circular path of
5 m radius on a frictionless horizontal table.
It is attached to a string that passes through
The tension T in the string provides the
centripetal force required to keep m1 moving
in X
a circular path. Applying
Fr = m ar to m1 , we obtain
v2
T = m1 ac = m1
r
and
oldmidterm 02 – JYOTHINDRAN, VISHNU – Due: Mar 5 2007, 4:00 am
X
Fy = m ay = 0 to m2 , so
which implies
m2 g − T = 0
m2 g
m2 g
T2
T
v2
= m1
r
4 π2 r
= m1
T2
4 π 2 m1 r
=
m2 g
r
m1 r
= 2π
m2 g
s
(2 kg) (5 m)
= 2π
(93 kg) (9.8 m/s2 )
= 0.658151 s .
Question 6, chap 3, sect 4.
part 1 of 1
10 points
~ and B
~ are given by
The vectors A
~ = 4.61 î + 3.34 ĵ
A
~ = −1.02 î + 1.03 ĵ
B
~ and B.
~
Find the angle between A
◦
Correct answer: 98.7968 (tolerance ± 1 %).
Explanation:
~ and B
~ are given by
The magnitudes of A
q
A = A2x + A2y = 5.69278
q
B = Bx2 + By2 = 1.44959 .
î · î = ĵ · ĵ = 1 and î · ĵ = 0, so
~ ·B
~ = 4.61 î + 3.34 ĵ · −1.02 î + 1.03 ĵ
A
= (4.61) (−1.02) î · î + (3.34) (1.03) ĵ · ĵ
h
i
+ (4.61) (1.03) + (3.34) (−1.02) î · ĵ
= −1.262 .
The dot product is also
~ ·B
~ = A B cos θ
A
~ ·B
~
A
cos θ =
AB
−1.262
=
(5.69278) (1.44959)
= −0.152929 ,
4
θ = arccos(−0.152929)
= 98.7968◦ .
Question 7, chap 3, sect 4.
part 1 of 1
10 points
~ has components
Vector A
Ax = −3.06,
Ay = 8.24,
Az = 3.86
~ has components
while vector B
Bx = 7.65,
By = −8.08,
Bz = 6.59.
What is the angle θAB between these vectors?
(Answer between 0◦ and 180◦ .)
Correct answer: 121.329 ◦ (tolerance ± 1 %).
Explanation:
Consider two formulæ for the scalar prod~ ·B
~ of two vectors:
uct A
~ ·B
~ = Ax B x + Ay B y + Az B z
A
(1)
in terms of the two vectors’ components, and
also
~ ·B
~ = |A|
~ |B|
~ cos θAB
A
(2)
in term of their magnitudes and the angle between them. Given the data, we immediately
calculate
q
A2x + A2y + A2z = 9.60004, (3)
q
~
|B| = Bx2 + By2 + Bz2 = 12.932, (4)
~ =
|A|
and using eq. (1),
~ ·B
~ = −64.5508.
A
(5)
Hence, according to eq. (2),
cos θAB =
~ ·B
~
A
= −0.519951
~ |B|
~
|A|
(6)
oldmidterm 02 – JYOTHINDRAN, VISHNU – Due: Mar 5 2007, 4:00 am
5
and therefore
v
N
F
θAB = arccos(−0.519951) = 121.329◦ . (7)
µN
θ
mg
Question 8, chap 7, sect 3.
part 1 of 1
10 points
A crate is pulled by a force (parallel to the
incline) up a rough incline. The crate has an
initial speed shown in the figure below. The
crate is pulled a distance of 7.14 m on the
incline by a 150 N force.
The acceleration of gravity is 9.8 m/s2 .
1. 2
/
7m
s
15
k
11
µ
0N
Wf ric + Wappl + Wgravity = ∆K
To find the work done by friction we need the
normal force on the block from Newton’s law
X
Fy = N − m g cos θ = 0
Thus
g
=
The work-energy theorem with nonconservative forces reads
0. 3
39
36◦
What is the change in kinetic energy of the
crate?
Correct answer: 407.493 J (tolerance ± 1 %).
Explanation:
⇒ N = m g cos θ .
Wf ric = −µ m g d cos θ
= −(0.339) (11 kg) (9.8 m/s2 )
× (7.14 m) cos 36◦
= −211.093 J .
The work due to the applied force is
Wappl = F d
= (150 N) (7.14 m)
= 1071 J ,
and the work due to gravity is
Wgrav = −m g d sin θ
Let : F = 150 N ,
d = 7.14 m ,
θ = 36◦ ,
m = 11 kg ,
g = 9.8 m/s2 ,
µ = 0.339 , and
v = 1.27 m/s .
= −(11 kg) (9.8 m/s2 )
× (7.14 m) sin 36◦
= −452.414 J ,
so that
∆K = Wf ric + Wappl + Wgrav
= (−211.093 J) + (1071 J)
+ (−452.414 J)
= 407.493 J .
oldmidterm 02 – JYOTHINDRAN, VISHNU – Due: Mar 5 2007, 4:00 am
Question 9, chap -1, sect -1.
part 1 of 1
10 points
A 2.56 kg block is pushed 1.68 m up a vertical wall with constant speed by a constant
force of magnitude F applied at an angle of
58.8◦ with the horizontal.
The acceleration of gravity is 9.8 m/s2 .
6
Since the block moves with constant velocity,
X
Fy = F sin θ − m g − fk = 0
F sin θ − m g − µ F cos θ = 0
F (sin θ − µ cos θ) = m g , so
mg
sin θ − µ cos θ
(2.56 kg) (9.8 m/s2 )
=
sin 58.8◦ − 0.521 cos 58.8◦
= 42.8509 N .
F =
Thus
2.56 kg
WF = (F sin θ) (∆y)
= (42.8509 N) (sin 58.8◦ )(1.68 m)
◦
= 61.5772 J ,
F
8
58.
since 1 J = 1 kg · m2 /s2 .
If the coefficient of kinetic friction between
the block and wall is 0.521, find the work done
by F .
Correct answer: 61.5772 J (tolerance ± 1 %).
Explanation:
Given : m = 2.56 kg ,
µ = 0.521 ,
θ = 58.8◦ , and
∆y = 1.68 m .
v
n
Question 10, chap 8, sect 5.
part 1 of 1
10 points
A car of weight 1150 N operating at a rate of
198 kW develops a maximum speed of 27 m/s
on a level, horizontal road.
Assuming that the resistive force (due to
friction and air resistance) remains constant,
what is the car’s maximum speed on an incline
of 1 in 20; i.e., if θ is the angle of the incline
with the horizontal, sin θ = 1/20 ?
Correct answer: 26.7899 m/s (tolerance ± 1
%).
Explanation:
If f is the resisting force on a horizontal
road, then the power P is
P = f vhorizontal .
fk
so that
F
◦
8
58.
mg
The block is in equilibrium horizontally, so
so that
X
Fx = F cos θ − n = 0 ,
n = F cos θ
P
vh
198000 W
=
27 m/s
= 7333.33 N .
f=
On the incline, the resisting force is
F = f + m g sin θ = f +
P
W
W
=
.
+
20
vh
20
oldmidterm 02 – JYOTHINDRAN, VISHNU – Due: Mar 5 2007, 4:00 am
And,
v
7
177 N/m
8 kg
7 m/s
Fv=P,
8 kg
d
v⊃ = 0
8 kg
so
vf
8 kg
P
v=
F
P
P
W
+
vh
20
198000 W
=
198000 W 1150 N
+
27 m/s
20
= 26.7899 m/s .
=
v⊢ = 0
D
8 kg
177 N/m
µ = 0.84
Find the compressed distance d.
Correct answer: 1.16192 m (tolerance ± 1
%).
Explanation:
The principle we are going to use to solve
this problem is that the change in the total
energy of the system is equal to the work done
by the nonconservative forces
Wnc = ∆K + ∆U .
(1)
The frictional force is
Question 11, chap 10, sect 99.
part 1 of 1
10 points
A 8 kg mass slides to the right on a surface
having a coefficient of friction 0.84 as shown
in the figure. The mass has a speed of 7 m/s
when contact is made with a spring that has
a spring constant 177 N/m. The mass comes
to rest after the spring has been compressed
a distance d. The mass is then forced toward
the left by the spring and continues to move
in that direction beyond the unstretched position. Finally the mass comes to rest a distance
D to the left of the unstretched spring.
The acceleration of gravity is 9.8 m/s2 .
f = µmg
= 0.84 (8 kg) (9.8 m/s2 )
= 65.856 N .
In our case, the nonconservative force is the
frictional force, therefore
Wnc = −f d
= −(65.856 N) d .
The change in kinetic energy is
1
∆K = 0 − m vi2
2
1
= − (8 kg) (7 m/s)2
2
= −196 J ,
oldmidterm 02 – JYOTHINDRAN, VISHNU – Due: Mar 5 2007, 4:00 am
and the change in potential energy is
1
k d2
2
1
= (177 N/m) d2
2
= (88.5 N/m) d2 , so
Wnc = ∆K + ∆U
−(65.856 N) d = −196 J + (88.5 N/m) d2
(88.5 N/m) d2 + (65.856 N) d + (−196 J) = 0
A d2 + B d + C = 0 .
∆U =
This is a quadratic equation, and since
p
2
B − 4 A C = (65.856 N)2
−4 (88.5 N/m) (−196 J)
= 271.516 N ,
1/2
the positive solution is
√
−B + B2 − 4 A C
d=
2A
−(65.856 N) + (271.516 N)
=
2 (88.5 N/m)
where h is the distance the pile driver falls,
i.e., h = (7.47 m) + (0.107 m). Solving for the
average force, F , we obtain
mgh
s
(966 kg)(9.8 m/s2 )(7.47 m + 0.107 m)
=
0.107 m
= 670373 N .
F =
Question 13, chap 8, sect 1.
part 1 of 1
10 points
Consider a frictionless roller coaster such as
depicted below.
The acceleration of gravity is 9.8 m/s2 .
A
C
hA
17 m
= 1.16192 m .
Question 12, chap 7, sect 3.
part 1 of 1
10 points
A 966 kg pile driver is used to drive a
steel I-beam into the ground. The pile driver
falls 7.47 m before contacting the beam, and
it drives the beam 10.7 cm into the ground
before coming to rest.
The acceleration of gravity is 9.8 m/s2 .
Using the work-energy theorem, calculate
the magnitude of the average force the beam
exerts on the pile driver while the pile driver
is brought to rest.
Correct answer: 670373 N (tolerance ± 1 %).
Explanation:
The work energy theorem tells us that the
change in potential energy of the falling pile
driver equals the work done on the I beam.
We can express this relation as
mgh = F ·s,
8
B
Passenger cars start at point A with zero
initial speed, accelerate as they go down to
point B, swing around the circular vertical
loop B → C → B of radius 17 m, then go
on towards further adventures (not shown).
When a car goes through the top of the loop
(point C), the passengers feel weightless (for
just a moment).
What is the height hA of the starting
point A above the loop’s bottom B?
Correct answer: 42.5 m (tolerance ± 1 %).
Explanation:
Let :
R = 17 m and
g = 9.8 m/s2 .
A passenger feels weightless when his acceleration ~a is precisely equal to the freefall acceleration ~g . At the top of the loop
oldmidterm 02 – JYOTHINDRAN, VISHNU – Due: Mar 5 2007, 4:00 am
(point C), passengers have no tangential acceleration while their normal acceleration is
directed straight down and has magnitude
aN =
In the absence of friction, the mechanical
energy of a car is conserved as it follows the
coaster’s route, thus
KA + UA = KC + UC
0.2 m
1848 N/m
x
1.7 kg
v2
.
R
Weightlessness happens when this acceleration equals g, therefore the cars should go
through point C at speed
p
vC = R g .
9
µ = 0.3
x=0
After the block is released, how far along
the plane will the block move before coming
to a stop?
Correct answer: 7.39496 m (tolerance ± 1
%).
Explanation:
Basic Concepts: Spring Potential Energy.
Frictional Forces.
Work-Energy Theorem.
=⇒
m 2
m 2
vA + m g hA =
v + m g 2 R.
2
2 C
Therefore, in light of the initial condition
vA = 0, we have
m 2
v
m g hA = m g 2 R +
2 C
m
= mg2R +
Rg
2
5
= mgR
2
5
hA = R
2
5
= (17 m)
2
= 42.5 m .
Question 14, chap 8, sect 1.
part 1 of 1
10 points
A(n) 1.7 kg block is pushed by an external
force against a spring with spring constant
1848 N/m until the spring is compressed by
0.2 m from its uncompressed length (x = 0).
The block rests on a horizontal plane that has
a coefficient of kinetic friction of 0.3. The
external force is then rapidly removed so that
the compressed spring can push the mass.
The acceleration of gravity is 9.8 m/s2 .
Remember: The block is not attached to
the spring.
Let : L = 0.2 m ,
m = 1.7 kg ,
k = 1848 N/m ,
µ = 0.3 .
and
Solution: Examining the vertical forces,
we observe that the normal force is N = m g
and so the friction force is f = µ m g . Call the
distance the block travels d . Then the work
W done by the non-conservative friction force
is
W = ∆K + ∆U
1
−f d = (0 − 0) + 0 − k L2
2
1
µ m g d = k L2 .
2
Solving for d gives
k L2
2µmg
(1848 N/m) (0.2 m)2
=
2 (0.3) (1.7 kg) (9.8 m/s2 )
= 7.39496 m .
d=
Question 15, chap -1, sect -1.
part 1 of 1
10 points
oldmidterm 02 – JYOTHINDRAN, VISHNU – Due: Mar 5 2007, 4:00 am
Suppose the incline is frictionless for the
system shown. The angle of inclination is 21◦ ,
the spring constant is 639 N/m and the mass
of the block is 5.22 kg. The block is released
from rest with the spring initially unstretched.
The acceleration of gravity is 9.8 m/s2 .
639 N/m
x
5. 22
µ
kg
=0
21◦
How far x does it move down the incline
before coming to rest?
Correct answer: 0.0573793 m (tolerance ± 1
%).
Explanation:
Let : θ = 21◦ ,
k = 639 N/m ,
m = 5.22 kg , and
µ = 0.
gx
9
=
ge
64
gx
3
2.
=
ge
8
gx
3
3.
=
correct
ge
16
gx
1
4.
=
ge
4
4
gx
=
5.
ge
9
gx
4
6.
=
ge
25
gx
9
7.
=
ge
49
gx
7
8.
=
ge
16
5
gx
=
9.
ge
49
gx
1
10.
=
ge
5
Explanation:
1.
Let : MX = 3 Me ,
RX = 4 Re .
Energy is conserved, so
Ui = m g x sin θ = Uf =
k x2
2
Hence, solving for x
2 m g sin θ
k
2 (5.22 kg) (9.8 m/s2 ) sin 21◦
=
639 N/m
x=
= 0.0573793 m .
10
Because
m ge =
G Me m
,
Re2
where Me is the mass of the earth and Re
G Me
is the radius. So that ge =
. Similar
Re2
equation applies to the planet X
gx =
G Mx
.
Rx2
Question 16, chap 9, sect 1.
part 1 of 1
10 points
The ratio between the surface gravitational
accelerations is
Planet X has four times the diameter and
three times the mass of the earth.
What is the ratio of gravitational acceleration at the surface of planet X to the gravitational acceleration at the surface of the Earth,
gx
?
ge
gx
Mx Re2
=
ge
Me Rx2
3 Me Re2
=
Me (4 Re )2
=
3
.
16
oldmidterm 02 – JYOTHINDRAN, VISHNU – Due: Mar 5 2007, 4:00 am
Question 17, chap 9, sect 99.
part 1 of 1
10 points
Two satellites have circular orbits about
the same planet.
The masses of the two satellites are respectively mm = m and m3m = 3 m. Both satellites have circular orbits with respective radii
rm = r and r3m = 2 r.
Explanation:
The only force acting on an orbiting satellite is the force of gravity,
F =
GM m
,
r2
where M is the planet’s mass and m is the
mass of the satellite itself. Consequently, the
satellite is in free fall with acceleration
g=
r
2r
g = ac =
2.
3.
4.
5.
6.
7.
8.
9.
10.
v3m
vm
v3m
vm
v3m
vm
v3m
vm
v3m
vm
v3m
vm
v3m
vm
v3m
vm
v3m
vm
v3m
vm
v2
,
r
and therefore
What is the ratio of the orbital speeds of
the two satellites?
1.
GM
r2
directed towards the planet’s center.
Note: This acceleration is independent on
the satellite’s mass m.
For a circular orbit, the free-fall acceleration equals the centripetal acceleration,
m
3m
11
1
= √ correct
2
1
=√
3
=3
=
=
=
1
2
√
3
1
9
=9
=2
=
=
1
3
√
v2
GM
=
=⇒ v =
r2
r
r
GM
,
r
regardless of the satellite’s mass m. Consequently, for two satellites in circular orbits
around the same planet,
r
GM
v3m
r
= r 3m
vm
GM
rm
r
rm
=
r3m
r
r
=
2r
1
= √ .
2
Question 18, chap 9, sect 4.
part 1 of 1
10 points
2
A “synchronous” satellite, which always remains above the same point on a planet’s
equator, is put in orbit about a planet similar
to Jupiter.
oldmidterm 02 – JYOTHINDRAN, VISHNU – Due: Mar 5 2007, 4:00 am
This planet rotates once every 9.2 h, has a
mass of 2.1 × 1027 kg and a radius of 6.99 ×
107 m.
Given that G = 6.67 × 10−11 N m2 /kg2 ,
calculate how far above Jupiter’s surface the
satellite must be.
Correct answer: 8.73976 × 107 m (tolerance
± 1 %).
Explanation:
Basic Concepts: Solution:
According to Kepler’s third law:
T2 =
4 π2 3
r
GM
where r is the radius of the satellite’s orbit.
Thus, solving for r
r=
M T2
G
4 π2
31
2
2
1
3
Now, the altitude h of the satellite (measured
from the surface of Jupiter) is
h=r−R
= (1.57298 × 108 m) − (6.99 × 107 m)
= 8.73976 × 107 m .
Question 19, chap 9, sect 5.
part 1 of 1
10 points
A rocket of mass m is to be launched from
planet X, which has a mass M and a radius R.
What is the minimum speed that the rocket
must have for it to escape into space?
GM
R2
r
2GM
correct
2.
R
r
GM m
3.
2R
1.
5.
r
GmM
R
2GM m
R
r
GM
R
GM
7.
(2R)2
r
GM
8.
2R
Explanation:
For the rocket to escape to infinity, its total
energy must be non-negative,
6.
Etotal = (1/2)mv 2 − GmM/R ≥ 0.
= {6.67 × 10
N m /kg }
1
(2.1 × 1027 kg) (33120 s)2 3
×
4 (3.1415926)2
= 1.57298 × 108 m .
−11
4.
r
12
The escape speed is the minimal speed for
which it happens;
r
2GM
v=
.
R