Download Final Exam

Physics 130
General Physics
Fall 2012
Final Exam
December 14, 2012
Name:
Instructions
1. This examination is closed book and closed notes. All your belongings
except a pen or pencil and a calculator should be put away and your
bookbag should be placed on the floor.
2. You will find one page of useful formulae on the last page of the exam.
3. Please show all your work in the space provided on each page. If you
need more space, feel free to use the back side of each page.
4. Academic dishonesty (i.e., copying or cheating in any way) will
result in a zero for the exam, and may cause you to fail the
class.
In order to receive maximum credit,
each solution should have:
1.
2.
3.
4.
5.
6.
7.
8.
9.
A labeled picture or diagram, if appropriate.
A list of given variables.
A list of the unknown quantities (i.e., what you are being asked to find).
One or more free-body or force-interaction diagrams, as appropriate,
with labeled 1D or 2D coordinate axes.
Algebraic expression for the net force along each dimension, as appropriate.
Algebraic expression for the conservation of energy or momentum equations, as appropriate.
An algebraic solution of the unknown variables in terms of the
known variables.
A final numerical solution, including units, with a box around it.
An answer to additional questions posed in the problem, if any.
1
Physics 130
General Physics
Fall 2012
1. A tennis player hits a ball 2.0 m above the ground. The ball leaves his racquet with a
speed of 20.0 m{s at an angle of 5˝ above the horizontal. The horizontal distance to the
net is 7 m and the net is 1 m high. Does the ball clear the net? If so, by how much? If
not, by how much does it miss?
Solution:
This is a 2D projectile motion problem. We are given the initial height of the ball
above the ground, yi “ 2 m, its initial velocity, vi “ 20.0 m{s, and the angle above
the horizontal with which the ball is hit, θ “ 5˝ . We are also given the horizontal
distance to the net, xf “ 7 m, and the vertical height of the net, yf “ 1 m. We are
asked whether the ball clears the net or not.
To solve the problem we first determine the initial velocity of the ball in the horizontal
(x) and vertical (y) dimensions:
vxi “ vi cos θ “ p20.0 m{sq ˆ cos p5˝ q “ 19.9 m{s,
vyi “ vi sin θ “ p20.0 m{sq ˆ sin p5˝ q “ 1.74 m{s.
(1)
(2)
Next, we use the horizontal kinematic equation of projectile motion to find the time
it takes the ball to travel the distance to the net:
xf “ xi ` vxi ∆t
“ vxi ∆t
xf
7m
ñ ∆t “
“ 0.351 s.
“
vxi
19.9 m{s
(3)
(4)
(5)
Finally, we use the vertical kinematic equation of projectile motion to find the vertical
height of the ball as it reaches the net:
1
yf “ yi ` vyi ∆t ´ gp∆tq2 ,
2
(6)
“ p2.0 mq ` p1.74 m{sq ˆ p0.351 sq ´
1
ˆ p9.8 m{s2 q ˆ p0.351 sq2
2
“ 2.01 m.
(7)
(8)
So yes, the ball clears the net by about one meter!
2
Physics 130
General Physics
Fall 2012
2. Sand moves without slipping at 6.0 m{s down a conveyer that is tilted at 15˝ . The sand
enters a pipe 3.0 m below the end of the conveyer belt. What is the horizontal distance
d between the conveyer belt and the pipe?
Solution:
This is a 2D projectile motion problem. We are given the initial velocity of the sand,
vi “ 6.0 m{s, the angle of the conveyer belt, θ “ 15˝ , and the vertical distance to the
top of the pipe, yi “ 3.0 m. We are asked to find the horizontal distance, xf “ d,
between the conveyer belt and the pipe.
To solve this problem we must recognize that once the sand leaves the conveyer
belt it is a free-falling projectile. In other words, it maintains the same constant
horizontal velocity it had when it left the belt, while its vertical velocity increases
due to gravity.
First, we determine the velocity of the sand in the horizontal (x) and vertical (y)
dimensions as it leaves the edge of the conveyer belt:
vxi “ vi cos θ “ p6.0 m{sq ˆ cos p15˝ q “ 5.80 m{s,
vyi “ ´vi sin θ “ ´p6.0 m{sq ˆ sin p15˝ q “ ´1.55 m{s,
(1)
(2)
where note the negative sign on the vyi velocity. Next, we use the vertical kinematic
equation of projectile motion to find the time it takes the sand to fall to the top of
the pipe (yf “ 0):
1
yf “ yi ` vyi ∆t ´ gp∆tq2 ,
(3)
2
1
0 “ yi ` vyi ∆t ´ gp∆tq2
(4)
2
This is a quadratic
equation for time of the form 0 “ c ` bz ` az 2 , with solution
?
z “ p´b ˘ b2 ´ 4acq{2a. Solving for time, we get:
b
2
´ 4yi ˆ p´g{2q
´vyi ˘ vyi
,
(5)
∆t “
2p´g{2q
3
Physics 130
General Physics
“
´vyi ˘
b
2
vyi
` 2yi g
,
´g
a
1.55 m{s ˘ p´1.55 m{sq2 ` 2 ˆ p9.8 m{s2 q ˆ p3.0 mq
,
“
´9.8 m{s2
1.55 m{s ˘ 7.82 m{s
“
,
´9.8 m{s2
“ 0.640 s.
Fall 2012
(6)
(7)
(8)
(9)
Note that in the last line we had to adopt the “minus” solution to ensure that time
was a positive number.
Finally, we can use the horizontal equation of motion to solve for the final distance,
noting that the initial horizontal position xi “ 0:
xf “ d “
“
“
“
xi ` vxi ∆t,
vxi ∆t,
p5.80 m{sq ˆ p0.640 sq,
3.7 m.
4
(10)
(11)
(12)
(13)
Physics 130
General Physics
Fall 2012
3. A rock stuck in the tread of a 60.0 cm diameter bicycle wheel has a tangential speed of
3.00 m{s. When the brakes are applied, the rock’s tangential deceleration is 1.00 m{s2 .
(a) What are the magnitudes of the rock’s angular velocity and angular acceleration at
t “ 1.50 s?
(b) At what time is the magnitude of the rock’s acceleration equal to g?
Solution:
This is a non-uniform angular acceleration problem. We are given the size of the
wheel, r “ 30 cm “ 0.30 m, the initial tangential speed of the rock, vi “ 3.00 m{s, and
the tangential deceleration of the rock when the brakes are applied, aT “ 1.00 m{s2 .
(a) In order to determine the final angular velocity, ωf of the rock after t “ 1.50 s,
we first have to determine the initial angular velocity:
wi “
3.00 m{s
vi
“
“ 10 rad{s.
r
0.30 m
(1)
The angular acceleration is a constant and given by:
α“
aT
´1.00 m{s2
“
“ ´3.33 rad{s2 .
r
0.30 m
(2)
Finally, using the appropriate rotational kinematic equation and substituting, we
obtain
ωf “ ωi ` α∆t
“ 10 rad{s ´ p3.33 rad{s2 q ˆ p1.5 sq
“ 5.01 rad{s.
5
(3)
(4)
(5)
Physics 130
General Physics
Fall 2012
(b) In order to find the time at which the magnitude of the angular acceleration of
the rock equals g, wea
first need to find the linear velocity of the rock at that time.
We want |~a| “ g “ a2T ` a2r , where aT is given in the problem and ar “ v 2 {r.
Substituting and solving for v we get
d
ˆ 2 ˙2
v
g “
(6)
a2T `
r
v4
r2
2 2
r pg ´ a2T q
r1{2 pg 2 ´ a2T q1{4
p0.30 mq1{2 ˆ rp9.8 m{s2 q2 ´ p1.00 m{s2 q2 s1{4
1.71 m{s.
g 2 “ a2T `
ñ v4 “
v “
“
“
(7)
(8)
(9)
(10)
(11)
To find the time at which the rock reaches this velocity, we use the 1D kinematic
equation for velocity and solve for the time:
vf “ vi ´ aT ∆t
vi ´ vf
ñ ∆t “
aT
3.00 m{s ´ 1.71 m{s
“
1.00 m{s2
“ 1.29 s.
6
(12)
(13)
(14)
(15)
Physics 130
General Physics
Fall 2012
4. You and your friend Peter are putting new shingles on a roof pitched at 25˝ . You’re
sitting on the very top of the roof when Peter, who is at the edge of the roof directly
below you, 5.0 m away, asks you for the box of nails. Rather than carry the 2.5 kg box
of nails down to Peter, you decide to give the box a push and have it slide down to him.
If the coefficient of kinetic friction between the box and the roof is 0.55, with what speed
should you push the box to have it gently come to rest right at the edge of the roof?
Solution:
The pictorial representation and free-body diagram are shown below:
This is a 1D dynamics problem. The relevant forces are gravity, FG , the normal
force, n, and the kinetic friction force, fk . Note that we do not include the initial
force that was applied to the box of nails to get it moving, but we do include the
fact that the box has some initial velocity.
The most natural coordinate system is one that is rotated by 25˝ and therefore
aligned with the roof. The interaction and free-body diagrams are shown above.
The net force along the x- and y-axis is
ÿ
pFnet qx “
Fx “ FG sin 25˝ ´ fk “ ma
(1)
ÿ
pFnet qy “
Fy “ n ´ FG cos 25˝ “ 0
(2)
ñ n “ FG cos 25˝
(3)
In the second line we used the fact that the shingles are not leaping off the roof to
set the acceleration in the y-direction equal to zero. The magnitude of the force of
gravity is FG “ mg, and the magnitude of the kinetic force of friction is
fk “ µk n
“ µk FG cos 25˝
“ µk mg cos 25˝ .
7
(4)
(5)
(6)
Physics 130
General Physics
Fall 2012
Substituting equation (6) into equation (1) and solving for the acceleration a we get
mg sin 25˝ ´ µk mg cos 25˝ “ ma
ñ a “ psin 25˝ ´ µk cos 25˝ qg
“ psin 25˝ ´ 0.55 ˆ cos 25˝ q ˆ 9.8 m{s2
“ ´0.743 m{s2
(7)
(8)
(9)
(10)
where the minus sign indicates the acceleration is directed up the incline. To find the
initial speed, vi , necessary to have the box come to rest (i.e., vf “ 0) after ∆x “ 5.0 m
can be found from the kinematic equation linking velocity and acceleration:
vf2 “ vi2 ` 2a∆x
?
ñ vi “
´2a∆x
b
´2p´0.743 m{s2 qp5.0 mq
“
“ 2.7 m{s.
8
(11)
(12)
(13)
(14)
Physics 130
General Physics
Fall 2012
5. The 1.0 kg block in the figure below is tied to the wall with a rope. It sits on top of the
2.0 kg block. The lower block is pulled to the right with a tension force of 20 N. The
coefficient of kinetic friction at both the lower and upper surfaces of the 2.0 kg block is
µk “ 0.40.
(a) What is the tension in the rope holding the 1.0 kg block to the wall?
(b) What is the acceleration of the 2.0 kg block?
Solution:
The free-body diagram for the problem is shown below:
To solve this problem we need to use both Newton’s third and second laws. The separate free-body diagrams for the two blocks show that there are two action/reaction
pairs. Notice how the top block (block 1) both pushes down on the bottom block
9
Physics 130
General Physics
Fall 2012
(block 2) with force ~n11 , and exerts a retarding friction force f~2,top on the top surface
of block 2.
(a) Block 1 is in static equilibrium (a1 “ 0 m{s2 ), but block 2 is accelerating to the
right. Newton’s second law for block 1 is
pFnet
pFnet
on 1 qx
on 1 qy
“ f1 ´ Trope “ 0 ñ Trope “ f1
“ n1 ´ m1 g “ 0 ñ n1 “ m1 g.
(1)
(2)
Although block 1 is stationary, there is a kinetic force of friction because there is
motion between block 1 and block 2. The friction model means
f1 “ µk n1 “ µk m1 g.
(3)
Substituting this result into equation (1) we get the tension of the rope:
Trope “ f1 “ µk m1 g
“ p0.40q ˆ p1.0 kgq ˆ p9.8 m{s2 q
“ 3.9 N.
(4)
(5)
(6)
(b) Newton’s second law for block 2 is
pFnet on 2 qx
Tpull ´ f2 top ´ f2 bot
“
m2
m2
1
n2 ´ n1 ´ m 2 g
pFnet on 2 qy
“
“0
“
m2
m2
ax ” a “
(7)
ay
(8)
Forces ~n1 and ~n11 are an action/reaction pair, so ~n11 “ ~n1 “ m1 g. Substituting into
equation (8) gives
n2 “ pm1 ` m2 qg.
(9)
This result is not surprising because the combined weight of both objects presses
down on the surface. The kinetic friction on the bottom surface of block 2 is then
f2
bot
“ µk n2 “ µk pm1 ` m2 qg.
Next, we recognize that the forces f~1 and f~2
f2
top
top
(10)
are an action/reaction pair, so
“ f1 “ µk m1 g.
(11)
Finally inserting these friction results into equation (7) gives
Tpull ´ µk m1 g ´ µk pm1 ` m2 qg
(12)
m2
20 N ´ p0.40qp1.0 kgqp9.8 m{s2 q ´ p0.40qp1.0 kg ` 2.0 kgqp9.8 m{s2 q
“
(13)
2.0 kg
“ 2.2 m{s2 .
(14)
a “
10
Physics 130
General Physics
Fall 2012
6. The lower block in the figure below is pulled on by a rope with a tension force of 20 N.
The coefficient of kinetic friction between the lower block and the surface is 0.30. The
coefficient of kinetic friction between the lower block and the upper block is also 0.30.
What is the acceleration of the 2.0 kg block?
Solution:
The pictorial representation and free-body diagrams are shown below:
The blocks accelerate with the same magnitude but in opposite directions. Thus the
acceleration constraint is a2 “ a “ ´a1 , where a will have a positive value because
of how we have defined the `x direction. There are two real action/reaction pairs.
The two tension forces will act as if they are action/reaction pairs because we are
assuming a massless rope and a frictionless pulley.
Make sure you understand why the friction forces point in the directions shown in
the free-body diagrams, especially force f~11 exerted on the bottom block (block 2) by
the top block (block 1).
11
Physics 130
General Physics
Fall 2012
We have quite a few pieces of information to include. First, Newton’s second law
applied to block 1 gives
pFnet
on 1 qx
pFnet
on 1 qy
“
“
“
ñ
f1 ´ T1
µk n1 ´ T1 “ m1 a1 “ ´m1 a
n1 ´ m 1 g “ 0
n1 “ m1 g.
(1)
(2)
(3)
(4)
Tpull ´ f11 ´ f2 ´ T2
Tpull ´ f11 ´ µk n2 ´ T2 “ m2 a2 “ m2 a
n2 ´ n11 ´ m2 g “ 0
n2 “ n11 ` m2 g.
(5)
(6)
(7)
(8)
For block 2 we have
pFnet
on 2 qx
pFnet
on 2 qy
“
“
“
ñ
Note that to simplify the two x-equations above we have already used the kinetic
friction model. Next, from Newton’s third law we have three additional constraints:
n11 “ n1 “ m1 g
f11 “ f1 “ µk n1 “ µk m1 g
T1 “ T2 “ T.
(9)
(10)
(11)
Knowing n11 we can now use the y-equation for block 2 to find n2 . Substituting
all these pieces into the two x-equations, we end up with two equations with two
unknowns:
µk m1 g ´ T “ ´m1 a
Tpull ´ T ´ µk m1 g ´ µk pm1 ` m2 qg “ m2 a.
(12)
(13)
Subtracting the first equation from the second we get
Tpull ´ T ´ µk m1 g ´ µk pm1 ` m2 qg ´ µk m1 g ` T “ m2 a ` m1 a
Tpull ´ 3µk m1 g ´ µk m2 g “ pm2 ` m1 qa
Tpull ´ µk p3m1 ` m2 qg “ pm2 ` m1 qa.
(14)
(15)
(16)
And finally solving for a we get
Tpull ´ µk p3m1 ` m2 qg
m1 ` m2
20 N ´ p0.30qp3 ˆ 1.0 kg ` 2.0 kgqp9.8 m{s2 q
“
1.0 kg ` 2.0 kg
2
“ 1.8 m{s .
ña “
12
(17)
(18)
(19)
Physics 130
General Physics
Fall 2012
7. The 1.0 kg physics book in the figure below is connected by a string to a 500 g coffee
cup. The book is given a push up the slope and released with a speed of 3.0 m{s. The
coefficients of friction are µs “ 0.50 and µk “ 0.20.
(a) How far does the book slide?
(b) At the highest point, does the book stick to the slope, or does it slide back down?
Solution:
The pictorial representation and free-body diagrams are shown below:
13
Physics 130
General Physics
Fall 2012
To solve this problem we use the particle model for the book (B) and the coffee cup
(C), the models of kinetic and static friction, and the constant-acceleration kinematic
equations.
(a) To find the distance x1 the book slides we must find its acceleration. Newton’s
second law applied to the book gives
ÿ
pFon B qy “ nB ´ pFG qB cos θ “ 0
(1)
ÿ
ñ nB “ mB g cos θ
(2)
“ ´T ´ µk nB ´ mB g sin θ
“ ´T ´ µk mB g cos θ ´ mB g sin θ “ mB aB .
(4)
(5)
pFon B qx “ ´T ´ fk ´ pFG qB sin θ
Similarly, for the coffee cup we have
ÿ
pFon C qy “ T ´ pFG qC “ T ´ mC g “ mC aC .
(3)
(6)
Equations (5) and (6) can be rewritten using the acceleration constrain aC “ aB “ a
as
´T ´ µk mB g cos θ ´ mB g sin θ “ mB a
T ´ mC g “ mC a.
(7)
(8)
Adding these two equations and solving for the acceleration gives
´µk mB g cos θ ´ mB g sin θ ´ mC g “ pmB ` mC qa
„

mB pµk cos θ ` sin θq ` mC
ña “ ´
g
mB ` mC
„

p1.0 kgq ˆ p0.20 cos 20˝ ` sin 20˝ q ` 0.5 kg
“ ´
ˆ p9.8 m{s2 q
1.0 kg ` 0.5 kg
“ ´6.73 m{s2 .
(9)
(10)
(11)
(12)
Finally, to solve for the distance x1 we use the following kinematic equation with
v1x “ 0, v0x “ 3.0 m{s, and x0 “ 0:
2
2
v1x
“ v0x
` 2apx1 ´ x0 q
2
0 “ v0x ` 2ax1
v2
ñ x1 “ ´ 0x
2a
´p3.0 m{sq2
“
2 ˆ p´6.73 m{s2 q
“ 0.67 m.
14
(13)
(14)
(15)
(16)
(17)
Physics 130
General Physics
Fall 2012
(b) In order to figure out if the book sticks to the slope or slides back down we have
to determine if the static friction force needed to keep the book in place, fs is larger
or smaller than the maximum static friction force
pfs qmax “ µs nB “ µs mB gcosθ
“ p0.50q ˆ p9.8 m{s2 q ˆ cos 20˝
“ 4.60 N.
(18)
(19)
(20)
When the cup is at rest, the string tension is T “ mC g. In this case, Newton’s first
law for the book becomes
ÿ
pFon B qx “ fs ´ T ´ mB g sin θ
(21)
ñ fs
“
“
“
“
fs ´ mC g ´ mB g sin θ “ 0
pmC ` mB sin θqg
p0.5 kg ` 1.0 kg sin 20˝ q ˆ p9.8 m{s2 q
8.25 N.
Because fs ą pfs qmax , the book slides back down.
15
(22)
(23)
(24)
(25)
Physics 130
General Physics
Fall 2012
8. A conical pendulum is formed by attaching a 500 g ball to a 1.0 m-long string, then
allowing the mass to move in a horizontal circle of radius 20 cm. The figure shows that
the string traces out the surface of a cone, hence the name.
(a) What is the tension in the string?
(b) What is the ball’s angular speed, in rpm?
Solution:
(a) What is the tension in the string?
The forces in the z-direction in the free body diagram are the component of the
tension in the z-direction and the force of gravity. To find the z-component of
the tension,
16
Physics 130
General Physics
Fall 2012
Tz
adjacent
“
hypotenuse
T
“ T cos θ
cos θ “
Tz
Apply Newton’s second law in the z-direction
ÿ
Fz “ T cos θ ´ mg “ 0
mg
T “
cos θ
(1)
(2)
(3)
(4)
To calculate cos θ,
adjacent
cos θ “
“
hypotenuse
ñ θ “ 11.5˝
?
L2 ´ r 2
“
L
a
p1mq2 ´ p0.2mq2
“ 0.98
1.0m
(5)
(6)
Plugging this in for tension,
p0.500 kgq ˆ p9.80 m{s2 q
mg
“
cos θ
0.98
T “ 5.0 N
T “
(7)
(8)
(b) What is the ball’s angular speed, in rpm?
Referring back to the tension triangle, the radial component of tension is
opposite
Tr
“
hypotenuse
T
“ T sin θ
sin θ “
Tr
17
(9)
(10)
Physics 130
General Physics
Fall 2012
Referring back to the triangle for the length of the pendulum,
r
opposite
“
hypotenuse
L
r “ L sin θ
sin θ “
Apply Newton’s second law in the r-direction
ÿ
Fr “ Tr “ T sin θ “ mω 2 r
c
T sin θ
ñω “
d mr
“
5.0 N ˆ sin 11.5˝
p0.500 kgq ˆ p0.2 mq
“ 3.16 rad{sec
ˆ
˙ ˆ
˙
rad
1 rev
60 sec
“ 3.16
ˆ
ˆ
sec
2π rad
1 min
“ 30 rpm
18
(11)
(12)
(13)
(14)
(15)
(16)
(17)
(18)
Physics 130
General Physics
Fall 2012
9. A block of mass m slides down a frictionless track, then around the inside of a circular
loop-the-loop of radius R. From what minimum height h must the block start to make
it around without falling off? Give your answer as a multiple of R.
Solution:
This is a two-part problem. In the first part, we will find the critical velocity for the
block to go over the top of the loop without falling off. Since there is no friction,
the sum of the kinetic and gravitational potential energy is conserved during the
block’s motion. We will use this conservation equation in the second part to find the
minimum height the block must start from to make it around the loop.
Place the origin of the coordinate system directly below the block’s starting position
on the frictionless track. From the free-body diagram we have
´Fg ´ n “ ´m
vc2
.
R
(1)
For the block to just stay on the track, n “ 0. Therefore, the critical velocity vc is
vc2
R
vc2
mg “ m
aR
ñ vc “
gR
Fg “ m
(2)
(3)
(4)
We can now use conservation of mechanical energy to find the minimum height h:
Using vf “ vc “
?
Kf ` Uf ` Ki ` Ui
1 2
1 2
mvf ` mgyf “
mv ` mgyi .
2
2 i
gR, yf “ 2R, vi “ 0, and yi “ h, we get
1
gR ` gp2Rq “ 0 ` gh
2
ñ h “ 2.5R.
19
(5)
(6)
(7)
(8)
Physics 130
General Physics
Fall 2012
10. A pendulum is formed from a small ball of mass m on a string of length L. As the figure
shows, a peg is height h “ L{3 above the pendulum’s lowest point. From what minimum
angle θ must the pendulum be released in order for the ball to go over the top of the
peg without the string going slack?
Solution: This is a two part problem. First, we need to find the velocity for the
ball to go over the peg without the string going slack. Then we need to find the
potential energy to match that velocity.
(a) Find the velocity of the ball so that the string doesn’t go slack. We’ll set our
origin on the peg. This means that once the string touches the peg, the ball
will be moving in circular motion with a radius of L{3.
When the ball is let go, it moves below the peg and then moves in circular
motion. When it reaches the top of the circle and is directly above the peg, we
can use Newton’s second law and the fact that the ball is moving in circular
motion.
ÿ
F “ ma “ ´m
´T ´ Fg
v2
“ ´m
R
v2
R
(1)
(2)
(3)
20
Physics 130
General Physics
Fall 2012
The critical speed is just as the tension equals zero. Factoring the negatives and
the masses, and setting Fg “ mg and T “ 0,
Setting R “ L{3,
v2
0 ` mg “ m c
aR
gR
vc “
vc “
c
gL
3
(4)
(5)
(6)
(b) Find the angle to release the ball. Now that we know the velocity for the string
to remain taut, we’ll use the conservation of energy to find the height to release
the ball.
Kf ` Uf ` Ki ` Ui
1 2
1 2
mvf ` mgyf “
mv ` mgyi
2
2 i
1 2
v ` gyf “ 0 ` gyi
2 c
(7)
(8)
(9)
Plugging in yf “ L{3 and vc2 “ gL{3,
gL gL
`
“ gyi
6
3
L L
`
“ yi
6
3
L
yi “
2
(10)
(11)
(12)
This result tells us that we need to release the ball at a height of L/2 above the
peg.
To find the angle, consider the third figure in the three-part figure above. We
need to find the distance of the ball below the point where the string is attached.
We can use the string length and result above.
h“L´
L
L L
´ “
2
3
6
(13)
To find the angle, we can use cosine,
L{6
1
“
L
6
ñ θ “ 80.4˝
cos θ “
21
(14)
(15)
Physics 130
General Physics
Fall 2012
11. A sled starts from rest at the top of the frictionless, hemispherical, snow-covered hill
shown below.
(a) Find an expression for the sled’s speed when it is at angle φ.
(b) Use Newton’s laws to find the maximum speed the sled can have at angle φ without
leaving the surface.
(c) At what angle φmax does the sled “fly off” the hill?
Solution:
(a) When the sled is at a position that relates to angle φ, the height is yf “ R cos φ.
Using conservation of energy with yi “ R, we get
Kf ` Uf “ Ki ` Ui
1 2
1 2
mvf ` mgyf “
mv ` mgyi
2
2 i
1 2
v ` gR cos φ “ 0 ` gyi
2 f
ñ vf2 “ 2gR ´ 2gR cos φ
“ 2gRp1 ´ cos φq
a
ñ vf “
2gRp1 ´ cos φq
22
(1)
(2)
(3)
(4)
(5)
(6)
Physics 130
General Physics
Fall 2012
(b) To find the maximum speed we use the free-body diagram:
ÿ
F “ ma “ ´m
FN ´ Fg cos φ “ ´m
v2
R
v2
R
(7)
(8)
The critical speed is just as the normal force equals zero. Factoring the negatives
and the masses, and setting Fg “ mg and FN “ 0,
vc2
R
a
“
gR cos φ
g cos φ “
(9)
vc
(10)
(c) Find the angle when the sled gets air! We can set the speeds from the two
previous parts equal to each other to find φ.
a
a
gR cos φ “
2gRp1 ´ cos φq
(11)
gR cos φ “ 2gRp1 ´ cos φq
(12)
cos φ “ 2 ´ 2 cos φ
(13)
3 cos φ “ 2
(14)
ˆ ˙
2
ñ φ “ cos´1
“ 48˝ .
(15)
3
23
Physics 130
General Physics
Fall 2012
12. Truck brakes can fail if they get too hot. In some mountainous areas, ramps of loose
gravel are constructed to stop runaway trucks that have lost their brakes. The combination of a slight upward slope and a large coefficient of rolling resistance as the truck
tires sink into the gravel brings the truck safely to a halt. Suppose a gravel ramp slopes
upward at 6.0˝ and the coefficient of rolling friction is 0.40. Use work and energy to find
the length of a ramp that will stop a 15, 000 kg truck that enters the ramp at 35 m/s
(« 75 mph).
Solution:
We’ll identify the truck and the loose gravel as the system. We need the gravel inside
the system because friction increases the temperature of the truck and the gravel.
We will also use the model of kinetic friction and the conservation of energy equation.
Kf ` Uf ` ∆Eth “ Ki ` Ui ` Wext
0 ` Uf ` ∆Eth “ Ki ` 0 ` 0
(1)
(2)
The thermal energy created by friction is
∆Eth “
“
“
“
fk p∆xq
pµk FN qp∆xq
pµk mg cos θqp∆xq
pµk mg cos θqp∆xq
(3)
(4)
(5)
(6)
Using geometry, the final gravitationl potential energy of the truck is
Uf “ mgyf
“ mgp∆xq sin θ
(7)
(8)
Finally, the initial kinetic energy is simply
Ki “
24
1 2
mv
2 i
(9)
Physics 130
General Physics
Fall 2012
Plugging everything into the conservation of energy equation, we get
1 2
mv
2 i
1 2
g∆xpsin θ ` µk cos θq “
v
2 i
mgp∆xq sin θ ` µk mg cos θp∆xq “
(10)
(11)
vi2
(12)
2gpsin θ ` µk cos θq
p35 m{sq2
(13)
“
2 ˆ p9.8 m{s2 q ˆ psin 6˝ ` 0.4 ˆ cos 6˝ q
“ 124 m
(14)
“ 0.12 km.
(15)
ñ ∆x “
25
Physics 130
General Physics
Fall 2012
13. The spring shown in the figure below is compressed 50 cm and used to launch a 100 kg
physics student. The track is frictionless until it starts up the incline. The student’s
coefficient of kinetic friction on the 30˝ incline is 0.15.
(a) What is the student’s speed just after losing contact with the spring?
(b) How far up the incline does the student go?
Solution:
(a) This is a conservation of energy problem. We want to know the initial velocity
immediately after the spring is no longer in contact with the student.
Kf ` Ugf ` Usf ` ∆Eth “ Ki ` Ugi ` Usi ` Wext
1 2
1 2
1
1
mvf ` mgyf ` kp∆xf q2 ` 0 “
mvi ` mgyi ` kp∆xi q2 ` 0
2
2
2
2
1 2
1
2
mv “
kp∆xi q
2 f
2
26
(1)
(2)
(3)
Physics 130
General Physics
ñ vf “
c
Fall 2012
k
∆xi “ 14 m{s
m
(4)
(b) Using the conservation of energy equation again, we can how far the student
travels up the incline. Let’s define the y-height when the student reaches the
highest point as yf “ ∆s sin 30˝ . Then we have:
Kf ` Ugf ` Usf ` ∆Eth “ Ki ` Ugi ` Usi ` Wext
1
1 2
1
1 2
mvf ` mgyf ` kp∆xf q2 ` 0J “
mvi ` mgyi ` kp∆xi q2 ` 0J
2
2
2
2
1
mg∆s sin 30˝ ` µk Fn ∆s “ mgyi ` p∆xq2
2
(5)
(6)
(7)
(8)
From the sum of the forces in the y-direction on the incline, the normal force is
Fn “ mg cos 30˝ . Plugging this in, we get
1
mg∆s sin 30˝ ` µk mg cos 30˝ ∆s “ mgyi ` p∆xq2
2
1
∆smgpsin 30˝ ` µk cos 30˝ q “ mgyi ` p∆xq2
2
mgyi ` 12 kp∆xq2
ñ ∆s “
mgpsin 30˝ ` µk cos 30˝ q
∆s “ 32.1 m
27
(9)
(10)
(11)
(12)
Physics 130
General Physics
Fall 2012
14. A hollow sphere is rolling along a horizontal floor at 5.0 m{s when it comes to a 30˝
incline. How far up the incline does it roll before reversing direction?
Solution:
Assume that the hollow sphere is a rigid rolling body and that the sphere rolls up
the incline without slipping. Also assume that the coefficient of rolling friction is
zero.
The initial kinetic energy, which is a combination of rotational and translational
energy, is transformed in gravitational potential energy. Choose the bottom of the
incline as the zero of the gravitational potential energy.
Starting from conservation of energy, we have
Kf ` Ugf “ Ki ` Ugi
1
1
1
1
M v12 ` Iω12 ` M gy1 “
M v02 ` Iω02 ` M gy0 .
2
2
2
2
(1)
(2)
Substituting v1 “ 0, ω1 “ 0, y0 “ 0, I “ 2{3M R2 (appropriate for a hollow sphere),
ω0 “ v0 {R, and solving for the final height, y1 , we get
˙
ˆ
1 2
1
2
2
(3)
M v0 `
M R ω02 ` 0
0 ` 0 ` M gy1 “
2
2 3
ˆ ˙
1 2 1 2 v02
gy1 “
(4)
v ` R
2 0 3
R2
5v02
ñ y1 “
(5)
6g
5 ˆ p5.0 m{sq2
(6)
“
6 ˆ p9.8 m{s2 q
“ 2.126 m.
(7)
28
Physics 130
General Physics
Fall 2012
The distance traveled along the incline is
y1
sin 30˝
2.126 m
“
0.5
“ 4.3 m.
s “
29
(8)
(9)
(10)
Physics 130
General Physics
Fall 2012
15. A 10 g bullet traveling at 400 m{s strikes a 10 kg, 1.0-m wide door at the edge opposite
the hinge. The bullet embeds itself in the door, causing the door to swing open. What
is the angular velocity of the door just after impact?
Solution:
In order to solve this problem, we use the fact that angular momentum is conserved
in the collision for the (bullet+door) system. As the bullet hits the door, its velocity
~v is perpendicular to ~r. Consequently, the initial angular momentum about the
rotation axis, with r “ L, is
Li “ mB vB L.
(1)
After the collision, with the bullet in the door, the moment of inertia about the
hinges is
I “ Idoor ` Ibullet
1
“
m D L2 ` m B L2 .
3
(2)
(3)
Therefore, the final angular momentum is
Lf “ Iω
˙
ˆ
1
2
2
mD L ` mB L ω.
“
3
30
(4)
(5)
Physics 130
General Physics
Fall 2012
Equating the initial and final angular momentum and solving for ω we get
ˆ
L “ Li
˙f
1
m D L2 ` m B L2 ω “ m B v B L
3
3mB vB
ñω “
LpmD ` 3mB q
3 ˆ 0.010 kg ˆ 400 m{s
“
1.0 m ˆ p10 kg ` 3 ˆ 0.010 kgq
“ 1.2 rad{s.
31
(6)
(7)
(8)
(9)
(10)
Physics 130
General Physics
Fall 2012
16. A 355 mL soda can is 6.2 cm in diameter and has a mass of 20 g. Such a soda can half
full of water is floating upright in water. What length of the can is above the water
level?
Solution:
The buoyant force, FB , on the can is given by Archimedes’ principle.
Let the length of the can above the water level be d, the total length of the can be
L, and the cross-sectional area of the can be A. The can is in static equilibrium, so
from the free-body diagram sketched above we have
ÿ
Fy “ FB ´ FG,can ´ FG,water “ 0
(1)
ρwater ApL ´ dqg “ pmcan ` mwater qg
pmcan ` mwater q
L´d “
Aρwater
pmcan ` mwater q
ñd “ L´
Aρwater
(2)
(3)
(4)
Recalling that one liter equals 10´3 m3 and the density of water is ρwater “ 1000 kg m´3 ,
the mass of the water in the can is
ˆ
˙
Vcan
mwater “ ρwater
(5)
2
ˆ
˙
355 ˆ 10´6 m3
´3
“ p1000 kg m q ˆ
(6)
2
“ 0.1775 kg.
(7)
To find the length of the can, L, we have:
Vcan “ AL
355 ˆ 10´6 m3
ñL “
πp0.031 mq2
“ 0.1176 m.
32
(8)
(9)
(10)
Physics 130
General Physics
Fall 2012
Finally, substituting everything into equation (4), we get:
d “ 0.1176 m ´
“ 0.0522 m
“ 5.2 cm.
p0.020 kg ` 0.1775 kgq
π ˆ p0.031 mq2 ˆ p1000 kg m´3 q
33
(11)
(12)
(13)
Physics 130
General Physics
Fall 2012
17. Water flowing out of a 16 mm diameter faucet fills a 2.0 L bottle in 10 s. At what
distance below the faucet has the water stream narrowed to 10 mm in diameter?
Solution:
We treat the water as an ideal fluid obeying Bernoulli’s equation. The pressure at
point 1 is p1 and the pressure at point 1 is p2 . Both p1 and p2 are atmospheric
pressure. The velocity and the area at point 1 are v1 and A1 , and at point 2 they
are v2 and A2 . Let d be the distance of point 2 below point 1.
First, we use the time it takes to fill a 2.0 L bottle to compute the flow rate, Q, the
rate at which water is flowing out of the faucet, remembering that one liter equals
10´3 m3 :
p2.0 Lq ˆ p10´3 m3 {Lq
Q “
10 s
“ 2.0 ˆ 10´4 m3 {s.
(1)
(2)
Next, we use this result to find the velocity, v1 , with which the water leaves the
faucet through the D1 “ 16 ˆ 10´3 m diameter aperture (at point 1):
Q “ v1 A1
Q
ñ v1 “
A1
Q
“
πpD1 {2q2
2.0 ˆ 10´4 m3 {s
“
π ˆ p8 ˆ 10´3 mq2
“ 1.0 m{s.
34
(3)
(4)
(5)
(6)
(7)
Physics 130
General Physics
Fall 2012
Now, from the continuity equation we have
v2 A2 “ v1 A1
A1
ñ v2 “ v1
A2
πpD1 {2q2
“ v1
πpD2 {2q2
ˆ ˙2
D1
“ v1
D2
ˆ
˙2
16 ˆ 10´3 m
“ 1.0 m{s
10 ˆ 10´3 m
“ 2.56 m{s.
(8)
(9)
(10)
(11)
(12)
(13)
Finally, we turn to Bernoulli’s equation to find the height, d:
1
1
p1 ` ρv12 ` ρgy1 “ p2 ` ρv22 ` ρgy2
2
2
1 2
ρgpy1 ´ y2 q “
ρpv2 ´ v12 q
2
1 2
pv2 ´ v12 q
gd “
2
pv22 ´ v12 q
d “
2g
p2.56 m{sq2 ´ p1.0 m{sq2
“
2 ˆ 9.8 m{s2
“ 0.283 m
« 28 cm.
35
(14)
(15)
(16)
(17)
(18)
(19)
(20)
Physics 130
General Physics
Fall 2012
18. A 100 g ice cube at ´10˝ C is placed in an aluminum cup whose initial temperature is
70˝ C. The system comes to an equilibrium temperature of 20˝ C. What is the mass of
the cup?
Solution:
There are two interacting systems: aluminum and ice. The system comes to thermal
equilibrium in four steps: (1) the ice temperature increases from ´10 ˝ C to ´0 ˝ C;
(2) the ice becomes water at 0 ˝ C; (3) the water temperature increases from 0 ˝ C to
20 ˝ C; and (4) the cup temperature decreases from 70 ˝ C to 20 ˝ C.
Since the aluminum and ice form a closed system, we have
Q “ Q1 ` Q2 ` Q3 ` Q4 “ 0.
(1)
Mice cice ∆T
p0.100 kgq ˆ r2090 J{pkg Kqs ˆ p10 Kq
2090 J.
Mice Lf
p0.100 kgq ˆ p3.33 ˆ 105 J{kgq
33, 300 J.
Mice cwater ∆T
p0.100 kgq ˆ r4190 J{pkg Kqs ˆ p20 Kq
8380 J.
MAl cAl ∆T
MAl ˆ r900 J{pkg Kqs ˆ p´50 Kq
´p45, 000 J{kgqMAl .
(2)
Each term is as follows:
Q1 “
“
“
Q2 “
“
“
Q3 “
“
“
Q4 “
“
“
(3)
(4)
(5)
Inserting everything into equation (1), we get
43, 770 J ´ p45, 000 J{kgqMAl “ 0
ñ MAl “ 0.97 kg.
36
(6)
Physics 130
General Physics
Fall 2012
19. A group of rebels wants to invade the castle and they made a trebuchet as their first line
of attack. The group optimized their trebuchet so that their projectile can be fired from
a great distance to hit the castle wall. Their trebuchet includes a counterweight mass of
931 kg, a moment arm, R2 “ 2 m, a projectile mass of 7 kg, a moment arm R1 “ 10 m,
an arm mass of 25 kg, and a release angle of φ “ 90˝ due to a stopper on the trebuchet
base that stops the projectile arm from going any further. The pivot point is 8 m above
the ground. Find the distance from the castle wall that the trebuchet needs to be placed
so that the projectile hits the wall.
Solution:
This problem uses the torque created by the difference in the force/moment-arm
ratios of the counterweight and projectile to generate an angular acceleration.
There are two torque terms for the trebuchet. The counterweight will create a
clockwise rotation (which by convention is negative) and the projectile will create a
positive counterclockwise rotation.
Because the projectile is in a cup attached to the arm, the force and the moment
arm are perpendicular to each other. For that torque term, R2 Fp sinpangleq “ R2 Fp .
However, the counterweight and its moment arm are not perpendicular to each other.
The figure below shows the angle between them. The torque component for the
counterweight is R1 FCW sin ψ.
ř
ř
Just like the F “ ma for linear motion, τ “ Iα for rotational motion where I is
the moment of inertia of the R2 which we’ll consider to be a rod rotating about it’s
end. Alpha is the angular acceleration.
37
Physics 130
General Physics
ÿ
τ “ Iα
Fp R2 ´ FCW R1 sinψ “ Iα
Fall 2012
(1)
(2)
(3)
The moment of inertia for a rod rotating about its end is I “ 31 M L2 . We’re interested
in finding the speed of the projectile as it leaves the the moment arm, R2 . So, we’ll
use the length of R2 for L and the mass of that portion of the rod in our calculation.
Let the total mass of the rod be M where the mass of the projectile portion of the
rod is N1 and the mass of the counterweight portion of the rod is N2 . To find N1 ,
M
N1
“
R1
L
R1 M
10m ˆ 25kg
N1 “ “
“
“ 20.8kg
L
12m
(4)
(5)
(6)
ÿ
τ “ Iα
Fp R2 ´ FCW R1 sin psi “ Iα
1
m2 gR2 ´ mCW gR1 sin psi “
M R2 α
3
(7)
(8)
(9)
(10)
Calculate the angular acceleration, α,
α “
3gpm2 R2 ´ m1 R1 sinψq
N 1 R2 2
38
(11)
Physics 130
General Physics
p3 ˆ 9.8m{s2 qp7kg ˆ 10m ´ 931kg ˆ 2m sin 143˝ q
20.8kg ˆ p10mq2
rad
α “ 14.8 2
s
α “
Fall 2012
(12)
(13)
(14)
To find the angular velocity, ωf when the projectile leaves the trebuchet, we need to
use kinematics. We’re told in the problem, that the projectile moment arm travels
θ “ 90˝ “ π2 radians before the projectile is released.
pωf q2 “ pωi q2 ` 2αθ
pωf q2 “ 2αθ “ 2αθ
c
?
rad π
ωf “
2αθ “ 2 ˆ 14.8 2 ˆ
s
2
rad
ωf “ 6.8 2
s
(15)
(16)
(17)
(18)
(19)
So far, our calculations have been for circular motion as the projectile is moved by
the trebuchet. Once the projectile leaves the trebuchet, we need to switch to linear
motion.
To find the linear velocity with which the projectile leaves the trebuchet, we convert
the angular velocity to linear velocity.
v “ ωf R2
39
(20)
Physics 130
General Physics
rad
ˆ 10m
s2
v “ 68m{s
v “ 6.8
Fall 2012
(21)
(22)
(23)
The original description tells us that the projectile travels through an angle of
90˝ while on the trebuchet. Using this information, we can find the angle at which
the projectile leaves the trebuchet and its height.
We need to use the kinematic equations for projectile motion to find the time it takes
for the projectile to hit the ground. Then use this time, to calculate the horizontal
range of the trebuchet.
vy “ v sin θ “ 68m{s ˆ sin 37˝ “ 41m{s
(24)
(25)
We’re able to calculate the starting height of the projectile using the figure above.
Because the trebuchet travels through 90˝ , we can use the same triangle for right
before the trebuchet is fired and right after the projectile is released. Because these
triangles are the same, the height at which the projectile leaves the trebuchet is 16
m.
1
yf “ yi ` vy ∆t ´ g∆t
2
1
0 “ 16m ` 41m{s ˆ ∆t ´ 9.8m{s2 ∆t
2
(26)
(27)
(28)
40
Physics 130
General Physics
Fall 2012
Rearranging this equation into the form of a quadratic equation,
4.9m{s2 ∆t ´ 41m{s ˆ ∆t ´ 16m “ 0
∆t “ 8.7 sec
(29)
(30)
(31)
To find the range, we plug this time into the range equation (i.e. no acceleration
in the x-direction). We also need to use the initial velocity of the projectile in the
x-direction. From the figure above, we can see that the velocity in the x-direction is
vx “ v cos 37˝ .
xf “ xi ` vxi ∆t “ v cos 37˝ ∆t
xf “ 68m{s ˆ cos 37˝ ˆ 8.7 sec
xf “ 472m
41
(32)
(33)
(34)
(35)
Physics 130
General Physics
Kinematics and Mechanics
Energy
1
xf “ xi ` vxi t ` ax t2
2
vxf “ vxi ` ax t
Kf ` Ugf “ Ki ` Ugi
∆K ` ∆U ` ∆Eth “ ∆Emech ` ∆Eth “ ∆Esys “ Wext
2
2
vxf
“ vxi
` 2ax pxf ´ xi q
1
yf “ yi ` vyi t ` ay t2
2
vyf “ vyi ` ay t
2
2
vyf
“ vyi
` 2ay pyf ´ yi q
1
θf “ θi ` ωi ∆t ` α∆t2
2
ωf “ ωi ` α∆t
ωf 2 “ ωi 2 ` 2α∆θ
s “ rθ
c “ 2πr
vt “ ωr
v2
“ ω2r
r
2πr
v“
T
ar “
Forces
F~net “ ΣF~ “ ma
fk “ µk FN
F~AonB “ ´F~BonA
Momentum
p~i “ p~f
p~ “ m~v
m1 ´ m2
pvix q1
m1 ` m2
2m1
pvix q1
pvfx q2 “
m1 ` m2
pvfx q1 “
Kf ` Uf ` ∆Eth “ Ki ` Ui ` Wext
1
K “ mv 2
2
U “ mgy
∆Eth “ fk ∆s
~ ¨ ~d
W “F
~ ¨ ~v
P “F
P “ ∆E {∆t
Fluids and Thermal Energy
F
P“
A
m
ρ“
V
Q “ vA
v1 A 1 “ v2 A 2
1
1
p1 ` ρv12 ` ρgy1 “ p2 ` ρv22 ` ρgy2
2
2
Qnet “ Q1 ` Q2 ` ... “ 0
Q “ MLf for freezing/melting
2
v
F~net “ ΣF~r “ ma “ m
r
Fg “ mg
0 ă fs ă“ µs FN
Fall 2012
Q “ Mc∆T
ÿ
I “
mr 2
~ “ Iα
~τ “ ~r ˆ F
~L “ m~v ˆ ~r
~ ˆB
~ “ ABsinθ
A
1
Krot “ I ω 2
2
1
Idisk “ mr 2
2
2
Ihollow sphere “ mr 2
3
1
I door “ mr 2 about its hinges
3
1
mr 2 about its end
I rod “
12
42
Physics 130
I rod “
General Physics
1
mr 2 about its center
3
Constants
g “ 9.8 m{s2
ρwater “ 1000 kg{m 3
ρair “ 1 .28 kg{m 3
J
kg K
J
cwater “ 4190
kg K
J
cAl “ 900
kg K
J
ice to water
Lf “ 333 , 000
kg
cice “ 2090
43
Fall 2012