Download 11-Apr-16 15 - Fulton Schools of Engineering Tutoring Centers

11-Apr-16
Example: A 4.00 kg particle moves along the x-axis. Its position varies with time
according to x(t) = t + 2t3, where x is measured in meters and t is in seconds.
Example continued
How much additional work must be done to stretch the spring to 0.80 m?
(a) What is the particle’s kinetic energy?
2
W   F  dr 

0.80 m
 Fdx 
0.40 m

0.80 m
K
1 2
kx
2
0.40 m



2
1 2 1  dx 
mv  m   2 1  6t 2 J
2
2  dt 
(b) What is the particle’s acceleration? What is the net force acting on the
particle?
1
20.0 N/m0.80 m 2  0.40 m 2  4.8 J
2
a
Note that the result is not 1.6 J.
d 2x
 12t m/s2
dt 2
Fnet  ma  48t N
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Example: A box of mass m is towed up a frictionless incline at constant speed. The
applied force F is parallel to the incline. What is the net work done on the box?
Example continued
(c) The power delivered to the particle at time t?

y
F

P  Fv  48t  1  6t 2  48t  288t 3 W
Nrb
Fpb
x
(d) The work done on the particle from t = 0 to t = 2 sec?

xf
tf
2


t
f
dx
W   F  dr   Fdx   F dt   Fv dt
dt
xi
ti
ti
web
Apply Newton’s 2nd
law:

  48t  288t3 dt  1248 J
0
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F
F
x
 Fpb  web sin   0
y
 N rb  web cos  0
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11-Apr-16
Example continued
Example continued
Fpb  F  mg sin
The magnitude of F is:
The work by the normal force is
WN  N rb x cos 90  0
If the box travels along the ramp a distance of x then the work by the
force F is
WF  Fx cos0  mgx sin
The net work done on the box is
Wnet  WF  Wg  WN
The work by gravity is
 mgx sin  mgx sin  0
0
Wg  web x cos  90  mgx sin
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Example: A raindrop of mass 3.3510-5 kg falls vertically at constant speed
under the influence of the forces of gravity and drag. In falling through 100
m…
Example: What is the net work done on the box in the previous example if the box
is not pulled at constant speed?
 F  F  w sin  ma
 F  ma  w sin
x
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eb
(a) What is the work done by gravity?
eb
y
Proceeding as before:
W   F  dr
Far
Wnet  WF  Wg  WN
x
 ma  mg sin  x  mgx sin   0
 ma x  Fnet x

yf
  mg yˆ   dy yˆ 
yi
wer
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 mgy  0.033 J
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11-Apr-16
Example: A 2.00 kg mass hangs suspended from a string. The ceiling is 3.00
m above the floor and the length of the string is 1.00 m.
Example continued
(b) What is the work done by drag?
ceiling
(a) What is the gravitational potential energy of
the mass if the ceiling is the reference point?
Wnet  Wdrag  Wg  0
Wdrag  Wg  mgy  0.033 Joules
U g  mgy


 2.00 kg  9.8 m/s2  1.00 m 
 19.6 J
floor
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Example: A cart starts from position 4 with v = 15.0 m/s to the left. Find the speed
of the cart at positions 1, 2, and 3. Ignore friction.
Example continued
(b) What is the gravitational potential energy of the mass if the floor is the
reference point?
U g  mgy


 2.00 kg  9.8 m/s2 2.00 m 
 39.2 J
E4  E3
(c) What is the gravitational potential energy of the mass if the ball’s
location is the reference point?
U g  mgy

1
1
mgy4  mv42  mgy3  mv32
2
2

 2.00 kg  9.8 m/s 0 m 
0J
2
U 4  K 4  U 3  K3
v3  v42  2 g  y4  y3   20.5 m/s
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Example continued
E4  E2
U 4  K4  U 2  K2
The work-energy theorem can be used to solve problems instead of using
Newton’s laws.
Or use
E3=E2
Determine the speed of a mass m that starts from rest and
slides along a frictionless ramp for some distance d using (a)
Newton’s laws and (b) the work-energy theorem.
1
1
mgy4  mv42  mgy2  mv22
2
2
v2  v  2 g  y4  y2   18.0 m/s
2
4
E4  E1
U 4  K 4  U 1  K1
1
1
mgy4  mv42  mgy1  mv12
2
2
Answer:
v  2gd sin
Or use
E3=E1
Repeat the calculation but take the coefficient of kinetic
friction to be k.
E2=E1
Answer:
v1  v42  2 g  y4  y1   24.8 m/s
v  2 gd sin  k cos 
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Example: Particle A is at the origin and has a mass of 30.0 grams. Particle B has a
mass of 10.0 grams. Where must particle B be located so that the center of mass
(marked with a red x) is located at the point (2.00 cm, 5.00 cm)?
Momentum and Collisions
y
Use conservation principles to calculate momentum and energy transfer between
two or more interacting bodies.
x
A
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x
xcm 
ma xa  mb xb
mb xb

ma  mb
ma  mb
ycm 
ma ya  mb yb
mb yb

ma  mb
ma  mb
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Example: The positions of three particles are (4.00 m, 0.00 m); (2.00 m, 4.00 m);
and (-1.00 m, -2.00 m). The masses are 4.00 kg, 6.00 kg, and 3.00 kg
respectively. What is the location of the center of mass?
Example continued
10.0 g xb  2.00 cm
10.0 g  30.0 g
xb  8.00 cm
xcm 
y
2
10.0 g yb  5.00 cm
30.0 g  10.0 g
yb  20.0 cm
ycm 
1
x
3
rb  8.00 xˆ  20.0 yˆ cm
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Example: In a railroad freight yard an empty freight car of mass m rolls along a
straight level track at 1.00 m/s and collides with an initially stationary, fully loaded,
boxcar of mass 4.00m. The two cars couple together upon collision.
Example continued
xcm 

m1 x1  m2 x2  m3 x3
m1  m2  m3
4.00 kg 4.00 m   6.00 kg 2.00 m   3.00 kg  1.00 m 
4.00  6.00  3.00 kg
(a) What is the speed of the two cars after the collision?
 1.92 m
ycm
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pi  p f
p1i  p2i  p1 f  p2 f
m y  m2 y2  m3 y3
 1 1
m1  m2  m3
m1v1  0  m1v  m2 v  m1  m2 v
4.00 kg 0.00 m   6.00 kg 4.00 m   3.00 kg  2.00 m 

4.00  6.00  3.00 kg
 m1 
v  
v1  0.20 m/s
 m1  m2 
 1.38 m
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Example: Body 1 of mass M has an original velocity of 6.00 m/s in the +xdirection toward a stationary body 2 of the same mass. After the collision, body
1 has vx=+1.00 m/s and vy=+2.00 m/s. What is the magnitude of body 2’s
velocity after the collision?
Example continued
(b) Suppose instead that both cars are at rest after the collision with what
speed was the loaded boxcar moving before the collision if the empty one had
v1i = 1.00 m/s.
pi  p f
Final
Initial
1
p1i  p2i  p1 f  p2 f
m1v1i  m2 v2i  0  0
1
v1i
m 
v2i   1 v1i  0.25 m/s
 m2 
2
2
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Example: Head-on 1D elastic collision. The values of v1i and v2i are known.
Determine the velocity of the pucks after the collision.
Example continued
y momentum:
x momentum:
pix  p fx
piy  p fy
p1ix  p2ix  p1 fx  p2 fx
p1iy  p2iy  p1 fy  p2 fy
m1v1ix  0  m1v1 fx  m2 v2 fx
Initial
1
0  0  m1v1 fy  m2 v2 fy
v1i
Final
v2i
v2f
1
v1f
2
2
Solve for v2fy:
Solve for v2fx:
v2 fx 
78
Take Fext = 0 so that pi = pf.
m1v1ix  m1v1 fx
v2 fy 
m2
 v1ix  v1 fx
 m1v1 fy
m2
pi  p f
 v1 fy
p1i  p2i  p1 f  p2 f
 2.00 m/s
m1v1i  m2 v2i  m1v1 f  m2 v2 f
 5.00 m/s
The mag. of v2 is
v2 f  v
2
2 fy
v
2
2 fx
 5.40 m/s
m1 v1i  v1 f   m2 v2 f  v2i 
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(1)
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Example continued
Example continued
The collision is elastic so Ki = Kf..
v1i  v1 f  v2i  v2 f
Divide (3) by (1) to get
Ki  K f
K1i  K 2i  K1 f  K 2 f
Take
mv m v mv m v
2
1 1i
2
2 2i
2
1 1f
2
2 2f
v2 f  v1i  v1 f  v2i
substitute into (1) and solve for v1f
 m  m2 
 2m2 
v1i  
v2i
v1 f   1
 m1  m2 
 m1  m2 
(2)
Rewrite (2):
m1v12i  m1v12f  m2v22 f  m2v22i

v
 
  m v
m1 v12i  v12f  m2 v22 f  v22i
m1 v1i  v1 f
(3)
1i
 v1 f
2
2i
 v2 f

v
Then solve for v2f:
2i
 2m1 
 m2  m1 
v2 f  
v1i   m  m v2i
2 
 m1  m2 
 1
 v2 f 
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Example: An elastic collision in 2D. Take particles A and B to be protons and vAi
= 3.50105 m/s and vBi = 0. After the collision, particle A travels at an angle  =
37 with the x-axis and particle B travels at an unknown angle . Determine the
velocities of the protons after the collision and also the value of .
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Take Fext = 0 so that pi = pf.
pix  p fx
x-component:
p Ai , x  pBi , x  p Af , x  p Bf , x
m Av Ai  0  m Av Af cos  mB vBf cos
piy  p fy
Final
Initial
A
A
y-component:
vAi
p Ai , y  p Bi , y  p Af , y  p Bf , y
0  0  m Av Af sin   mB vBf sin 
B
Ki  K f
B
Kinetic energy:
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K Ai  K Bi  K Af  K Bf
2
2
mAv Ai
 mAv Af
 mB vBf2
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Example continued
Example continued
vAi  vAf cos  vBf cos
(1)
0  vAf sin  vBf sin
(2)
Summary, so far:
2
2
v Ai
 v Af
 vBf2
(3)
Now use (3) to determine vBf:
2
2
vBf  v Ai
 v Af
vAi  vAf cos  vBf cos
Rewrite (1) and (2):


2
 v Ai
1  cos2   v Ai sin   2.11105 m/s
vAf sin  vBf sin
Now use (2) to solve for :
Square each and add together (and use a trig identity):
v  v  2v Ai v Af cos  v
2
Ai
2
Af
2
Bf
v Af  v Ai cos  2.80 105 m/s
Use (3) to simplify to:
sin  
v Af
vBf
sin   0.8
  53
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Example continued
Example: The Ballistic Pendulum. A bullet of mass mb is fired at a block of wood
of mass mp. The bullet embeds in the wood and the block rises a height h before
coming to rest. What was the speed of the bullet as it entered the block?
What is v?
Even though energy is not conserved during the collision, it will
be conserved during the upward swing.
This is in inelastic collision.
h
vb
mb
mp
pi  p f
Ei  E f
pbi  p pi  pbf  p pf
Ki  U i  K f  U f
mb vb  0  mb  m p v
1
mb  m p v 2  0  mb  m p  gh
2
v  2 gh
 mb  m p 
v
vb  
 mb 
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Example continued
Example: A 3.00 gram particle is moving toward a 7.00 gram particle with a
speed of 3.00 m/s.
How much kinetic energy is lost in the collision?
(a) With what speed does each particle approach the center of mass?
K  K f  K i

1
mb  m p v 2  1 mbvb2
2
2
vcm 
2
1
1  mb  m p  2
 v
 mb  m p  v 2  mb 
2
2  mb 

v1m1  v2 m2
 0.900 m/s
m1  m2
v1  v1  vcm 
1
mb  m p   m p v 2
2
 mb 
 3.00 m/s  0.900 m/s  2.10 m/s
v2  v2  vcm
 0.00 m/s  0.900 m/s  0.900 m/s
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Example continued
90
Momentum in the center of mass frame
(b) What is the momentum of each particle relative to the center of mass?
ptotal  p1  p2  m1v1  m2v2
 m1 v1  vcm   m2 v2  vcm 
p1  m1v1  0.0063kg m/s
p2  m2v2  0.0063kg m/s

Note: the total momentum relative to the center of mass is zero.
m1m2
v1  v2   m1m2 v2  v1   0
m1  m2
m1  m2
The momentum of a system in the center of
mass frame is zero.
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Example: A centrifuge in a medical laboratory rotates at an angular speed of
3600 rev/min. When switched off, it rotates 50 times before coming to rest. Find
the constant angular acceleration of the centrifuge.
Rotational Momentum and Inertia
  3600
Systems of particles that involve rotation require separate equations to account
for total system energy and momentum.
rev  2 rad  1 min 


  120 rad/sec
min  1 rev  60 sec 
 2 rad 
  50 rev
  100 rad
 1 rev 
 2  02  2  0
 
93
Example: A pulsar is a rapidly spinning neutron star that emits a radio beam like a
lighthouse emits a light beam. We receive a radio pulse for each rotation of the
star. The period T of rotation is found by measuring the time between pulses. The
pulsar in the Crab Nebula has a period of T = 0.033 sec that is increasing at the
rate of 1.2610-5 sec/year.
 72 rad/sec2
94
Example continued
(b) If the pulsar’s angular acceleration is constant, how many years from now
will the pulsar stop rotating?
   0  t  0
2

T0
t 0 
 9  1010 sec  2800 years


(a) What is the pulsar’s angular acceleration?

02
2
(c) The pulsar originated in a supernova explosion in the year 1054 A.D. What
was the initial period for the pulsar?
d d  2 
2 dT
   2
 2 109 rad/sec2
dt dt  T 
T dt
dT
sec 
1 year

-13 sec
 1.26 105

  4 10
dt
year  3.15107 sec 
sec
95
t  953 years  3.0 1010 sec
2
 0    t 
 t  250 rad/sec
T
2
T
 0.025 sec  25 msec
0
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Example continued
The KE of the pulsar decreases over time. The rate of energy loss is
(d) What is the rotational kinetic energy of the pulsar?
K rot 
1 2
I
2
Assume the pulsar to be a solid sphere of
uniform density. M = 1.4Msun and R = 10
km.
P
2
MR 2
5
 1 1038 kg m 2
For comparison, the luminosity of the sun
is 3.861026 W.
I
The moment of inertia for a sphere is
dE d  1 2 
d
  I   I
 4 1031 W
dt dt  2
dt

  190 rad/sec
K rot 
1 2
I  2  1042 J
2
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Example: (a) Find the moment of inertia of the system below. The masses are m 1
and m2 and they are separated by a distance r. Assume the rod connecting the
masses is massless.
Example: What is the moment of inertia of a thin disk when you add a small
mass near the rim? The disk has a rotation axis though its center and points out
of the page.

R
r1 and r2 are the distances between
mass 1 and the rotation axis and
mass 2 and the rotation axis (the
dashed, vertical line) respectively.
I new  I disk  mr 2

1
M disk R 2  mR 2
2
m1
99
r1
r2
m2
100
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Parallel Axis theorem
Example continued
Take m1 = 2.00 kg, m2 = 1.00 kg, r1= 0.33 m ,
and r2 = 0.67 m.
I new  I cm  md 2
2
I   mi ri 2  m1r12  m2 r22  0.67 kg m 2
Where m is the mass of the object and d is the distance the axis
is moved. Note: the new axis must be parallel to the original
axis!
i 1
(b) What is the moment of inertia if the axis is moved so that is passes through
m1?
2
I   mi ri 2  m2 r22  m2 r 2  1.00 kg m 2
i 1
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Example: A centrifuge in a medical laboratory rotates at an angular speed of
3600 rev/min. When switched off, it rotates 50 times before coming to rest. Find
the constant angular acceleration of the centrifuge.
Example: Rank the I1, I2, I3, and I4 from greatest to smallest.
a
2
1
102
I new  I cm  md 2
3
4
I 3  I cm 
b
a
I 4  I cm  m 
2
2

1
m a2  b2
12

 a   b 
I 2  I cm  m    
 2   2 
2
2
  3600
rev  2 rad  1 min 


  120 rad/sec
min  1 rev  60 sec 
 2 rad 
  50 rev
  100 rad
 1 rev 



 2  02  2  0
I1  I cm  mL2
 
Ranking: I1 > I2 > I4 > I3
103
02
2
 72 rad/sec2
104
26
11-Apr-16
Example: A pulsar is a rapidly spinning neutron star that emits a radio beam like a
lighthouse emits a light beam. We receive a radio pulse for each rotation of the
star. The period T of rotation is found by measuring the time between pulses. The
pulsar in the Crab Nebula has a period of T = 0.033 sec that is increasing at the
rate of 1.2610-5 sec/year.
Example continued
(b) If the pulsar’s angular acceleration is constant, how many years from now
will the pulsar stop rotating?
   0  t  0
2

T0
t 0 
 9  1010 sec  2800 years


(a) What is the pulsar’s angular acceleration?

(c) The pulsar originated in a supernova explosion in the year 1054 A.D. What
was the initial period for the pulsar?
d d  2 
2 dT
   2
 2 109 rad/sec2
dt dt  T 
T dt
dT
sec 
1 year

-13 sec
 1.26 105

  4 10
dt
year  3.15107 sec 
sec
105
t  953 years  3.0 1010 sec
2
 0    t 
 t  250 rad/sec
T
2
T
 0.025 sec  25 msec
0
106
Example continued
The KE of the pulsar decreases over time. The rate of energy loss is
(d) What is the rotational kinetic energy of the pulsar?
K rot 
1 2
I
2
Assume the pulsar to be a solid sphere of
uniform density. M = 1.4Msun and R = 10
km.
The moment of inertia for a sphere is
P
2
MR 2
5
 1 1038 kg m 2
dE d  1 2 
d
  I   I
 4 1031 W
dt dt  2
dt

For comparison, the luminosity of the sun
is 3.861026 W.
I
  190 rad/sec
K rot 
1 2
I  2  1042 J
2
107
108
27
11-Apr-16
Example: Calculate the torque due to the three forces shown about the left end of
the bar (the red X). The length of the bar is 4m and F2 acts in the middle of the
bar.
Top view of door
F

r
Hinge
end
F2=30 N

30
F3=20 N
Line of action
of the force
X
Lever arm
r
r
r  r sin 
sin  
The torque is:
F1=25 N
  r F
 rF sin
10
45
Same as
before
109
Example continued
110
Example continued
Lever arm
for F2
F2=30 N
30
F3=20 N
The torques are:
10
X
45
1  0
 2  1.73 m 30 N   51.9 Nm
 3  0.695 m 20 N   13.9 Nm
F1=25 N
Lever arm
for F3
The lever arms are:
The net torque is + 65.8 Nm and is the sum of the above results.
r1  0
r2  2m sin 60  1.73 m
r3  4m sin10  0.695 m
111
112
28