PHYSICS 231 INTRODUCTORY PHYSICS I Lecture 9

PHYSICS 231
INTRODUCTORY PHYSICS I
Lecture 9
Last Lecture
r
r
• Momentum: p = mv
r "pr
F=
"t
•
Newton’s 2nd:
•
Conserved for isolated objects
• Impulse = F!t = !p
•
•
! short times
Useful for
Collisions (p always conserved)
•
Elastic (E conserved too)
•
Inelastic (E not conserved)
Example 6.7
A proton (mp=1.67x10-27 kg) elastically collides with a
target proton which then moves straight forward. If
the initial velocity of the projectile proton is 3.0x106
m/s, and the target proton bounces forward, what are
a) the final velocity of the projectile proton?
b) the final velocity of the target proton?
0.0
3.0x106 m/s
Elastic collision in 1-dimension
1. Conservation of Energy:
1
2
m1v1i2 + 12 m2v 2i2 = 12 m1v12f + 12 m2v 22 f
(1)
2. Conservation of Momentum:
m1v1i + m2v 2i = m1v1 f + m2v 2 f
(2)
!
• Rearrange both equations and divide:
!
m1 (v1i2 " v12f )
= m2 (v 22 f " v 2i2 )
(1)
m1 (v1i " v1 f )(v1i + v1 f ) = m2 (v 2 f " v 2i )(v 2 f + v 2i )
m1 (v1i " v1 f )
= m2 (v 2 f " v 2i )
v1i + v1 f = v 2 f + v 2i
!
" v1i # v 2i = #(v1 f # v 2 f )
(2)
Elastic collision in 1-dimension
Final equations for head-on elastic collision:
m1v1i + m2v 2i = m1v1 f + m2v 2 f
v1i " v 2i = "(v1 f " v 2 f )
!
•
Relative velocity changes sign
•
Equivalent to Conservation of Energy
Example 6.8
An proton (mp=1.67x10-27 kg) elastically collides with
a target deuteron (mD=2mp) which then moves straight
forward. If the initial velocity of the projectile
proton is 3.0x106 m/s, and the target deuteron
bounces forward, what are
a) the final velocity of the projectile proton?
b) the final velocity of the target deuteron?
vp =-1.0x106 m/s
vd = 2.0x106 m/s
Head-on collisions with heavier objects always lead to
reflections
Example 6.9a
The mass M1 enters from the left with velocity v0 and
strikes the mass M2=M1 which is initially at rest. The
collision is perfectly elastic.
a) Just after the collision v2 ______ v0.
A) >
B) <
C) =
Example 6.9b
The mass M1 enters from the left with velocity v0 and
strikes the mass M2=M1 which is initially at rest. The
collision is perfectly elastic.
Just after the collision v1 ______ 0.
A) >
B) <
C) =
Example 6.9c
The mass M1 enters from the left with velocity v0 and
strikes the mass M2=M1 which is initially at rest. The
collision is perfectly elastic.
Just after the collision P2 ______ M1v0.
A) >
B) <
C) =
Example 6.9d
The mass M1 enters from the left with velocity v0 and
strikes the mass M2=M1 which is initially at rest. The
collision is perfectly elastic.
At maximum compression, the energy stored
in the spring is ________ (1/2)M1v02
A) >
B) <
C) =
Example 6.9e
The mass M1 enters from the left with velocity v0 and
strikes the mass M2<M1 which is initially at rest. The
collision is perfectly elastic.
Just after the collision v2 ______ v0.
A) >
B) <
C) =
Example 6.9f
The mass M1 enters from the left with velocity v0 and
strikes the mass M2<M1 which is initially at rest. The
collision is perfectly elastic.
Just after the collision v1 ______ 0.
A) >
B) <
C) =
Example 6.9g
The mass M1 enters from the left with velocity v0 and
strikes the mass M2<M1 which is initially at rest. The
collision is perfectly elastic.
Just after the collision P2 ______ M1v0.
A) >
B) <
C) =
Example 6.9h
The mass M1 enters from the left with velocity v0 and
strikes the mass M2<M1 which is initially at rest. The
collision is perfectly elastic.
At maximum compression, the energy stored
in the spring is ________ (1/2)M1v02
A) >
B) <
C) =
Example 6.9i
The mass M1 enters from the left with velocity v0 and
strikes the mass M2>M1 which is initially at rest. The
collision is perfectly elastic.
Just after the collision v2 ______ v0.
A) >
B) <
C) =
Example 6.9j
The mass M1 enters from the left with velocity v0 and
strikes the mass M2>M1 which is initially at rest. The
collision is perfectly elastic.
Just after the collision v1 ______ 0.
A) >
B) <
C) =
Example 6.9k
The mass M1 enters from the left with velocity v0 and
strikes the mass M2>M1 which is initially at rest. The
collision is perfectly elastic.
Just after the collision P2 ______ M1v0.
A) >
B) <
C) =
Example 6.9l
The mass M1 enters from the left with velocity v0 and
strikes the mass M2>M1 which is initially at rest. The
collision is perfectly elastic.
At maximum compression, the energy stored
in the spring is ________ (1/2)M1v02
A) >
B) <
C) =
Example
6.10
Ballistic Pendulum: used to measure speed
of bullet. 0.5-kg block of wood swings up
by height h = 65 cm after stopping 8.0-g
bullet.
What was bullet’s velocity?
227 m/s
Example
6.11
A 5-g bullet traveling at 500 m/s embeds in a 1.495
kg block of wood resting on the edge of a 0.9-m high
table. How far does the block land from the edge of
the table?
71.4 cm
Example 6.12
Tarzan (M=80 kg) swings on a 12m vine by letting go from an angle
of 60 degrees from the vertical.
At the bottom of his swing, he
picks up Jane (m=50 kg). To what
angle do Tarzan and Jane swing?
35.8 degrees
(indepedent of L or g)
Example 6.12a
Tarzan (M=80 kg) swings on a 12m vine by letting go from an angle
of 60 degrees from the vertical.
At the bottom of his swing, he
picks up Jane (m=50 kg). To what
angle do Tarzan and Jane swing?
To calculate Tarzan’s speed
just before he picks up Jane,
you should apply:
A) Conservation of Energy
B) Conservation of Momentum
Example 6.12b
Tarzan (M=80 kg) swings on a 12m vine by letting go from an angle
of 60 degrees from the vertical.
At the bottom of his swing, he
picks up Jane (m=50 kg). To what
angle do Tarzan and Jane swing?
To calculate Tarzan&Jane’s
speed just after their
collision (given the previous
answer) you should apply:
A) Conservation of Energy
B) Conservation of Momentum
Example 6.12c
Tarzan (M=80 kg) swings on a 12m vine by letting go from an angle
of 60 degrees from the vertical.
At the bottom of his swing, he
picks up Jane (m=50 kg). To what
angle do Tarzan and Jane swing?
To calculate Tarzan&Jane’s final
height (given the previous
answer) you should apply:
A) Conservation of Energy
B) Conservation of Momentum
Chapter 7
Rotational Motion
Universal Law of Gravitation
Kepler’s Laws
Angular Displacement
•
•
Circular motion about a fixed AXIS
Use polar coordinates: (r,! )
• Distance r doesn’t change
•
Three possible units for !:
• Revolutions
• Degrees (1 rev. = 360 deg.)
• Radians (1 rev. = 2! rad.)
!
Angular Displacement, cont.
• Distance traveled by a point:
(distance r from axis)
s = 2! r N (N counts revolutions)
= r" (" is in radians)
" (rad) = 2# N(revolutions)
Example 7.1
An automobile wheel has a radius of 42 cm. If a
car drives 10 km, through what angle has the
wheel rotated?
a) In revolutions
b) In radians
c) In degrees
a) N = 3789
b) " = 2.38x104 radians
c) " = 1.36x106 degrees
Angular Speed
#$ $ f % $ i
"=
=
#t
t f % ti
•
!
•
(in rad/s)
Can also be given in
• Revolutions/s
• Degrees/s
Linear (tangential) Speed at r
"s r"#
vt =
=
"t
"t
!
v t = r"
(" in rad/s)
Example 7.2
A race car engine can turn at a maximum rate of 12,000
rpm. (revolutions per minute).
a) What is the angular velocity in radians per second.
b) If helicopter blades were attached to the crankshaft
while it turns with this angular velocity, what is the
maximum radius of a blade such that the speed of the
blade tips stays below the speed of sound.
DATA: The speed of sound is 343 m/s
a) 1256 rad/s
b) 27 cm
Angular Acceleration
• Denoted by #
!=
" f # "i
t
• $ in rad/s
• # rad/s!
• Every point on rigid object has same $ and #
Rotational/Linear Correspondence:
!" # !x
$ 0 # v0
$ f # vf
% #a
t #t
Rotational/Linear Correspondence, cont’d
Rotational Motion
Linear Motion
#0 + # f )
(
!" =
t
v0 + v f )
(
!x =
t
! f = ! 0 + "t
v f = v0 + at
1 2
!" = # 0 t + $ t
2
1 2
!" = # f t $ % t
2
2
!f
2
=
! 02
2
+ "#$
2
1 2
!x = v0 t + at
2
1 2
!x = v f t " at
2
2
vf
v02
=
+ a!x
2
2
Constant a
Constant #
2
Example 7.3
A pottery wheel is accelerated uniformly from rest
to a rotation speed of 10 rpm in 30 seconds.
a.) What was the angular acceleration? (in rad/s2)
b.) How many revolutions did the wheel undergo
during that time?
a) 0.0349 rad/s2
b) 2.50 revolutions
Linear movement of a rotating point
• Distance
x = r!"
• Speed
v = r!
• Acceleration
a = r!
Angles must be in radians!
Different points
have different
linear speeds!
Special Case - Rolling Motion
• Wheel (radius r) rolls without slipping
• Angular motion of wheel gives linear motion of car
• Distance
x = r!"
• Speed
v = r!
• Acceleration
a = r!
Example 7.4
A coin of radius 1.5 cm is initially rolling with a
rotational speed of 3.0 radians per second, and
comes to a rest after experiencing a slowing down of
# = 0.05 rad/s2.
a.) Over what angle (in radians) did the coin rotate?
b.) What linear distance did the coin move?
a) 90 rad
b) 135 cm