Introducing Torque
(If is constant)
I
r F
r F
r F
I mi ri
i
I r dm
2
2
A 2.0 kg bucket hangs
from a cylindrical pulley
of radius 0.10 m and
mass 1.0 kg.
(A) What is the linear
acceleration of the
bucket?.
(B) What is the angular
acceleration of the
PHYSICS 191
LECTURE OUTLINES
Spring 2016
PHYS 191, Arts & Science Program, The Petroleum Institute
L1.0: Getting Acquainted
What we are doing today:
Course Introduction:
Why Studio Physics?
Syllabus & Schedule
Course Notebooks and how they will help you l
L2.1 Motion in 1D
What we are doing today:
Introduction to motion concepts, basic relationships
Deadlines:
Read: Instructions for A1.2
Start reading Ch. 2, Straight Line Motion ASAP
Physics 191, Studio Physics, The Petroleum Institute
Physics 191, Studio
The angular displacement
of the earth in 12 hours is
A.
B.
C.
D.
/2
2
12
Physics 191, Studio Physics, The Petroleum Institute
The angular velocity of the second
hand on a clock is approximately
A.
B.
C.
D.
10 radian/sec
1.0 radian/sec
0.1 radian/sec
0.01
Potential Energy Functions
1. Gravitational Potential ()
=
=
()
=
=
= = ( )
If we set = 0 at = , then
() =
2. Elastic Potential ()
=
=
=
If = 0 at = 0, then
() =
PHYS191, Studio Physics, The Petroleum Institute
=
1
2
2
Conservation of M
Types of Friction
1. Forces due to Surface Contact
Normal Force
Friction Forces (Dry or Lubricated)
Kinetic Friction
Static Friction
2. Forces from a Fluid
Buoyancy
Resistance to Relative Motion (Drag)
3. Other
E.g. Rolling Resistance (tire deformation,
PHYSICS 191
LECTURE OUTLINES
Topic 7
Spring 2016
Physics 191, Studio Physics, The Petroleum Institute
L7.1 The W-KE Theorem
What we are doing today:
Introducing: Mechanical Work and Kinetic Energy
Deadlines:
Read: Ch.7, Kinetic Energy & Work & Ch.3.8
Phys
Force and Motion
What are some example FORCES?
So what is a force, really?
Physics 191, Studio Physics, The Petroleum Institute
Newtons Laws
1. N1L
2. N2L
3. N3L
Physics 191, Studio Physics, The Petroleum Institute
Weight:
The name given to the gravitatio
Motion in 2D and 3D
1. Examples
2. Need Coordinate System
3. Need to Use Vectors
Physics 191, Studio Physics, The Petroleum Institute
r
2
1
Find the unknown angles and r.
r
h
30
1
Find the unknown angle, h, and r.
Vector Representations
1. Vector Componen
A4.1 Vectors & 2D Motion
A ball dropped and a ball given an initial horizontal velocity will have
the same height at any later time.
Projectile Motion
It can be understood by analyzing the horizontal
and vertical motions separately.
Physics 191, Studio Ph
Students Material Experiment 07 Faradays Law of Induction
Experiment 07
Material
7-1
Faradays Law of Induction
Sensor-CASSY
Power-CASSY
CASSY Lab
Coil, 250 turns
Coil, 1000 turns
2 round bar magnets with clips
Stand base, V-shape, 28 cm
Stand rod, 50 cm
T
Students Material Experiment 06 Force in a magnetic field
Experiment 06
6-1
Force in a magnetic field
Material
Multimeter:
Arzana: Analog 30
Bu Hasa: METRAmax 3
Sensor-CASSY
CASSY Lab
30 Ampere-box
Force sensor with sensor plug
Support for conductor lo
alsalhi (raa2359) OH-06 bradley (19102) This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page nd all choices before answering. This is Online Homework No. 6 (OH-06). All answers should be submitted onli
23. (a) and (b) We note that the charge on C3 is q3 = 12 C 8.0 C = 4.0 C. Since the charge on C4 is q4 = 8.0 C, then the voltage across it is q4/C4 = 2.0 V. Consequently, the voltage V3 across C3 is 2.0 V C3 = q3/V3 = 2.0 F. Now C3 and C4 are in parallel
24. For maximum capacitance the two groups of plates must face each other with maximum area. In this case the whole capacitor consists of (n 1) identical single capacitors connected in parallel. Each capacitor has surface area A and plate separation d so
25. We note that the total equivalent capacitance is C123 = [(C3)1 + (C1 + C2)1]1 = 6 F. (a) Thus, the charge that passed point a is C123 Vbatt = (6 F)(12 V) = 72 C. Dividing this by the value e = 1.60 1019 C gives the number of electrons: 4.5 1014, which
26. The charges on capacitors 2 and 3 are the same, so these capacitors may be replaced by an equivalent capacitance determined from
1 1 1 C2 + C3 = + = . Ceq C2 C3 C2 C3
Thus, Ceq = C2C3/(C2 + C3). The charge on the equivalent capacitor is the same as th
27. (a) In this situation, capacitors 1 and 3 are in series, which means their charges are necessarily the same: q1 = q3 = C1C3V (1.00 F ) ( 3.00 F ) (12.0V ) = = 9.00 C. C1 + C3 1.00 F+3.00 F
(b) Capacitors 2 and 4 are also in series: q2 = q4 = (c) q3 =
28. Initially the capacitors C1, C2, and C3 form a combination equivalent to a single capacitor which we denote C123. This obeys the equation C + C2 + C3 1 1 1 =+ =1 . C123 C1 C2 + C3 C1 (C2 + C3 ) Hence, using q = C123V and the fact that q = q1 = C1 V1 ,
29. The total energy is the sum of the energies stored in the individual capacitors. Since they are connected in parallel, the potential difference V across the capacitors is the same and the total energy is U= 1 1 2 ( C1 + C2 )V 2 = ( 2.0 106 F + 4.0 106
10. The equivalent capacitance is given by Ceq = q/V, where q is the total charge on all the capacitors and V is the potential difference across any one of them. For N identical capacitors in parallel, Ceq = NC, where C is the capacitance of one of them.
12. (a) The potential difference across C1 is V1 = 10.0 V. Thus, q1 = C1V1 = (10.0 F)(10.0 V) = 1.00 104 C. (b) Let C = 10.0 F. We first consider the three-capacitor combination consisting of C2 and its two closest neighbors, each of capacitance C. The eq
14. The two 6.0 F capacitors are in parallel and are consequently equivalent to Ceq = 12 F . Thus, the total charge stored (before the squeezing) is
qtotal = CeqV = (12 F ) (10.0V) = 120 C.
(a) and (b) As a result of the squeezing, one of the capacitors i
15. The charge initially on the charged capacitor is given by q = C1V0, where C1 = 100 pF is the capacitance and V0 = 50 V is the initial potential difference. After the battery is disconnected and the second capacitor wired in parallel to the first, the
17. (a) First, the equivalent capacitance of the two 4.00 F capacitors connected in series is given by 4.00 F/2 = 2.00 F. This combination is then connected in parallel with two other 2.00-F capacitors (one on each side), resulting in an equivalent capaci
18. We determine each capacitance from the slope of the appropriate line in the graph. Thus, C1 = (12 C)/(2.0 V) = 6.0 F. Similarly, C2 = 4.0 F and C3 = 2.0 F. The total equivalent capacitance is given by C + C2 + C3 1 1 1 =+ =1 , C123 C1 C2 + C3 C1 (C2 +
19. (a) After the switches are closed, the potential differences across the capacitors are the same and the two capacitors are in parallel. The potential difference from a to b is given by Vab = Q/Ceq, where Q is the net charge on the combination and Ceq