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Physics Unit 4 - andrews.edu
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Physics
Unit 4
Physics Unit 4
1
This Slideshow was developed to accompany the textbook
OpenStax Physics
Available for free at https://openstaxcollege.org/textbooks/college-physics
By OpenStax College and Rice University
2013 edition
Some examples and diagrams are taken from the textbook.
Slides created by Richard Wright, Andrews Academy [email protected]
2
In this lesson you willβ¦
β’ State the first condition of equilibrium.
β’ Explain static equilibrium.
β’ Explain dynamic equilibrium.
β’ State the second condition that is necessary to achieve equilibrium.
β’ Explain torque and the factors on which it depends.
β’ Describe the role of torque in rotational mechanics.
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Statics
Study of forces in equilibrium
Equilibrium means no acceleration
First condition of equilibrium
πππ‘ πΉ = 0
πΉπ₯ = 0 and πΉπ¦ = 0
They can still rotate, soβ¦
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Think of opening a door
Which opens the door the best?Picture a
Big force β large torqueForce away from pivot β large torqueForce directed β₯ to door β large torque
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Ο = F Γ rThis means we use the component of the force that is perpendicular
to the lever arm Ο = Fβ₯ r Ο = F r sin ΞΈ
ΞΈ is the angle between the force and the radius
Unit: NmCCW β +CW β -
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You are meeting the parents of your new βspecialβ friend for the first time. After being at their house for a couple of hours, you walk out to discover the little brother has let all the air out of one of your tires. Not knowing the reason for the flat tire, you decide to change it. You have a 50-cm long lug-wrench attached to a lugnut as shown. If 900 Nm of torque is needed, how much force is needed?
F = 2078 N
Less force required if pushed at 90Β°
120Β°
π = πΉπ sin π900 ππ = πΉ(0.5π)(sin 120Β°)
πΉ = 2078 π
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Second condition of equilibrium
Net torque = 0
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A 5 m, 10 kg seesaw is balanced by a little girl (25 kg) and her father (80 kg) at opposite ends as shown below. How far from the seesawβs center of mass must the fulcrum be placed?1.20 m
How much forcemust the fulcrum support?1029 N
80 kg25 kg x m
10 kg
5 m
βπ = 025 ππ β 9.8 π/π 2 2.5 π + π₯ + 10 ππ β 9.8 π/π 2 π₯β 80 ππ β 9.8 π/π 2 2.5 π β π₯ = 0
612.5 ππ + 245 π π₯ + 98 π π₯ β 1960 ππ + 784 π π₯ = 0β1347.5 ππ + 1127 π π₯ = 0
1127 π π₯ = 1347.5 πππ₯ = 1.20 π
βπΉ = βππ βππ + πΉπ = 0
β 25 ππ 9.8π
π 2β 80 ππ 9.8
π
π 2+ πΉπ = 0
β1029 π + πΉπ = 0
πΉπ = 1029 π
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Twist out the answers to these torque questions
Read 9.3, 9.4
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In this lesson you willβ¦
β’ State the types of equilibrium.
β’ Describe stable and unstable equilibriums.
β’ Describe neutral equilibrium.
β’ Discuss the applications of Statics in real life.
β’ State and discuss various problem-solving strategies in Statics.
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Stable equilibrium
When displaced from equilibrium, the system experiences a net force or torque in a direction opposite to the direction of the displacement.
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Unstable equilibrium
When displaced from equilibrium, the net force or torque is in same direction of the displacement
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Neutral Equilibrium
Equilibrium is independent of displacement from its original position
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Problem-Solving Strategy for Static Equilibrium
1. Is it in equilibrium? (no acceleration or accelerated rotation)
2. Draw free body diagram
3. Apply βπΉ = 0 and/or βπ = 0
a. Choose a pivot point to simplify the problem
4. Check your solution for reasonableness.
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The system is in equilibrium. A mass of 225 kg hangs from the end of the uniform strut whose mass is 45.0 kg. Find (a) the tension T in the cable and the (b) horizontal and (c) vertical force components exerted on the strut by the hinge.
Free body diagramnext slide
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π = 225 ππ,π = 45.0 ππ, π = ? , πΉβπ₯ = ? , πΉβπ¦ = ?
π = 6627 π
πΉβπ₯ = 5739 π
πΉβπ¦ = 5959 π
Use hinge as pivot point to eliminate need for πΉββπ = 0
πΉβ 0 + π πΏ sin 15Β° β πππΏ
2sin 45Β° β ππ πΏ sin 45Β° = 0
0 + ππΏ sin 15Β° β 45 ππ 9.8m
s2L
2sin 45Β° β 225 ππ 9.8
π
π 2πΏ sin 45Β° = 0
ππΏ sin 15Β° β 155.917 π πΏ β 1559.170 π πΏ = 0ππΏ sin 15Β° = 1715.087 π πΏ
π =1715.087 π
sin 15Β°= 6627 π
Find πΉβπ₯βπΉπ₯ = 0
πΉβπ₯ β π cos 30Β° + 0 + 0 = 0πΉβπ₯ = 6627 π cos 30Β° = 5739 π
Find πΉβπ¦βπΉπ¦ = 0
πΉβπ¦ β π sin 30Β° β ππ βππ = 0
πΉβπ¦ = π sin 30Β° + ππ +ππ
πΉβπ¦ = 6627 π sin 30Β° + 45 ππ 9.8π
π 2+ 225 ππ 9.8
π
π 2= 5959 π
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Pick a stable position while you apply your knowledge.
Read 9.5, 9.6
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In this lesson you willβ¦
β’ Describe different simple machines.
β’ Calculate the mechanical advantage.
β’ Explain the forces exerted by muscles.
β’ State how a bad posture causes back strain.
β’ Discuss the benefits of skeletal muscles attached close to joints.
β’ Discuss various complexities in the real system of muscles, bones, and joints.
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Machines make work easier
Energy is conserved so same amount of energy with or without machine
Mechanical Advantage (MA)
ππ΄ =πΉπ
πΉπ
πΉπ = πππππ ππ’π‘πΉπ = πππππ ππ
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LeverUses torques with pivot at N
πΉπππ = πΉπππ
πΉπ
πΉπ=
ππ
ππ
When F β, π β
ππ΄ =πΉπ
πΉπ=
ππ
ππ
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Other simple machinesWheel and Axle
Lever Inclined Plane
Ramp β less force to slide up, but longer distance
ScrewInclined plane wrapped
around a shaftWedge
Two inclined planes
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Pulley
Grooved wheel
Changed direction of force
In combination, can decrease force
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What is the mechanical advantage of the inclined plane?
MA = 10
What is the weight of the cart assuming no friction?
πΉπ = 500 N
ππ΄ =πΉππΉπ
=ππππ
ππ΄ =10 π
1π= 10
πΉπ50 π
=10 π
1 ππΉπ = 500 π
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Muscles only contract, so they come in pairs
Muscles are attached to bones close to the joints using tendons
This makes the muscles supply larger force than is lifted
Input force > output forceMA < 1
Huge forces can be created this way.
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Machines canβt help you much here, exercise your mental muscles instead
Read 10.1, 10.2
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In this lesson you willβ¦
β’ Describe uniform circular motion.
β’ Explain non-uniform circular motion.
β’ Calculate angular acceleration of an object.
β’ Observe the link between linear and angular acceleration.
β’ Observe the kinematics of rotational motion.
β’ Derive rotational kinematic equations.
β’ Evaluate problem solving strategies for rotational kinematics.
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Rotational motion Describes spinning motion
π is like xπ₯ = ππβ position
π is like v
π =Ξπ
Ξπ‘
π£ = ππβ velocity πΌ is like a
πΌ =Ξπ
Ξπ‘
ππ‘ = ππΌβ acceleration
CCW is +CW is -
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Two components to accelerationCentripetal
Toward centerChanges direction
only since perpendicular to v
ππ =π£2
π
Tangental (linear)Tangent to circleChanges speed only
since parallel to v
ππ‘ = ππΌ
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Equations of kinematics for rotational motion are same as for linear motion
π = ππ‘
π = π0 + πΌπ‘
π = π0π‘ +1
2πΌπ‘2
π2 = π02 + 2πΌπ
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Reasoning Strategy
1. Examine the situation to determine if rotational motion involved
2. Identify the unknowns (a drawing can be useful)
3. Identify the knowns
4. Pick the appropriate equation based on the knowns/unknowns
5. Substitute the values into the equation and solve
6. Check to see if your answer is reasonable
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A figure skater is spinning at 0.5 rev/s and then pulls her arms in and increases her speed to 10 rev/s in 1.5 s. What was her angular acceleration?
39.8 rad/s2
π0 = 0.5πππ£
π
2π πππ
πππ£= π
πππ
π
π = 10πππ£
π
2π πππ
πππ£= 20π
πππ
π
π‘ = 1.5 π π = π0 + πΌπ‘
20ππππ
π = π
πππ
π + πΌ 1.5 π
19ππππ
π = πΌ 1.5 π
πΌ = 39.8πππ
π 2
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34
A ceiling fan has 4 evenly spaced blades of negligible width. As you are putting on your shirt, you raise your hand. It brushes a blade and then is hit by the next blade. If the blades were rotating at 4 rev/s and stops in 0.01 s as it hits your hand, what angular displacement did the fan move after it hit your hand?
π = 0.02 rev = 0.126 rad = 7.2Β°
π0 = 4πππ£
π , π‘ = 0.01 π
π = ππ‘
π =π + π0
2π‘
π =0 + 4
πππ£π
20.01 π = 0.02 πππ£ = 0.126 πππ = 7.2Β°
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Spin up your mind and toss out some answers
Read 10.3, 10.4
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In this lesson you willβ¦β’ Understand the relationship between force, mass and acceleration.β’ Study the turning effect of force.β’ Study the analogy between force and torque, mass and moment of inertia, and linear acceleration and angularacceleration.β’ Derive the equation for rotational work.β’ Calculate rotational kinetic energy.β’ Demonstrate the Law of Conservation of Energy.
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π = πΉππ
πΉπ = πππ‘ π = πππ‘π
ππ‘ = ππΌ
π = ππ2πΌ πΌ = ππ2βMoment of inertia of a
particle
π = πΌπΌNewtonβs second law for rotationΞ± is in rad/s2
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Moment of Inertia (I) measures how much an object wants to keep rotating (or not start rotating)
Use calculus to find πΌ =βππ2
Unit:kg m2
Page 328 lists I for many different mass distributions
39
Work for rotation
π = πΉΞπ
π = πΉπΞπ
π
π = ππ
Kinetic Energy
πΎπΈπππ‘ =1
2πΌπ2
Conservation of Mechanical Energy
ππΈπ + πΎπΈπ = ππΈπ + πΎπΈπRemember that the KE can
include both translational and rotational.
πΎπΈ =1
2ππ£2
I is the same as m for rotation
Conservation of ME is when closed system
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Zorch, an archenemy of Superman, decides to slow Earthβs rotation to once per 28.0 h by exerting an opposing force at and parallel to the equator. Superman is not immediately concerned, because he knows Zorch can only exert a force of 4.00Γ107 N (a little greater than a Saturn V rocketβs thrust). How long must Zorch push with this force to accomplish his goal? (This period gives Superman time to devote to other villains.)
1.26 Γ 1011 π¦π
πΉ = β4 Γ 107 π, π = 6.38 Γ 106 π,ππΈ = 5.98 Γ 1024 ππ
π0 =1 πππ£
24 β
2π πππ
1 πππ£
1 β
3600 π =
π
43200
πππ
π β 7.272 Γ 10β5
πππ
π
π€π =1 πππ£
28 β
2π
1 πππ£
1 β
3600 π =
π
50400
πππ
π β 6.233 Γ 10β5
πππ
π
π = πΌπΌ
πΌ =Ξπ
Ξπ‘=ππ β π0
π‘
π = πΌππ β π0
π‘
π‘ =πΌ ππ β π0
π
πΌ =2
5ππ 2, π = πΉπ
π‘ =
25ππ 2 ππ β π0
πΉπ
π‘ =2 5.98 Γ 1024 ππ 6.38 Γ 106 π 6.233 Γ 10β5
ππππ β 7.272 Γ 10β5
ππππ
5 β4.00 Γ 107 ππ‘ = 3.964 Γ 1018 π = 1.101 Γ 1015 β = 4.588 Γ 1013 π = 1.26 Γ 1011 π¦π
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A solid sphere (m = 2 kg and r = 0.25 m) and a thin spherical shell (m = 2 kg and r = 0.25 m) roll down a ramp that is 0.5 m high. What is the velocity of each sphere as it reaches the bottom of the ramp?
Solid: 2.65 m/sShell: 2.42 m/s
Notice masses canceled so mass didnβt matter
Solid Sphere: π = 2 ππ, π = 0.25 π, β0 = 0.5 π, βπ = 0, π£0 = 0,π0 = 0
ππΈπ + πΎπΈπ‘ππππ 0 + πΎπΈπππ‘0 = ππΈπ + πΎπΈπ‘ππππ π + πΎπΈπππ‘π
ππβ0 + 0 + 0 = 0 +1
2ππ£π
2 +1
2πΌππ
2
ππβ0 =1
2ππ£π
2 +1
2
2
5ππ2 ππ
2
πβ0 =1
2π£π2 +
1
5π2
π£π
π
2
πβ0 =1
2π£π2 +
1
5π£π2
πβ0 =7
10π£π2
π£π =10
7πβ0
π£π =10
79.80
π
π 20.5 π = 2.65
π
π
Shell: π = 2 ππ, π = 0.25 π, β0 = 0.5 π, βπ = 0, π£0 = 0, π0 = 0
ππΈπ + πΎπΈπ‘ππππ 0 + πΎπΈπππ‘0 = ππΈπ + πΎπΈπ‘ππππ π + πΎπΈπππ‘π
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ππβ0 + 0 + 0 = 0 +1
2ππ£π
2 +1
2πΌππ
2
ππβ0 =1
2ππ£π
2 +1
2
2
3ππ2 ππ
2
πβ0 =1
2π£π2 +
1
3π2
π£π
π
2
πβ0 =1
2π£π2 +
1
3π£π2
πβ0 =5
6π£π2
π£π =6
5πβ0
π£π =6
59.80
π
π 20.5 π = 2.42
π
π
42
Donβt spin your wheels as you use energy to solve these problems
Read 10.5
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In this lesson you willβ¦
β’ Understand the analogy between angular momentum and linear momentum.
β’ Observe the relationship between torque and angular momentum.
β’ Apply the law of conservation of angular momentum.
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Linear momentump = mv
Angular momentumL = IΟ
Unit:
kg m2/s
Ο must be in rad/s
When you rotate something you exert a torque.
More torque = faster change in angular momentum
ππππ‘ =ΞπΏ
Ξπ‘
Like πΉ =Ξπ
Ξπ‘
Demo with the rotation rods
I is like m for rotational motion
45
Linear momentum of a system is conserved if πΉπππ‘ = 0
p0 = pf
Angular momentum of a system is also conserved if ππππ‘ = 0
L0 = Lf
46
A 10-kg solid disk with r = 0.40 m is spinning at 8 rad/s. A 9-kg solid disk with r = 0.30 m is dropped onto the first disk. If the first disk was initially not rotating, what is the angular velocity after the disks are together?
Ο = 5.31 rad/s
What was the torque applied by the first disk onto the second if the collision takes 0.01 s?
π = 215 Nm
Disk 1:
π0 = 8πππ
π , π = 0.4 π,π = 10 ππ
πΌ =1
2ππ 2 =
1
210 ππ 0.4 π 2 = 0.8 ππ β π2
Disk 2:
π0 = 0πππ
π , π = 0.3 π,π = 9 ππ
πΌ =1
2ππ 2 =
1
29 ππ 0.3 π 2 = 0.405 ππ π2
πΏ = πΌππΏ0 = πΏπ
0.8 πππ2 8πππ
π + 0.405 πππ2 0 = 0.8 πππ2 + 0.405 πππ2 π
6.4 πππ2 = 1.205 πππ2 ππ = 5.31 πππ/π
π =ΞπΏ
Ξπ‘
π =0.405 πππ2 5.31
ππππ β 0
0.01 π = 215.055 ππ
47
Angular Momentum conserved if net external torque is zero
Linear Momentum conserved if net external force is zero
Kinetic Energy conserved if elastic collision
48
Direction of angular quantities
Right-hand Rule
Hold hand out with thumb out along axis
Curl your fingers in direction of motion (you may have to turn your hand upside down)
vector in direction of thumb
49
A person is holding a spinning bicycle wheel while he stands on a stationary frictionless turntable. What will happen if he suddenly flips the bicycle wheel over so that it is spinning in the opposite direction?
Consider the system of turntable, person, and wheel. The total angular momentum before is L upward. Afterward, the total angular momentum must be the same. If the wheel is upside down, its angular momentum is βL so the angular momentum of the person must be +2L. So the person rotates in the direction the wheel was initially spinning.
50
Gyroscopes Two forces acting on a spinning
gyroscope. The torque produced is perpendicular to the angular momentum, thus the direction of the torque is changed, but not its magnitude. The gyroscope precessesaround a vertical axis, since the torque is always horizontal and perpendicular to L .
If the gyroscope is not spinning, it acquires angular momentum in the direction of the torque ( L = ΞL ), and it rotates around a horizontal axis, falling over just as we would expect.
Use right-hand rule to find direction of torque.
Earth itself acts like a gigantic gyroscope. Its angular momentum is along its axis and points at Polaris, the North Star. But Earth is slowly precessing (once in about 26,000 years) due to the torque of the Sun and the Moon on its nonspherical shape.
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Let you momentum carry you through these problems
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