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Transcript
```Physics
Unit 4
 This Slideshow was developed to accompany the textbook
 OpenStax Physics
 By OpenStax College and Rice University
 2013 edition
 Some examples and diagrams are taken from the textbook.
Slides created by
[email protected]
Statics
Study of forces in equilibrium
Equilibrium means no acceleration
First condition of equilibrium
𝑛𝑒𝑡 𝐹 = 0
𝐹𝑥 = 0 and 𝐹𝑦 = 0
They can still rotate, so…
 Think of opening a door
 Which opens the door the best?
Picture a
 Big force  large torque
 Force away from pivot  large torque
 Force directed ⊥ to door  large torque
τ=F×r
This means we use the component of the force that is perpendicular
to the lever arm
 τ = F r
 τ = F r sin θ
θ is the angle between the force and the radius
 Unit: Nm
CCW  +
CW  -
 You are meeting the parents of your new “special” 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 90°
120°
Second condition of equilibrium
Net torque = 0
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 seesaw’s center of mass must the fulcrum be
placed?
80 kg
1.20 m
xm
25 kg
How much force
must the fulcrum
10 kg
support?
5m
1029 N
these torque questions
9P1-5
9CQ6-8
1) 46.8 Nm
2) 23.1 Nm, 17.0 ft lb
3) 23.3 Nm
4) 568 N
5) 1.36 m, 686 N
Stable equilibrium
When displaced from
equilibrium, the
system experiences a
net force or torque in a
direction opposite to
the direction of the
displacement.
Unstable equilibrium
When displaced from
equilibrium, the net
force or torque is in
same direction of the
displacement
Neutral Equilibrium
Equilibrium is
independent of
displacement from its
original position
Problem-Solving Strategy for Static Equilibrium
1. Is it in equilibrium? (no acceleration or accelerated
rotation)
2. Draw free body diagram
3. Apply ∑𝐹 = 0 and/or ∑𝜏 = 0
a. Choose a pivot point to simplify the problem
4. Check your solution for reasonableness.
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
diagram
next slide
𝑚 = 225 𝑘𝑔, 𝑀 = 45.0 𝑘𝑔, 𝑇 = ? , 𝐹ℎ𝑥 = ? , 𝐹ℎ𝑦 = ?
𝑇 = 6627 𝑁
𝐹ℎ𝑥 = 5739 𝑁
𝐹ℎ𝑦 = 5959 𝑁
 Pick a stable position while
 9P6-7, 9, 11-13, 16-17
 9CQ10-12, 15, 17, 19
 6) 1.43 × 103 𝑁
 7) 1.8 m
 9) 392 N, 196 N
 11) 11.0 N, 0.450
1
 12) of the
6
weight, 2.0 × 104
N up
 13) 7.20 × 103 𝑁, 65.2°
 16) 1.11 × 103 𝑁 along each
leg
 17) 126 N, 751 N
Machines make work easier
Energy is conserved so same amount of energy with
or without machine
𝑀𝐴 =
𝐹𝑜
𝐹𝑖
 Lever
 Uses torques with pivot at N
 𝐹𝑖 𝑙𝑖 = 𝐹𝑜 𝑙𝑜

𝐹𝑜
𝐹𝑖
=
𝑙𝑖
𝑙𝑜
 When F ↑, 𝑙 ↓
 𝑀𝐴 =
𝐹𝑜
𝐹𝑖
=
𝑑𝑖
𝑑𝑜
 Other simple machines
 Wheel and Axle
Lever
 Inclined Plane
Ramp – less force to slide up,
but longer distance
 Screw
Inclined plane wrapped
around a shaft
 Wedge
Two inclined planes
Pulley
Grooved wheel
Changed direction
of force
In combination,
can decrease force
What is the mechanical advantage of the inclined plane?
MA = 10
What is the weight of the cart assuming no friction?
𝐹𝑜 = 500 N
 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 force
 MA < 1
 9P19-24, 29, 31-32, 34-35
 10CQ1-4
 19) 25.0, 50.0 N
 20) 1.78 m
 21) 18.5, 29.1 N, 510 N
 22) 0.0400
 23) 1.30 × 103 N
 24) 564 N
 29) 1.72 × 103 N
 31) 470 N
 32) 25 N down, 75 N up
 34) 2.25 × 103 N
 35) 1.2 × 102 N up,
84 N down
 Rotational motion
 Describes spinning motion
 𝜃 is like x
𝑥 = 𝑟𝜃  position
 𝜔 is like v
Δ𝜃
𝜔 =
Δ𝑡
𝑣 = 𝑟𝜔  velocity
 𝛼 is like a
Δ𝜔
𝛼 =
Δ𝑡
𝑎𝑡 = 𝑟𝛼  acceleration
Two components to
acceleration
Centripetal
Toward center
Changes direction
only since
perpendicular to v
𝑎𝑐 =
𝑣2
𝑟
Tangental (linear)
Tangent to circle
Changes speed only
since parallel to v
𝑎𝑡 = 𝑟𝛼
Equations of kinematics
for rotational motion are
same as for linear motion
𝜃 = 𝜔𝑡
𝜔 = 𝜔0 + 𝛼𝑡
1
𝜃 = 𝜔0 𝑡 + 𝛼𝑡 2
2
𝜔2 = 𝜔02 + 2𝛼𝜃
 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/unknwons
5. Substitute the values into the equation and solve
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?
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?
𝜃 = 0.02 rev = 0.126 rad = 7.2°
 Spin up your mind and toss out some
 10P1-2, 5-8
 10CQ5-8, 10
 1) 0.737 rev/s
8.29 m/s2,
1.04 × 107 m/s2,
1.06 × 106 𝑔




6) 405 m
7) 45.7 s, 116 rev
28.7 rev,
3.80 s,
50.7 m,
26.6 m/s,
reasonable
 𝜏 = 𝐹𝑇 𝑟
 𝐹𝑇 = 𝑚𝑎𝑡
 𝜏 = 𝑚𝑎𝑡 𝑟
 𝑎𝑡 = 𝑟𝛼
 𝜏 = 𝑚𝑟 2 𝛼
 𝐼 = 𝑚𝑟 2  Moment of inertia of a
particle
 𝜏 = 𝐼𝛼
 Newton’s second law for rotation
 Moment of Inertia (I)
measures how much an
object wants to keep
rotating (or not start
rotating)
 Use calculus to find 𝐼 =
∑𝑚𝑟 2
 Unit:
 kg m2
 Page 328 lists I for many
different mass distributions
 Work for rotation
𝑊 = 𝐹Δ𝑠
𝑊 =
Δ𝑠
𝐹𝑟
𝑟
𝑊 = 𝜏𝜃
 Kinetic Energy
1
2
𝐾𝐸𝑟𝑜𝑡 = 𝐼𝜔2
 Conservation of Mechanical
Energy
𝑃𝐸𝑖 + 𝐾𝐸𝑖 = 𝑃𝐸𝑓 + 𝐾𝐸𝑓
Remember that the KE can
include both translational
and rotational.
Zorch, an archenemy of Superman, decides to slow Earth’s
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
4.00×107 N (a little greater than a Saturn V rocket’s 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 × 1011 𝑦𝑟
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/s
Shell: 2.42 m/s
Notice masses canceled so mass didn’t matter
 Don’t spin your wheels as you use
energy to solve these problems
 10P11-15, 22-26, 28, 30
 10CQ13-14, 16, 18-20, 25-27




11) 0.363 𝑘𝑔 ⋅ 𝑚2 , 2.34 𝑘𝑔 ⋅ 𝑚2
12) 0.500 𝑘𝑔 ⋅ 𝑚2
13) 1.18 × 103 𝑁
 14) 50.4 𝑁 ⋅ 𝑚, 17.1 𝑟𝑎𝑑/𝑠 2 ,
17.0 𝑟𝑎𝑑/𝑠 2
 15) 97.9 𝑟𝑎𝑑/𝑠 2 , 32.3 𝑚/𝑠, 0.817 s
 22) 7.00 𝑚/𝑠
 23) 2.57 × 1029 𝐽, 2.65 × 1033 𝐽
 24) 8.09 × 103 𝐽
 25) 434 𝐽
𝑚
 26) 9.66 𝑟𝑎𝑑/𝑠, 10.1
𝑠
 28) 2.18 𝑚, 3.27 𝑚
 30) 41.1 𝑟𝑎𝑑/𝑠 2 ,
15.7 𝐽
 Linear momentum
 p = mv
 Angular momentum
 L = Iω
 Unit:
 kg m2/s
 ω must be in rad/s
When you rotate something
you exert a torque.
More torque = faster change
in angular momentum
𝜏𝑛𝑒𝑡 =
Δ𝐿
Δ𝑡
Like 𝐹 =
Δ𝑝
Δ𝑡
Linear momentum of a system is conserved if 𝐹𝑛𝑒𝑡 = 0
p0 = pf
Angular momentum of a system is also conserved if
𝜏𝑛𝑒𝑡 = 0
L0 = Lf
 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?
 What was the torque applied by the first disk onto the second if the collision
takes 0.01 s?
 𝜏 = 215 Nm
Angular Momentum conserved if net external torque is
zero
Linear Momentum conserved if net external force is zero
Kinetic Energy conserved if elastic collision
 Direction of angular quantities
Right-hand Rule
Hold hand out with thumb
out along axis
of motion (you may have to
vector in direction of thumb
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?
 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 precesses
around 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 = ΔL ), and it
rotates around a horizontal axis, falling
over just as we would expect.
 Let you momentum carry you
through these problems
 10P36-41
 10CQ26-27, 29-30
 36) 2.66 × 1040 𝑘𝑔 ⋅
33
7.07 × 10
𝑘𝑔 ⋅
𝑚2
𝑠
𝑚2
𝑠


 39) 2.30 𝑟𝑎𝑑/𝑠
 40) 25.3 rpm

,
𝑚2
37) 2.89 × 10 𝑘𝑔 ⋅ ,
𝑠
𝑚2
29
2.37 × 10 𝑘𝑔 ⋅
𝑠
2
38) 22.5 𝑘𝑔 ⋅ 𝑚 /𝑠
34
𝑚2
41) 15.1 𝑘𝑔 ⋅ ,
𝑠
2
1.92 𝑘𝑔 ⋅ 𝑚 ,
−0.503 𝑁 ⋅ 𝑚
```
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