Chapter 7 Rotational Motion. Slide 7-3 © 2015 Pearson Education, Inc. Curvilinear coordinates.
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Transcript of Chapter 7 Rotational Motion. Slide 7-3 © 2015 Pearson Education, Inc. Curvilinear coordinates.
Chapter 7
• Rotational Motion
Slide 7-3
© 2015 Pearson Education, Inc.
𝑠=𝑟 𝜃
Curvilinear coordinates
© 2015 Pearson Education, Inc.
∆ 𝑥=𝑣𝑖 𝑡+12𝑎𝑡 2
∆ 𝜃=𝜔 𝑖 𝑡+12𝛼𝑡 2 𝜔 𝑓
2 =𝜔𝑖2+2𝛼 ∆𝜃
𝑣 𝑓2 =𝑣𝑖
2+2𝑎∆ 𝑥
Kinematic equations are the same, just different variables
© 2015 Pearson Education, Inc.
Angular Velocity
𝑣=𝑟 𝜔 𝑣
“omega”
© 2015 Pearson Education, Inc.
Ball rolling across frictionless floor
This slope is
𝜔=∆𝜃∆ 𝑡
Checking UnderstandingTwo coins rotate on a turntable. Coin B is twice as far from the axis as coin A.
A. The angular velocity of A is twice that of B.
B. The angular velocity of A equals that of B.
C. The angular velocity of A is half that of B.
Slide 7-13
Answer
All points on the turntable rotate through the same angle in the same time. All points have the same period.
Two coins rotate on a turntable. Coin B is twice as far from the axis as coin A.
A. The angular velocity of A is twice that of B.
B. The angular velocity of A equals that of B.
C. The angular velocity of A is half that of B.
Slide 7-14
Angular acceleration α measures how rapidly the angular velocity is changing:
Slide 7-17
𝛼=𝑎t
𝑟
Angular Acceleration
𝑎t
Tangential acceleration
Slide 7-18
Linear and Circular motion compared
Two coins rotate on a turntable. Coin B is twice as far from the axis as coin A.
A. The speed of A is twice that of B.
B. The speed of A equals that of B.
C. The speed of A is half that of B.
Checking Understanding
Slide 7-15
Answer
Twice the radius means twice the speed
Two coins rotate on a turntable. Coin B is twice as far from the axis as coin A.
A. The speed of A is twice that of B.
B. The speed of A equals that of B.
C. The speed of A is half that of B.
Slide 7-16
v r
© 2015 Pearson Education, Inc.
3 types of related motion
© 2015 Pearson Education, Inc.
Tangential velocity of circular motion can become linear motion
© 2015 Pearson Education, Inc.
Tangential velocity of circular motion can become linear motion
© 2015 Pearson Education, Inc.
Tangential velocity of circular motion can become linear motion
© 2015 Pearson Education, Inc.
Combination motion
𝑣𝜔
Center of mass follows original trajectory
Slide 7-19
The equations have the same form
Example ProblemA high-speed drill rotating CCW takes 2.5 s to speed up to 2400 rpm.
A. What is the drill’s angular acceleration?B. How many revolutions does it make as it reaches top speed?
Slide 7-21
∆ 𝜃=𝜔 𝑖 𝑡+12𝛼𝑡 2 𝜔 𝑓
2 =𝜔𝑖2+2𝛼 ∆𝜃
2 ∆𝜃𝑡 2 =𝛼
𝜔 𝑓2
2𝛼=∆𝜃
© 2015 Pearson Education, Inc.
Comparing rotational motion plots
Slide 7-22
Centripetal and Tangential acceleration
© 2015 Pearson Education, Inc.
The difference of nonuniform circular
motion
The speed is changing
Center of Gravity
=
Slide 7-29
Calculating the Center-of-Gravity Position
Slide 7-30
© 2015 Pearson Education, Inc.
Center of mass practice
𝑥cg=𝑚1 𝑥1+𝑚2 𝑥2
𝑚1+𝑚2
=5 kg ∙ m15 kg
=13
m
© 2015 Pearson Education, Inc.
© 2015 Pearson Education, Inc.
𝜏=𝑟𝐹 sin 𝜃 Angle between lever arm and force
Torque –when force causes circular motion
Checking UnderstandingWhich point could be the center of gravity of this L-shaped piece?
Slide 7-32
Which point could be the center of gravity of this L-shaped piece?
Answer
(a)
Slide 7-33
Interpreting Torque
rF rF sin
Torque is due to the component of the force perpendicular to the radial line.
Slide 7-25
Signs and Strengths of the Torque
Slide 7-27
The four forces below are equal in magnitude. Which force would be most effective in opening the door?
A. B. C. D. E. Either or
$$ 1
$$ 2
$$ 3
$$ 4
Either
$$ 1 o
r $$
3
98%
0% 0%2%0%
Example torque Problem Revolutionaries attempt to pull down a statue of the Great Leader by pulling on a rope tied to the top of his head. The statue is 17 m tall, and they pull with a force of 4200 N at an angle of 65° to the horizontal. What is the torque they exert on the statue? If they are standing to the right of the statue, is the torque positive or negative?
Slide 7-28
65°
17 m
F = 4200 N
𝜏=17 m ∙ 4200 N ∙ sin 𝜃
𝜃≠65°
𝜏=17 m ∙ 4200 N ∙ sin 25°
𝜏=𝑟𝐹 sin 𝜃
pivot
Negative torque, but why?
Rotating it in the CW directionr = 17m
Which force vector on point P would keep the wheel from spinning?
A. AB. CC. DD. E
A C D E
98%
0%2%0%
© 2015 Pearson Education, Inc.
Torque practice
𝜏=𝑟𝐹 sin𝜗
𝜏=𝑟𝑚𝑔 sin𝜗
𝜏= (1.6 m ) (3.2 kg )(− 9.8m
s2 )sin ( 6 5° )
𝜗
Which torques are equal?
A. B = C = D = E only
B. A = B and C = D = E
C. None are equal
D. B = E and C = D
B = C = D =
E only
A = B and C = D = E
None are e
qual
B = E a
nd C = D
0%
98%
0%2%
© 2015 Pearson Education, Inc.
Torque practice with centers of mass
What is the Net Torque is exerted by the gymnast about an axis through the rings?
© 2015 Pearson Education, Inc.
𝑟1
𝐹 1
𝑟2
𝐹 2
∑𝜏=0Torque equilibrium
Reading Quiz2. Which factor does the torque on an object not depend on?
A. The magnitude of the applied force.
B. The object’s angular velocity.
C. The angle at which the force is applied.
D. The distance from the axis to the point at which the force is applied.
Slide 7-7
Answer2. Which factor does the torque on an object not depend on?
A. The magnitude of the applied force.
B. The object’s angular velocity.
C. The angle at which the force is applied.
D. The distance from the axis to the point at which the force is applied.
Slide 7-8
Example ProblemAn object consists of the three balls shown, connected by massless rods. Find the x- and y-positions of the object’s center of gravity.
Slide 7-31
𝑥cg=𝑚1 𝑥1+𝑚2 𝑥2+𝑚3𝑥3
𝑚1+𝑚2+𝑚3
𝑦 cg=𝑚1𝑦1+𝑚2 𝑦2+𝑚3 𝑦3
𝑚1+𝑚2+𝑚3
An object consists of the three balls shown, connected by massless rods. Find the x- and y-positions of the object’s center of gravity.
Slide 7-31
𝑥cg=𝑚1 (0 )+𝑚2 ( 0 )+𝑚3 (1 m )
𝑚1+𝑚2+𝑚3
=2 kg ∙ m4 kg
=12
m
𝑦 cg=𝑚1 (1 m )+𝑚2 (0 )+𝑚3 (0 )
𝑚1+𝑚2+𝑚3
=1 kg ∙ m4 kg
=14
m
The center of mass for these 3 bodies
The moment of Inertia the rotational equivalent of mass
Newton’s Second Law for Rotation
/ II = moment of inertia. Objects with larger moments of inertia are harder to get rotating.
I miri2
Slide 7-34
Rotational and Linear Dynamics Compared
Slide 7-36
Which moment of inertia is greatest?
A. AB. BC. CD. D
𝑀𝑚
A B C D
25% 25%25%25%
Which force vector applied to point P will stop this rolling ball?
A. AB. BC. CD. DE. E
A B C D E
20% 20%20%20%20%
© 2015 Pearson Education, Inc.
Formulas more common moments of inertia 𝐼
Which red vector is your best bet for getting this bolt as tight as possible?
A. AB. BC. CD. D
A B C D
25% 25%25%25%
Reading Quiz1. Moment of inertia is
A. the rotational equivalent of mass.
B. the point at which all forces appear to act.
C. the time at which inertia occurs.
D. an alternative term for moment arm.
Slide 7-5
Answer1. Moment of inertia is
A. the rotational equivalent of mass.
B. the point at which all forces appear to act.
C. the time at which inertia occurs.
D. an alternative term for moment arm.
Slide 7-6
© 2015 Pearson Education, Inc.
Torque causing rotation
∑𝜏 ≠ 0
∑𝜏=¿ 𝐼 𝛼 ¿What happens to these masses when you let go?
What happens to this pulley system?
A. It does not moveB. The 10N force accelerates
the mass upwardC. The force of gravity on the
mass results in a net force upward
D. The mass moves upward at a constant speed
It does
not move
The 1
0N force
accelera
tes t.
.
The f
orce of g
ravit
y on th
e ...
The m
ass m
oves u
pward at a
...
25% 25%25%25%
© 2015 Pearson Education, Inc.
Do we always have to use torque?
∑ 𝐹1=¿𝐹𝑔 1+𝑇1=𝑚1𝑎1¿∑ 𝐹2=¿𝐹𝑔 2+𝑇 2=𝑚2𝑎2¿
𝑇 1=−𝑇2
Newton’s Third
𝑇 1
𝑇 2
𝑇 1
𝐹 𝑔
𝐹 𝑔
Starting from rest, how long does it take to hit the ground?
© 2015 Pearson Education, Inc.
Use forces to solve
𝑚1𝑔+𝑇1=𝑚1𝑎
𝑚2|𝑔|−𝑇 1=𝑚2𝑎
𝑎2=𝑎1=𝑎
𝑇 1
𝑇 2𝐹 𝑔
𝐹 𝑔
𝑎1 𝑎2
Get rid of by solving for it and substituting into the other equation
𝑎=− 3.27m
s2
Starting from rest, how long does it take to hit the ground?
∆ 𝑦=𝑣 𝑖𝑦𝑡+12𝑎𝑡 2
𝑡=√ 2 ∆ 𝑦𝑎
𝑡=.78 s
Reading Quiz4. A net torque applied to an object causes
A. a linear acceleration of the object.
B. the object to rotate at a constant rate.
C. the angular velocity of the object to change.
D. the moment of inertia of the object to change.
Slide 7-11
Answer4. A net torque applied to an object causes
A. a linear acceleration of the object.
B. the object to rotate at a constant rate.
C. the angular velocity of the object to change.
D. the moment of inertia of the object to change.
Slide 7-12
© 2015 Pearson Education, Inc.
Friction in rotational motionDraw the normal force for the wheel against the break
Draw the frictional force from the break
𝐹 𝑓
𝑁 𝑏
Is this beam balanced?A. YesB. No, it will spin CWC. No, it will spin CCWD. Not enough information
Yes
No, it w
ill sp
in CW
No, it w
ill sp
in CCW
Not enough
inform
ation
25% 25%25%25%
© 2015 Pearson Education, Inc.
Now the pulley has friction𝐼𝑝=
12𝑚𝑝𝑟
2
∑𝜏=¿𝜏1+𝜏2= 𝐼𝑝𝛼 ¿
© 2015 Pearson Education, Inc.
Now the pulley has friction and mass𝐼𝑝=
12𝑚𝑝𝑟
2
.02m ∙ 20N − .02m ∙ 30 N=(.0008 kg ∙m 2)𝛼
4 kg
Additional Example ProblemA baseball bat has a mass of 0.82 kg and is 0.86 m long. It’s held vertically and then allowed to fall. What is the bat’s angular acceleration when it has reached 20° from the vertical? (Model the bat as a uniform cylinder).
Slide 7-43