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11/16/2021
1
Rotational Motion Overview (Chapter 8, 9, 10)Chapters
[Pure] Rotational
Motion (8.1-4, 8.6-7):
Rotation about a fixed
Axis
Angular Kinematic Variables ( Ō¦š, š, Ō¦š¼) ā Axis of Rotation (8.1-2)
Torque š (8.3)
Moment of Inertia š¼ (8.4-5) and Rotational Kinetic Energy
Conservation of Energy including rotational motion (8.6)
Conservation of Angular Momentum šæ (8.7)
Rolling Motion (9.1-9.5)
= rotation + translation
Understanding Rolling Motions (9.1-9.3)
Conservation of Energy: rotation and translation (9.5)
Newtonās 2nd Law: rotation and translation: acceleration (9.4)
Equilibrium (10.1-10.3) Ī£ Ō¦š¹ = 0 and Ī£Ō¦š = 0 (10.1, 10.2)
Equilibrium Lab, Applications (10.3)
Overview: Rotational Kinetic Energy
ā¢ Use conservation of energy to solve rotational motion problems.ā¢ This requires knowing the Kinetic Energy in rotational motion.ā¢ Moment of Inertia.ā¢ Moment of Inertia depends on the axis of rotation.
ā¢ Moment of Inertial for rigid bodies.ā¢ Moment of inertia for a rodā¢ Moment of inertial for a rectangleā¢ Moment of inertia for a cylinderā¢ Moment of inertia for a sphere
ā¢ Example Problems along the way
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šæ = 1.0 š, š = 1 šš.
Rotational KE for rod of length šæ, mass š.
From the Rotational KE, we can find the Moment of Inertia about the axis of rotation.
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šæ
š1
š2
š2
š£ = ?
ā¢ A rod with mass š1 = 4 šš and length šæ = 1.0 š is hung horizontally on the wall with the pivot at the centre of the rod.
ā¢ A second mass š2 = 0.2 šš is attached to one end.
ā¢ Starting horizontally at rest, the rod rotates due to the added mass. What is the speed of the mass when it reaches the bottom?
š1
Overview: Rotational Kinetic Energy
ā¢ Use conservation of energy to solve rotational motion problems.ā¢ This requires knowing the Kinetic Energy in rotational motion.ā¢ Moment of Inertia.ā¢ Moment of Inertia depends on the axis of rotation.
ā¢ Moment of Inertial for rigid bodies.ā¢ Moment of inertia for a rodā¢ Moment of inertial for a rectangleā¢ Moment of inertia for a cylinderā¢ Moment of inertia for a sphere
ā¢ Example Problems along the way
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Moment of inertia not in table.
ā¢ A rectangle has sides š = 0.25 š,š = 0.75 š with mass š = 2.0 šš.
ā¢ Given that the moment of inertia of a
rectangle is 1
12š(š2 + š2) about the
centre of a rectangle, what are the moment of inertia at point A?
Ć Ć A
š
š
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Overview: Rotational Kinetic Energy
ā¢ Use conservation of energy to solve rotational motion problems.ā¢ This requires knowing the Kinetic Energy in rotational motion.ā¢ Moment of Inertia.ā¢ Moment of Inertia depends on the axis of rotation.
ā¢ Moment of Inertial for rigid bodies.ā¢ Moment of inertia for a rodā¢ Moment of inertial for a rectangleā¢ Moment of inertia for a cylinderā¢ Moment of inertia for a sphere
ā¢ Example Problems along the way
What is the moment of inertial of this thin walled hollow cylinder (mass š, radius š , length šæ)?
š¼šš„šš = š1š12 +š2š2
2 +š3š32 + ā¦
A. šš 2
B.1
2šš 2
C.1
4šš 2
D.1
8šš 2
E. 1
12šš 2
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š1
š2
Rotational KE example problem
ā¢ Two masses, š1 = 2 šš,š2 = 1 šš are connected by a string with negligible weight.
ā¢ The string is then put around a pulley with mass š = 6 šš and radius š = 0.5 š.
ā¢ The pulley is made like a bicycle wheel where the mass is concentrated on the rim only.
ā¢ Initially, mass š1is at rest at height ā = 2 šwhile š2 is resting on the ground.
ā¢ When released, what is the speed of š1 just before it hits the ground?
š = 6 ššš = 0.5 š
š1 = 2 šš
š2
= 1 šš
š1
š2
Massless wheel for the moment (š = 0)
ā¢ Īš¾šø =1
2š1š£
2 +1
2š2š£
2
ā¢ What is āĪššø = ššøš ā ššøš?
A. š1šā
B. š2šā
C. š1šā +š2šā
D. š1šā āš2šā
Conservation of Energy Īš¾šø = āĪššøš = 0
š1 = 2 šš
š2
= 1 šš
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š1
š2
Massless wheel for the moment (š = 0)
ā¢ Īš¾šø =1
2š1š£
2 +1
2š2š£
2
ā¢ āĪššø = ššøš ā ššøš = š1šā āš2šā
Conservation of Energy Īš¾šø = āĪššø
1
2š1š£
2 +1
2š2š£
2 =š1šā āš2šā
š = 0
š1 = 2 šš
š2
= 1 šš
š1
š2
Massless wheel for the moment (š = 6 šš)
ā¢ āĪššø = ššøš ā ššøš = š1šā āš2šā
ā¢ Īš¾šø =1
2š1š£
2 +1
2š2š£
2 + Īš¾šøššš”šš”ššš
Conservation of Energy Īš¾šø = āĪššøš = 6 ššš = 0.5 š
š1 = 2 šš
š2
= 1 šš
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What is the moment of inertial of this solid cylinder (mass š, radius š , length šæ)?
š¼ = šš 2
š¼ = ?
Replace the pulley in the previous problem with a solid disk. What is the moment of inertial š¼ of a solid disk (mass š, radius š ) ?
š1
š2
š1
š2
Previously š¼ =šš 2
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Overview: Rotational Kinetic Energy
ā¢ Use conservation of energy to solve rotational motion problems.ā¢ This requires knowing the Kinetic Energy in rotational motion.ā¢ Moment of Inertia.ā¢ Moment of Inertia depends on the axis of rotation.
ā¢ Moment of Inertial for rigid bodies.ā¢ Moment of inertia for a rodā¢ Moment of inertial for a rectangleā¢ Moment of inertia for a cylinderā¢ Moment of inertia for a sphere
ā¢ Example Problems along the way
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1. Torque ā Concepts (magnitudes only)
We need a torque to start rotating the objectLinear Angular
KE 1
2šš£2
1
2š¼š2
displacement š„ š
velocity š£ š
acceleration š š¼
Inertia š š¼
Force Torque
š¹ š
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Definition of torque š
šš
šš
Same force Ō¦š¹ applied at two different locations yield different torques
Easier to rotate the nut when the force is applied at šš than at šš
Applied force here is perpendicular to šš and šš.
Magnitudes only:
Definition of torque š
Ō¦š
šŌ¦š¹
The radial component š¹ cos š does not rotate the wrench/nut. Only the tangential component of the force š¹ sin š does rotate it.
š¹ sin š
š¹ cos š
When Ō¦š¹ is not perpendicular to Ō¦š
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Definition of torque š
Ō¦š
šŌ¦š¹
š
Keep the full magnitude of
vector Ō¦š¹, but take the perpendicular component (šā„) of the displacement vector Ō¦š.
We need a torque to start rotating the objectLinear Angular
KE 1
2šš£2
1
2š¼š2
š„ š
š£ š
š š¼
š š¼
Force Torque
š¹
Newtonās Law
š¹ = šš
Ō¦š
š Ō¦š¹
š¹ā„ = š¹ sin š
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We need a torque to start rotating the object
1. Tangential acceleration šā„
šā„ = šš¼
š¼ is the angular accel.
2. š¹ā„ = ššā„
Starting with š = šš¹ā„ which is correct?
A. š = šš¼
B. š = ššš¼
C. š = šš2š¼
Ō¦š
šŌ¦š¹
š¹ā„ = š¹ sin š
šā„
š¶
š
We need a torque to start rotating the objectLinear Angular
KE 1
2šš£2
1
2š¼š2
š„ š
š£ š
š š¼
š š¼
Force Torque
š¹
Newtonās Law
š¹ = šš
Ō¦š
š Ō¦š¹
š¹ā„ = š¹ sin š
š¶
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2. Torque ā Vector Math
From š = š¼š¼ to Ō¦š = š¼ Ō¦š¼
We need a torque to start rotating the objectLinear Angular
KE 1
2šš£2
1
2š¼š2
š„ š
š£ š
š š¼
š š¼
Force Torque
š¹
Newtonās Law
š¹ = šš
Ō¦š
š Ō¦š¹
š¹ā„ = š¹ sin š
š¶
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š
š„
š¦
š§
ā¢ Consider 1D rotation
ā¢ Displacement š
ā¢ Ang. Velocity Ļ = Īš/Īš”
ā¢ Ang. Accel. š¼ = Īš/Īš”
ā¢ object rotates about
Rotational Motion
ā¢ Consider 1D rotation
ā¢ Displacement š
ā¢ Ang. Velocity Ļ = Īš/Īš”
ā¢ Ang. Accel. š¼ =Īš
Īš”
ā¢ object rotates about +š§ axis
ā¢ They are all in the direction of the axis of rotation: in this example,
ā¢ Ō¦š = šš
ā¢ š = šš
ā¢ Ō¦š¼ = š¼š
Rotational Motion
š
š„
š¦
š§
Vector Ō¦š = š¼ Ō¦š¼Magnitude: š = š¼š¼
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We need a torque to start rotating the objectLinear Angular
KE 1
2šš£2
1
2š¼š2
š„ š
š£ š
š š¼
š š¼
Force Torque
š¹ š = šš¹ sinš
Newtonās Law
Ō¦š¹ = š Ō¦š
Ō¦š
š Ō¦š¹
š¹ā„ = š¹ sin š
š¶
We need a torque to start rotating the object
Ō¦šš Ō¦š¹
š¹ā„ = š¹ sin š
Recall
Axis of rotation:+š
Torque, angular accel, velocity, displacement are all in the + š direction.
š = šš š¬š¢š§š½ can be rewritten as
š„
š¦
š§š = šš¹ā„ = ššš¬š¢š§š½
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