Post on 05-Apr-2022
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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 ( Τ¦π, π, Τ¦πΌ) β Axis of Rotation (8.1-2)
Torque π (8.3)
Moment of Inertia πΌ (8.4-5) and Rotational Kinetic Energy
Conservation of Energy including rotational motion (8.6)
Conservation of Angular Momentum πΏ (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)
Newtonβs 2nd Law: rotation and translation: acceleration (9.4)
Equilibrium (10.1-10.3) Ξ£ Τ¦πΉ = 0 and Ξ£Τ¦π = 0 (10.1, 10.2)
Equilibrium Lab, Applications (10.3)
Overview: Rotational Kinetic Energy
β’ Use conservation of energy to solve rotational motion problems.β’ This requires knowing the Kinetic Energy in rotational motion.β’ Moment of Inertia.β’ Moment of Inertia depends on the axis of rotation.
β’ Moment of Inertial for rigid bodies.β’ Moment of inertia for a rodβ’ Moment of inertial for a rectangleβ’ Moment of inertia for a cylinderβ’ Moment of inertia for a sphere
β’ Example Problems along the way
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πΏ = 1.0 π, π = 1 ππ.
Rotational KE for rod of length πΏ, mass π.
From the Rotational KE, we can find the Moment of Inertia about the axis of rotation.
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πΏ
π1
π2
π2
π£ = ?
β’ A rod with mass π1 = 4 ππ and length πΏ = 1.0 π is hung horizontally on the wall with the pivot at the centre of the rod.
β’ A second mass π2 = 0.2 ππ is attached to one end.
β’ Starting horizontally at rest, the rod rotates due to the added mass. What is the speed of the mass when it reaches the bottom?
π1
Overview: Rotational Kinetic Energy
β’ Use conservation of energy to solve rotational motion problems.β’ This requires knowing the Kinetic Energy in rotational motion.β’ Moment of Inertia.β’ Moment of Inertia depends on the axis of rotation.
β’ Moment of Inertial for rigid bodies.β’ Moment of inertia for a rodβ’ Moment of inertial for a rectangleβ’ Moment of inertia for a cylinderβ’ Moment of inertia for a sphere
β’ Example Problems along the way
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Moment of inertia not in table.
β’ A rectangle has sides π = 0.25 π,π = 0.75 π with mass π = 2.0 ππ.
β’ Given that the moment of inertia of a
rectangle is 1
12π(π2 + π2) about the
centre of a rectangle, what are the moment of inertia at point A?
Γ Γ A
π
π
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Overview: Rotational Kinetic Energy
β’ Use conservation of energy to solve rotational motion problems.β’ This requires knowing the Kinetic Energy in rotational motion.β’ Moment of Inertia.β’ Moment of Inertia depends on the axis of rotation.
β’ Moment of Inertial for rigid bodies.β’ Moment of inertia for a rodβ’ Moment of inertial for a rectangleβ’ Moment of inertia for a cylinderβ’ Moment of inertia for a sphere
β’ Example Problems along the way
What is the moment of inertial of this thin walled hollow cylinder (mass π, radius π , length πΏ)?
πΌππ₯ππ = π1π12 +π2π2
2 +π3π32 + β¦
A. ππ 2
B.1
2ππ 2
C.1
4ππ 2
D.1
8ππ 2
E. 1
12ππ 2
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π1
π2
Rotational KE example problem
β’ Two masses, π1 = 2 ππ,π2 = 1 ππ are connected by a string with negligible weight.
β’ The string is then put around a pulley with mass π = 6 ππ and radius π = 0.5 π.
β’ The pulley is made like a bicycle wheel where the mass is concentrated on the rim only.
β’ Initially, mass π1is at rest at height β = 2 πwhile π2 is resting on the ground.
β’ When released, what is the speed of π1 just before it hits the ground?
π = 6 πππ = 0.5 π
π1 = 2 ππ
π2
= 1 ππ
π1
π2
Massless wheel for the moment (π = 0)
β’ ΞπΎπΈ =1
2π1π£
2 +1
2π2π£
2
β’ What is βΞππΈ = ππΈπ β ππΈπ?
A. π1πβ
B. π2πβ
C. π1πβ +π2πβ
D. π1πβ βπ2πβ
Conservation of Energy ΞπΎπΈ = βΞππΈπ = 0
π1 = 2 ππ
π2
= 1 ππ
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π1
π2
Massless wheel for the moment (π = 0)
β’ ΞπΎπΈ =1
2π1π£
2 +1
2π2π£
2
β’ βΞππΈ = ππΈπ β ππΈπ = π1πβ βπ2πβ
Conservation of Energy ΞπΎπΈ = βΞππΈ
1
2π1π£
2 +1
2π2π£
2 =π1πβ βπ2πβ
π = 0
π1 = 2 ππ
π2
= 1 ππ
π1
π2
Massless wheel for the moment (π = 6 ππ)
β’ βΞππΈ = ππΈπ β ππΈπ = π1πβ βπ2πβ
β’ ΞπΎπΈ =1
2π1π£
2 +1
2π2π£
2 + ΞπΎπΈπππ‘ππ‘πππ
Conservation of Energy ΞπΎπΈ = βΞππΈπ = 6 πππ = 0.5 π
π1 = 2 ππ
π2
= 1 ππ
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What is the moment of inertial of this solid cylinder (mass π, radius π , length πΏ)?
πΌ = ππ 2
πΌ = ?
Replace the pulley in the previous problem with a solid disk. What is the moment of inertial πΌ of a solid disk (mass π, radius π ) ?
π1
π2
π1
π2
Previously πΌ =ππ 2
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Overview: Rotational Kinetic Energy
β’ Use conservation of energy to solve rotational motion problems.β’ This requires knowing the Kinetic Energy in rotational motion.β’ Moment of Inertia.β’ Moment of Inertia depends on the axis of rotation.
β’ Moment of Inertial for rigid bodies.β’ Moment of inertia for a rodβ’ Moment of inertial for a rectangleβ’ Moment of inertia for a cylinderβ’ Moment of inertia for a sphere
β’ Example Problems along the way
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1. Torque β Concepts (magnitudes only)
We need a torque to start rotating the objectLinear Angular
KE 1
2ππ£2
1
2πΌπ2
displacement π₯ π
velocity π£ π
acceleration π πΌ
Inertia π πΌ
Force Torque
πΉ π
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Definition of torque π
ππ
ππ
Same force Τ¦πΉ applied at two different locations yield different torques
Easier to rotate the nut when the force is applied at ππ than at ππ
Applied force here is perpendicular to ππ and ππ.
Magnitudes only:
Definition of torque π
Τ¦π
πΤ¦πΉ
The radial component πΉ cos π does not rotate the wrench/nut. Only the tangential component of the force πΉ sin π does rotate it.
πΉ sin π
πΉ cos π
When Τ¦πΉ is not perpendicular to Τ¦π
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Definition of torque π
Τ¦π
πΤ¦πΉ
π
Keep the full magnitude of
vector Τ¦πΉ, but take the perpendicular component (πβ₯) of the displacement vector Τ¦π.
We need a torque to start rotating the objectLinear Angular
KE 1
2ππ£2
1
2πΌπ2
π₯ π
π£ π
π πΌ
π πΌ
Force Torque
πΉ
Newtonβs Law
πΉ = ππ
Τ¦π
π Τ¦πΉ
πΉβ₯ = πΉ sin π
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We need a torque to start rotating the object
1. Tangential acceleration πβ₯
πβ₯ = ππΌ
πΌ is the angular accel.
2. πΉβ₯ = ππβ₯
Starting with π = ππΉβ₯ which is correct?
A. π = ππΌ
B. π = πππΌ
C. π = ππ2πΌ
Τ¦π
πΤ¦πΉ
πΉβ₯ = πΉ sin π
πβ₯
πΆ
π
We need a torque to start rotating the objectLinear Angular
KE 1
2ππ£2
1
2πΌπ2
π₯ π
π£ π
π πΌ
π πΌ
Force Torque
πΉ
Newtonβs Law
πΉ = ππ
Τ¦π
π Τ¦πΉ
πΉβ₯ = πΉ sin π
πΆ
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2. Torque β Vector Math
From π = πΌπΌ to Τ¦π = πΌ Τ¦πΌ
We need a torque to start rotating the objectLinear Angular
KE 1
2ππ£2
1
2πΌπ2
π₯ π
π£ π
π πΌ
π πΌ
Force Torque
πΉ
Newtonβs Law
πΉ = ππ
Τ¦π
π Τ¦πΉ
πΉβ₯ = πΉ sin π
πΆ
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π
π₯
π¦
π§
β’ Consider 1D rotation
β’ Displacement π
β’ Ang. Velocity Ο = Ξπ/Ξπ‘
β’ Ang. Accel. πΌ = Ξπ/Ξπ‘
β’ object rotates about
Rotational Motion
β’ Consider 1D rotation
β’ Displacement π
β’ Ang. Velocity Ο = Ξπ/Ξπ‘
β’ Ang. Accel. πΌ =Ξπ
Ξπ‘
β’ object rotates about +π§ axis
β’ They are all in the direction of the axis of rotation: in this example,
β’ Τ¦π = ππ
β’ π = ππ
β’ Τ¦πΌ = πΌπ
Rotational Motion
π
π₯
π¦
π§
Vector Τ¦π = πΌ Τ¦πΌMagnitude: π = πΌπΌ
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We need a torque to start rotating the objectLinear Angular
KE 1
2ππ£2
1
2πΌπ2
π₯ π
π£ π
π πΌ
π πΌ
Force Torque
πΉ π = ππΉ sinπ
Newtonβs Law
Τ¦πΉ = π Τ¦π
Τ¦π
π Τ¦πΉ
πΉβ₯ = πΉ sin π
πΆ
We need a torque to start rotating the object
Τ¦ππ Τ¦πΉ
πΉβ₯ = πΉ sin π
Recall
Axis of rotation:+π
Torque, angular accel, velocity, displacement are all in the + π direction.
π = ππ π¬π’π§π½ can be rewritten as
π₯
π¦
π§π = ππΉβ₯ = πππ¬π’π§π½
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