April 17, Week 13 Today: Chapter 10, Angular Momentum Homework - - PowerPoint PPT Presentation

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april 17 week 13

April 17, Week 13 Today: Chapter 10, Angular Momentum Homework - - PowerPoint PPT Presentation

Dec 29, 2023 217 likes •757 views

April 17, Week 13 Today: Chapter 10, Angular Momentum Homework Assignment #10 - Due April 19. Mastering Physics: 7 problems from chapter 9 Written Question: 10.86 On Friday, we will begin chapter 13. From now on, Thursday office hours will be

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Torque April 17, 2013 - p. 1/8

April 17, Week 13

Today: Chapter 10, Angular Momentum Homework Assignment #10 - Due April 19.

Mastering Physics: 7 problems from chapter 9 Written Question: 10.86

On Friday, we will begin chapter 13. From now on, Thursday office hours will be held in room 109 of Regener Hall

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Torque April 17, 2013 - p. 2/8

Newton’s Second Law for Rotation

Newton’s Second law can be modified for rotation.

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Torque April 17, 2013 - p. 2/8

Newton’s Second Law for Rotation

Newton’s Second law can be modified for rotation. Original Version: − → F = M− → a

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Torque April 17, 2013 - p. 2/8

Newton’s Second Law for Rotation

Newton’s Second law can be modified for rotation. Original Version: − → F = M− → a Rotational Version: − → τ =?

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Newton’s Second Law for Rotation

Newton’s Second law can be modified for rotation. Original Version: − → F = M− → a Rotational Version: − → τ =? − → τ

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SLIDE 6

Torque April 17, 2013 - p. 2/8

Newton’s Second Law for Rotation

Newton’s Second law can be modified for rotation. Original Version: − → F = M− → a Rotational Version: − → τ =? − → τ − → α

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Newton’s Second Law for Rotation

Newton’s Second law can be modified for rotation. Original Version: − → F = M− → a Rotational Version: − → τ =? − → τ − → α I

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Torque April 17, 2013 - p. 2/8

Newton’s Second Law for Rotation

Newton’s Second law can be modified for rotation. Original Version: − → F = M− → a Rotational Version: − → τ =? − → τ − → α I Newton’s Second Law for Rotation: − → τ = I− → α

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Torque April 17, 2013 - p. 2/8

Newton’s Second Law for Rotation

Newton’s Second law can be modified for rotation. Original Version: − → F = M− → a Rotational Version: − → τ =? − → τ − → α I Newton’s Second Law for Rotation: − → τ = I− → α

(Only true for spinning motion with the origin of your coordinates at the axis of rotation.)

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Rotational Dynamics Exercise

A solid cylinder with moment of inertia 25 kg · m2 and radius 0.5 m has a rope wrapped around it. The rope is pulled and the cylinder spins about its center with angular acceleration 1 rev/s2. What is the tension in the rope? Ignore friction. − → T r

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Torque April 17, 2013 - p. 3/8

Rotational Dynamics Exercise

A solid cylinder with moment of inertia 25 kg · m2 and radius 0.5 m has a rope wrapped around it. The rope is pulled and the cylinder spins about its center with angular acceleration 1 rev/s2. What is the tension in the rope? Ignore friction. − → T r (a) T = 25 N

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Torque April 17, 2013 - p. 3/8

Rotational Dynamics Exercise

A solid cylinder with moment of inertia 25 kg · m2 and radius 0.5 m has a rope wrapped around it. The rope is pulled and the cylinder spins about its center with angular acceleration 1 rev/s2. What is the tension in the rope? Ignore friction. − → T r (a) T = 25 N (b) T = 50 N

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Torque April 17, 2013 - p. 3/8

Rotational Dynamics Exercise

A solid cylinder with moment of inertia 25 kg · m2 and radius 0.5 m has a rope wrapped around it. The rope is pulled and the cylinder spins about its center with angular acceleration 1 rev/s2. What is the tension in the rope? Ignore friction. − → T r (a) T = 25 N (b) T = 50 N (c) T = 25π N = 78.5 N

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Torque April 17, 2013 - p. 3/8

Rotational Dynamics Exercise

A solid cylinder with moment of inertia 25 kg · m2 and radius 0.5 m has a rope wrapped around it. The rope is pulled and the cylinder spins about its center with angular acceleration 1 rev/s2. What is the tension in the rope? Ignore friction. − → T r (a) T = 25 N (b) T = 50 N (c) T = 25π N = 78.5 N (d) T = 50π N = 157 N

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Torque April 17, 2013 - p. 3/8

Rotational Dynamics Exercise

A solid cylinder with moment of inertia 25 kg · m2 and radius 0.5 m has a rope wrapped around it. The rope is pulled and the cylinder spins about its center with angular acceleration 1 rev/s2. What is the tension in the rope? Ignore friction. − → T r (a) T = 25 N (b) T = 50 N (c) T = 25π N = 78.5 N (d) T = 50π N = 157 N (e) T = 100π N = 314 N

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Torque April 17, 2013 - p. 3/8

Rotational Dynamics Exercise

A solid cylinder with moment of inertia 25 kg · m2 and radius 0.5 m has a rope wrapped around it. The rope is pulled and the cylinder spins about its center with angular acceleration 1 rev/s2. What is the tension in the rope? Ignore friction. − → T r (a) T = 25 N (b) T = 50 N (c) T = 25π N = 78.5 N (d) T = 50π N = 157 N (e) T = 100π N = 314 N Tension only force exerting torque ⇒ τT = Iα

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Torque April 17, 2013 - p. 3/8

Rotational Dynamics Exercise

A solid cylinder with moment of inertia 25 kg · m2 and radius 0.5 m has a rope wrapped around it. The rope is pulled and the cylinder spins about its center with angular acceleration 1 rev/s2. What is the tension in the rope? Ignore friction. − → T r (a) T = 25 N (b) T = 50 N (c) T = 25π N = 78.5 N (d) T = 50π N = 157 N (e) T = 100π N = 314 N Tension only force exerting torque ⇒ τT = Iα

Iα = (25 kg · m2)(2π rad/s2) = 50π N · m

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Torque April 17, 2013 - p. 3/8

Rotational Dynamics Exercise

A solid cylinder with moment of inertia 25 kg · m2 and radius 0.5 m has a rope wrapped around it. The rope is pulled and the cylinder spins about its center with angular acceleration 1 rev/s2. What is the tension in the rope? Ignore friction. r − → T (a) T = 25 N (b) T = 50 N (c) T = 25π N = 78.5 N (d) T = 50π N = 157 N (e) T = 100π N = 314 N Tension only force exerting torque ⇒ τT = Iα

Iα = (25 kg · m2)(2π rad/s2) = 50π N · m − → T at 90◦ to − → r ⇒ τT = rT

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Torque April 17, 2013 - p. 3/8

Rotational Dynamics Exercise

A solid cylinder with moment of inertia 25 kg · m2 and radius 0.5 m has a rope wrapped around it. The rope is pulled and the cylinder spins about its center with angular acceleration 1 rev/s2. What is the tension in the rope? Ignore friction. r − → T (a) T = 25 N (b) T = 50 N (c) T = 25π N = 78.5 N (d) T = 50π N = 157 N (e) T = 100π N = 314 N Tension only force exerting torque ⇒ τT = Iα

Iα = (25 kg · m2)(2π rad/s2) = 50π N · m − → T at 90◦ to − → r ⇒ τT = rT T = 50π N · m 0.5 m

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Torque April 17, 2013 - p. 3/8

Rotational Dynamics Exercise

A solid cylinder with moment of inertia 25 kg · m2 and radius 0.5 m has a rope wrapped around it. The rope is pulled and the cylinder spins about its center with angular acceleration 1 rev/s2. What is the tension in the rope? Ignore friction. r − → T (a) T = 25 N (b) T = 50 N (c) T = 25π N = 78.5 N (d) T = 50π N = 157 N (e) T = 100π N = 314 N Tension only force exerting torque ⇒ τT = Iα

Iα = (25 kg · m2)(2π rad/s2) = 50π N · m − → T at 90◦ to − → r ⇒ τT = rT T = 50π N · m 0.5 m

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Angular Momentum

Rotating rigid bodies have an angular momentum.

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Torque April 17, 2013 - p. 4/8

Angular Momentum

Rotating rigid bodies have an angular momentum. Linear Momentum, p:

F = ma = m

dv dt

=

d(mv) dt

= dp

dt

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Angular Momentum

Rotating rigid bodies have an angular momentum. Linear Momentum, p:

F = ma = m

dv dt

=

d(mv) dt

= dp

dt In rotation, torque plays the role of force.

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Torque April 17, 2013 - p. 4/8

Angular Momentum

Rotating rigid bodies have an angular momentum. Linear Momentum, p:

F = ma = m

dv dt

=

d(mv) dt

= dp

dt In rotation, torque plays the role of force.

τ = Iα

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Torque April 17, 2013 - p. 4/8

Angular Momentum

Rotating rigid bodies have an angular momentum. Linear Momentum, p:

F = ma = m

dv dt

=

d(mv) dt

= dp

dt In rotation, torque plays the role of force.

τ = Iα = I

dω dt

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SLIDE 26

Torque April 17, 2013 - p. 4/8

Angular Momentum

Rotating rigid bodies have an angular momentum. Linear Momentum, p:

F = ma = m

dv dt

=

d(mv) dt

= dp

dt In rotation, torque plays the role of force.

τ = Iα = I

dω dt

=

d(Iω) dt

= dL

dt

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Torque April 17, 2013 - p. 4/8

Angular Momentum

Rotating rigid bodies have an angular momentum. Linear Momentum, p:

F = ma = m

dv dt

=

d(mv) dt

= dp

dt In rotation, torque plays the role of force.

τ = Iα = I

dω dt

=

d(Iω) dt

= dL

dt Angular Momentum: L = Iω

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Angular Momentum II

Angular Momentum: L = Iω

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Angular Momentum II

Angular Momentum: L = Iω Newton’s Second Law for Rotation: τ = dL dt Angular momentum measures how much torque is needed to change the rotation of an object.

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Angular Momentum II

Angular Momentum: L = Iω Newton’s Second Law for Rotation: τ = dL dt Angular momentum measures how much torque is needed to change the rotation of an object. Unit: kg · m2/s ← No fancy name!

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Angular Momentum II

Angular Momentum: L = Iω Newton’s Second Law for Rotation: τ = dL dt Angular momentum measures how much torque is needed to change the rotation of an object. Unit: kg · m2/s ← No fancy name! Note: For a point particle (an object with a single value of v) going around a circle of radius r, L = mvr . (Comes from I = mr2 for a particle and v = ωr.)

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Conservation of Angular Momentum

In the absence of external torques, the total angular momentum of a system cannot change.

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Conservation of Angular Momentum

In the absence of external torques, the total angular momentum of a system cannot change. A B

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Conservation of Angular Momentum

In the absence of external torques, the total angular momentum of a system cannot change. A B τB τB = Torque on B due to A

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Conservation of Angular Momentum

In the absence of external torques, the total angular momentum of a system cannot change. A B τB τA τB = Torque on B due to A τA = Torque on A due to B

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Conservation of Angular Momentum

In the absence of external torques, the total angular momentum of a system cannot change. A B τB τA τB = Torque on B due to A τA = Torque on A due to B 3rd Law for rotation: τA = −τB

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Conservation of Angular Momentum

In the absence of external torques, the total angular momentum of a system cannot change. A B τB τA τB = Torque on B due to A τA = Torque on A due to B 3rd Law for rotation: τA = −τB τA+τB = 0

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Conservation of Angular Momentum

In the absence of external torques, the total angular momentum of a system cannot change. A B τB τA τB = Torque on B due to A τA = Torque on A due to B 3rd Law for rotation: τA = −τB τA+τB = 0 τA = dLA dt

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Conservation of Angular Momentum

In the absence of external torques, the total angular momentum of a system cannot change. A B τB τA τB = Torque on B due to A τA = Torque on A due to B 3rd Law for rotation: τA = −τB τA+τB = 0 τA = dLA dt τB = dLB dt

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Torque April 17, 2013 - p. 6/8

Conservation of Angular Momentum

In the absence of external torques, the total angular momentum of a system cannot change. A B τB τA τB = Torque on B due to A τA = Torque on A due to B 3rd Law for rotation: τA = −τB τA+τB = 0 τA = dLA dt τB = dLB dt dLA dt + dLB dt = 0

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Conservation of Angular Momentum

In the absence of external torques, the total angular momentum of a system cannot change. A B τB τA τB = Torque on B due to A τA = Torque on A due to B 3rd Law for rotation: τA = −τB τA+τB = 0 τA = dLA dt τB = dLB dt dLA dt + dLB dt = 0 ∆ (LA + LB) = 0

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Single Object Conservation

Conservation of Angular Momentum: L = Iω ⇒ IAi ωAi + IBi ωBi = IAf ωAf + IBf ωBf

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Single Object Conservation

Conservation of Angular Momentum: L = Iω ⇒ IAi ωAi + IBi ωBi = IAf ωAf + IBf ωBf Conservation of angular momentum can occur in a single

bject!

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Torque April 17, 2013 - p. 7/8

Single Object Conservation

Conservation of Angular Momentum: L = Iω ⇒ IAi ωAi + IBi ωBi = IAf ωAf + IBf ωBf Conservation of angular momentum can occur in a single

bject!

A change in shape causes a change in the moment of inertia.

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Single Object Conservation

Conservation of Angular Momentum: L = Iω ⇒ IAi ωAi + IBi ωBi = IAf ωAf + IBf ωBf Conservation of angular momentum can occur in a single

bject!

A change in shape causes a change in the moment of inertia. Iiωi = Ifωf

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Angular Momentum Exercise

An isolated hoop of mass M and radius R is rotating about its center with angular speed 100 RPM. What would the hoop’s angular speed become if its radius suddenly doubled without changing its mass? Hint: The moment of inertia for a hoop is I = MR2. R ω

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Torque April 17, 2013 - p. 8/8

Angular Momentum Exercise

An isolated hoop of mass M and radius R is rotating about its center with angular speed 100 RPM. What would the hoop’s angular speed become if its radius suddenly doubled without changing its mass? Hint: The moment of inertia for a hoop is I = MR2. R ω (a) 25 RPM

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Torque April 17, 2013 - p. 8/8

Angular Momentum Exercise

An isolated hoop of mass M and radius R is rotating about its center with angular speed 100 RPM. What would the hoop’s angular speed become if its radius suddenly doubled without changing its mass? Hint: The moment of inertia for a hoop is I = MR2. R ω (a) 25 RPM (b) 50 RPM

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Torque April 17, 2013 - p. 8/8

Angular Momentum Exercise

An isolated hoop of mass M and radius R is rotating about its center with angular speed 100 RPM. What would the hoop’s angular speed become if its radius suddenly doubled without changing its mass? Hint: The moment of inertia for a hoop is I = MR2. R ω (a) 25 RPM (b) 50 RPM (c) 100 RPM

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Torque April 17, 2013 - p. 8/8

Angular Momentum Exercise

An isolated hoop of mass M and radius R is rotating about its center with angular speed 100 RPM. What would the hoop’s angular speed become if its radius suddenly doubled without changing its mass? Hint: The moment of inertia for a hoop is I = MR2. R ω (a) 25 RPM (b) 50 RPM (c) 100 RPM (d) 200 RPM

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Torque April 17, 2013 - p. 8/8

Angular Momentum Exercise

An isolated hoop of mass M and radius R is rotating about its center with angular speed 100 RPM. What would the hoop’s angular speed become if its radius suddenly doubled without changing its mass? Hint: The moment of inertia for a hoop is I = MR2. R ω (a) 25 RPM (b) 50 RPM (c) 100 RPM (d) 200 RPM (e) 400 RPM

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SLIDE 52

Torque April 17, 2013 - p. 8/8

Angular Momentum Exercise

An isolated hoop of mass M and radius R is rotating about its center with angular speed 100 RPM. What would the hoop’s angular speed become if its radius suddenly doubled without changing its mass? Hint: The moment of inertia for a hoop is I = MR2. R ω (a) 25 RPM (b) 50 RPM (c) 100 RPM (d) 200 RPM (e) 400 RPM

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Torque April 17, 2013 - p. 8/8

Angular Momentum Exercise

An isolated hoop of mass M and radius R is rotating about its center with angular speed 100 RPM. What would the hoop’s angular speed become if its radius suddenly doubled without changing its mass? Hint: The moment of inertia for a hoop is I = MR2. R ω (a) 25 RPM (b) 50 RPM (c) 100 RPM (d) 200 RPM (e) 400 RPM Doubling radius ⇒ If = 4Ii ⇒ ωf = 1 4ωi