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

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 Torque April 17, 2013 - p. 1/8

Newton’s Second Law for Rotation Newton’s Second law can be modified for rotation. 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 Torque April 17, 2013 - p. 2/8

Newton’s Second Law for Rotation Newton’s Second law can be modified for rotation. Rotational Version: � − Original Version: � − → → F = M − → τ =? a Torque April 17, 2013 - p. 2/8

Newton’s Second Law for Rotation Newton’s Second law can be modified for rotation. Rotational Version: � − Original Version: � − → → F = M − → τ =? a − → τ Torque April 17, 2013 - p. 2/8

Newton’s Second Law for Rotation Newton’s Second law can be modified for rotation. Rotational Version: � − Original Version: � − → → F = M − → τ =? a − → − → τ α Torque April 17, 2013 - p. 2/8

Newton’s Second Law for Rotation Newton’s Second law can be modified for rotation. Rotational Version: � − Original Version: � − → → F = M − → τ =? a − → − → I τ α Torque April 17, 2013 - p. 2/8

Newton’s Second Law for Rotation Newton’s Second law can be modified for rotation. Rotational Version: � − Original Version: � − → → F = M − → τ =? a → − − → I τ α Newton’s Second Law for Rotation: � − → τ = I − → α Torque April 17, 2013 - p. 2/8

Newton’s Second Law for Rotation Newton’s Second law can be modified for rotation. Rotational Version: � − Original Version: � − → → F = M − → τ =? a − → − → I τ α 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.) Torque April 17, 2013 - p. 2/8

Rotational Dynamics Exercise A solid cylinder with moment of inertia 25 kg · m 2 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/s 2 . What is the tension in the rope? Ignore friction. r − → T Torque April 17, 2013 - p. 3/8

Rotational Dynamics Exercise A solid cylinder with moment of inertia 25 kg · m 2 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/s 2 . What is the tension in the rope? Ignore friction. (a) T = 25 N r − → T Torque April 17, 2013 - p. 3/8

Rotational Dynamics Exercise A solid cylinder with moment of inertia 25 kg · m 2 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/s 2 . What is the tension in the rope? Ignore friction. (a) T = 25 N r (b) T = 50 N − → T Torque April 17, 2013 - p. 3/8

Rotational Dynamics Exercise A solid cylinder with moment of inertia 25 kg · m 2 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/s 2 . What is the tension in the rope? Ignore friction. (a) T = 25 N r (b) T = 50 N (c) T = 25 π N = 78 . 5 N − → T Torque April 17, 2013 - p. 3/8

Rotational Dynamics Exercise A solid cylinder with moment of inertia 25 kg · m 2 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/s 2 . What is the tension in the rope? Ignore friction. (a) T = 25 N r (b) T = 50 N (c) T = 25 π N = 78 . 5 N − → (d) T = 50 π N = 157 N T Torque April 17, 2013 - p. 3/8

Rotational Dynamics Exercise A solid cylinder with moment of inertia 25 kg · m 2 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/s 2 . What is the tension in the rope? Ignore friction. (a) T = 25 N r (b) T = 50 N (c) T = 25 π N = 78 . 5 N − → (d) T = 50 π N = 157 N T (e) T = 100 π N = 314 N Torque April 17, 2013 - p. 3/8

Rotational Dynamics Exercise A solid cylinder with moment of inertia 25 kg · m 2 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/s 2 . What is the tension in the rope? Ignore friction. (a) T = 25 N Tension only force exerting torque ⇒ r τ T = Iα (b) T = 50 N (c) T = 25 π N = 78 . 5 N − → (d) T = 50 π N = 157 N T (e) T = 100 π N = 314 N Torque April 17, 2013 - p. 3/8

Rotational Dynamics Exercise A solid cylinder with moment of inertia 25 kg · m 2 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/s 2 . What is the tension in the rope? Ignore friction. (a) T = 25 N Tension only force exerting torque ⇒ r τ T = Iα (b) T = 50 N Iα = (25 kg · m 2 )(2 π rad/s 2 ) (c) T = 25 π N = 78 . 5 N = 50 π N · m → − (d) T = 50 π N = 157 N T (e) T = 100 π N = 314 N Torque April 17, 2013 - p. 3/8

Rotational Dynamics Exercise A solid cylinder with moment of inertia 25 kg · m 2 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/s 2 . What is the tension in the rope? Ignore friction. (a) T = 25 N Tension only force exerting torque ⇒ r τ T = Iα (b) T = 50 N Iα = (25 kg · m 2 )(2 π rad/s 2 ) − → (c) T = 25 π N = 78 . 5 N T = 50 π N · m → − T at 90 ◦ to − (d) T = 50 π N = 157 N → r ⇒ τ T = rT (e) T = 100 π N = 314 N Torque April 17, 2013 - p. 3/8

Rotational Dynamics Exercise A solid cylinder with moment of inertia 25 kg · m 2 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/s 2 . What is the tension in the rope? Ignore friction. (a) T = 25 N Tension only force exerting torque ⇒ r τ T = Iα (b) T = 50 N Iα = (25 kg · m 2 )(2 π rad/s 2 ) − → (c) T = 25 π N = 78 . 5 N T = 50 π N · m → − T at 90 ◦ to − (d) T = 50 π N = 157 N → r ⇒ τ T = rT T = 50 π N · m (e) T = 100 π N = 314 N 0 . 5 m Torque April 17, 2013 - p. 3/8

Rotational Dynamics Exercise A solid cylinder with moment of inertia 25 kg · m 2 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/s 2 . What is the tension in the rope? Ignore friction. (a) T = 25 N Tension only force exerting torque ⇒ r τ T = Iα (b) T = 50 N Iα = (25 kg · m 2 )(2 π rad/s 2 ) − → (c) T = 25 π N = 78 . 5 N T = 50 π N · m → − T at 90 ◦ to − (d) T = 50 π N = 157 N → r ⇒ τ T = rT T = 50 π N · m (e) T = 100 π N = 314 N 0 . 5 m Torque April 17, 2013 - p. 3/8

Angular Momentum Rotating rigid bodies have an angular momentum. Torque April 17, 2013 - p. 4/8

Angular Momentum Rotating rigid bodies have an angular momentum. Linear Momentum, p : � dv � � d ( mv ) � = dp � F = ma = m = dt dt dt Torque April 17, 2013 - p. 4/8

Angular Momentum Rotating rigid bodies have an angular momentum. Linear Momentum, p : � dv � � d ( mv ) � = dp � F = ma = m = dt dt dt In rotation, torque plays the role of force. Torque April 17, 2013 - p. 4/8

Angular Momentum Rotating rigid bodies have an angular momentum. Linear Momentum, p : � dv � � d ( mv ) � = dp � F = ma = m = dt dt dt In rotation, torque plays the role of force. � τ = Iα Torque April 17, 2013 - p. 4/8

Angular Momentum Rotating rigid bodies have an angular momentum. Linear Momentum, p : � dv � � d ( mv ) � = dp � F = ma = m = dt dt dt In rotation, torque plays the role of force. � dω � � τ = Iα = I dt Torque April 17, 2013 - p. 4/8

Angular Momentum Rotating rigid bodies have an angular momentum. Linear Momentum, p : � dv � � d ( mv ) � = dp � F = ma = m = dt dt dt In rotation, torque plays the role of force. � dω � � d ( Iω ) � = dL � τ = Iα = I = dt dt dt Torque April 17, 2013 - p. 4/8

Angular Momentum Rotating rigid bodies have an angular momentum. Linear Momentum, p : � dv � � d ( mv ) � = dp � F = ma = m = dt dt dt In rotation, torque plays the role of force. � dω � � d ( Iω ) � = dL � τ = Iα = I = dt dt dt Angular Momentum: L = Iω Torque April 17, 2013 - p. 4/8

Angular Momentum II Angular Momentum: L = Iω Torque April 17, 2013 - p. 5/8

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. Torque April 17, 2013 - p. 5/8

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 · m 2 /s ← No fancy name! Torque April 17, 2013 - p. 5/8