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April 17, Week 13
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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
Transcript of April 17, Week 13
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 ofRegener Hall
Torque April 17, 2013 - p. 2/8
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.
Original Version:∑−→
F = M−→a Rotational Version:∑
−→τ =?
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
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
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
−→αI
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
−→αI
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.
Original Version:∑−→
F = M−→a Rotational Version:∑
−→τ =?
−→τ
−→αI
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. 3/8
Rotational Dynamics Exercise
A solid cylinder with moment of inertia 25 kg ·m2 and radius0.5m has a rope wrapped around it. The rope is pulled and thecylinder spins about its center with angular acceleration1 rev/s2. What is the tension in the rope? Ignore friction.
−→
T
r
Torque April 17, 2013 - p. 3/8
Rotational Dynamics Exercise
A solid cylinder with moment of inertia 25 kg ·m2 and radius0.5m has a rope wrapped around it. The rope is pulled and thecylinder spins about its center with angular acceleration1 rev/s2. What is the tension in the rope? Ignore friction.
−→
T
r
(a) T = 25N
Torque April 17, 2013 - p. 3/8
Rotational Dynamics Exercise
A solid cylinder with moment of inertia 25 kg ·m2 and radius0.5m has a rope wrapped around it. The rope is pulled and thecylinder spins about its center with angular acceleration1 rev/s2. What is the tension in the rope? Ignore friction.
−→
T
r
(a) T = 25N
(b) T = 50N
Torque April 17, 2013 - p. 3/8
Rotational Dynamics Exercise
A solid cylinder with moment of inertia 25 kg ·m2 and radius0.5m has a rope wrapped around it. The rope is pulled and thecylinder spins about its center with angular acceleration1 rev/s2. What is the tension in the rope? Ignore friction.
−→
T
r
(a) T = 25N
(b) T = 50N
(c) T = 25πN = 78.5N
Torque April 17, 2013 - p. 3/8
Rotational Dynamics Exercise
A solid cylinder with moment of inertia 25 kg ·m2 and radius0.5m has a rope wrapped around it. The rope is pulled and thecylinder spins about its center with angular acceleration1 rev/s2. What is the tension in the rope? Ignore friction.
−→
T
r
(a) T = 25N
(b) T = 50N
(c) T = 25πN = 78.5N
(d) T = 50πN = 157N
Torque April 17, 2013 - p. 3/8
Rotational Dynamics Exercise
A solid cylinder with moment of inertia 25 kg ·m2 and radius0.5m has a rope wrapped around it. The rope is pulled and thecylinder spins about its center with angular acceleration1 rev/s2. What is the tension in the rope? Ignore friction.
−→
T
r
(a) T = 25N
(b) T = 50N
(c) T = 25πN = 78.5N
(d) T = 50πN = 157N
(e) T = 100π N = 314N
Torque April 17, 2013 - p. 3/8
Rotational Dynamics Exercise
A solid cylinder with moment of inertia 25 kg ·m2 and radius0.5m has a rope wrapped around it. The rope is pulled and thecylinder spins about its center with angular acceleration1 rev/s2. What is the tension in the rope? Ignore friction.
−→
T
r
(a) T = 25N
(b) T = 50N
(c) T = 25πN = 78.5N
(d) T = 50πN = 157N
(e) T = 100π N = 314N
Tension only forceexerting torque ⇒
τT = Iα
Torque April 17, 2013 - p. 3/8
Rotational Dynamics Exercise
A solid cylinder with moment of inertia 25 kg ·m2 and radius0.5m has a rope wrapped around it. The rope is pulled and thecylinder spins about its center with angular acceleration1 rev/s2. What is the tension in the rope? Ignore friction.
−→
T
r
(a) T = 25N
(b) T = 50N
(c) T = 25πN = 78.5N
(d) T = 50πN = 157N
(e) T = 100π N = 314N
Tension only forceexerting torque ⇒
τT = Iα
Iα = (25 kg ·m2)(2π rad/s2)
= 50πN ·m
Torque April 17, 2013 - p. 3/8
Rotational Dynamics Exercise
A solid cylinder with moment of inertia 25 kg ·m2 and radius0.5m has a rope wrapped around it. The rope is pulled and thecylinder spins about its center with angular acceleration1 rev/s2. What is the tension in the rope? Ignore friction.
r
−→
T
(a) T = 25N
(b) T = 50N
(c) T = 25πN = 78.5N
(d) T = 50πN = 157N
(e) T = 100π N = 314N
Tension only forceexerting torque ⇒
τT = Iα
Iα = (25 kg ·m2)(2π rad/s2)
= 50πN ·m
−→
T at 90◦ to −→r ⇒ τT = rT
Torque April 17, 2013 - p. 3/8
Rotational Dynamics Exercise
A solid cylinder with moment of inertia 25 kg ·m2 and radius0.5m has a rope wrapped around it. The rope is pulled and thecylinder spins about its center with angular acceleration1 rev/s2. What is the tension in the rope? Ignore friction.
r
−→
T
(a) T = 25N
(b) T = 50N
(c) T = 25πN = 78.5N
(d) T = 50πN = 157N
(e) T = 100π N = 314N
Tension only forceexerting torque ⇒
τT = Iα
Iα = (25 kg ·m2)(2π rad/s2)
= 50πN ·m
−→
T at 90◦ to −→r ⇒ τT = rT
T =50πN ·m
0.5m
Torque April 17, 2013 - p. 3/8
Rotational Dynamics Exercise
A solid cylinder with moment of inertia 25 kg ·m2 and radius0.5m has a rope wrapped around it. The rope is pulled and thecylinder spins about its center with angular acceleration1 rev/s2. What is the tension in the rope? Ignore friction.
r
−→
T
(a) T = 25N
(b) T = 50N
(c) T = 25πN = 78.5N
(d) T = 50πN = 157N
(e) T = 100π N = 314N
Tension only forceexerting torque ⇒
τT = Iα
Iα = (25 kg ·m2)(2π rad/s2)
= 50πN ·m
−→
T at 90◦ to −→r ⇒ τT = rT
T =50πN ·m
0.5m
Torque April 17, 2013 - p. 4/8
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:
∑
F = ma = m
(
dv
dt
)
=
(
d(mv)
dt
)
=dp
dt
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.
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
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
)
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
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
Torque April 17, 2013 - p. 5/8
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 tochange 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 tochange the rotation of an object.
Unit: kg ·m2/s← No fancy name!
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 tochange 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.)
Torque April 17, 2013 - p. 6/8
Conservation of Angular Momentum
In the absence of external torques, the total angularmomentum of a system cannot change.
Torque April 17, 2013 - p. 6/8
Conservation of Angular Momentum
In the absence of external torques, the total angularmomentum of a system cannot change.
A B
Torque April 17, 2013 - p. 6/8
Conservation of Angular Momentum
In the absence of external torques, the total angularmomentum of a system cannot change.
A B
τB
τB = Torque on B due to A
Torque April 17, 2013 - p. 6/8
Conservation of Angular Momentum
In the absence of external torques, the total angularmomentum of a system cannot change.
A B
τB
τA
τB = Torque on B due to A
τA = Torque on A due to B
Torque April 17, 2013 - p. 6/8
Conservation of Angular Momentum
In the absence of external torques, the total angularmomentum 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
Torque April 17, 2013 - p. 6/8
Conservation of Angular Momentum
In the absence of external torques, the total angularmomentum 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
Torque April 17, 2013 - p. 6/8
Conservation of Angular Momentum
In the absence of external torques, the total angularmomentum 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
Torque April 17, 2013 - p. 6/8
Conservation of Angular Momentum
In the absence of external torques, the total angularmomentum 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
Torque April 17, 2013 - p. 6/8
Conservation of Angular Momentum
In the absence of external torques, the total angularmomentum 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
Torque April 17, 2013 - p. 6/8
Conservation of Angular Momentum
In the absence of external torques, the total angularmomentum 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
Torque April 17, 2013 - p. 7/8
Single Object Conservation
Conservation of Angular Momentum:
L = Iω ⇒ IAi ωAi + IBi ωBi = IAf ωAf + IBf ωBf
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 singleobject!
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 singleobject!
A change in shape causes a change in the moment of inertia.
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 singleobject!
A change in shape causes a change in the moment of inertia.
Iiωi = Ifωf
Torque April 17, 2013 - p. 8/8
Angular Momentum Exercise
An isolated hoop of mass M and radius R is rotating about itscenter with angular speed 100RPM . What would the hoop’sangular speed become if its radius suddenly doubled withoutchanging its mass? Hint: The moment of inertia for a hoop isI = MR2.
R
ω
Torque April 17, 2013 - p. 8/8
Angular Momentum Exercise
An isolated hoop of mass M and radius R is rotating about itscenter with angular speed 100RPM . What would the hoop’sangular speed become if its radius suddenly doubled withoutchanging its mass? Hint: The moment of inertia for a hoop isI = MR2.
R
ω(a) 25RPM
Torque April 17, 2013 - p. 8/8
Angular Momentum Exercise
An isolated hoop of mass M and radius R is rotating about itscenter with angular speed 100RPM . What would the hoop’sangular speed become if its radius suddenly doubled withoutchanging its mass? Hint: The moment of inertia for a hoop isI = MR2.
R
ω(a) 25RPM
(b) 50RPM
Torque April 17, 2013 - p. 8/8
Angular Momentum Exercise
An isolated hoop of mass M and radius R is rotating about itscenter with angular speed 100RPM . What would the hoop’sangular speed become if its radius suddenly doubled withoutchanging its mass? Hint: The moment of inertia for a hoop isI = MR2.
R
ω(a) 25RPM
(b) 50RPM
(c) 100RPM
Torque April 17, 2013 - p. 8/8
Angular Momentum Exercise
An isolated hoop of mass M and radius R is rotating about itscenter with angular speed 100RPM . What would the hoop’sangular speed become if its radius suddenly doubled withoutchanging its mass? Hint: The moment of inertia for a hoop isI = MR2.
R
ω(a) 25RPM
(b) 50RPM
(c) 100RPM
(d) 200RPM
Torque April 17, 2013 - p. 8/8
Angular Momentum Exercise
An isolated hoop of mass M and radius R is rotating about itscenter with angular speed 100RPM . What would the hoop’sangular speed become if its radius suddenly doubled withoutchanging its mass? Hint: The moment of inertia for a hoop isI = MR2.
R
ω(a) 25RPM
(b) 50RPM
(c) 100RPM
(d) 200RPM
(e) 400RPM
Torque April 17, 2013 - p. 8/8
Angular Momentum Exercise
An isolated hoop of mass M and radius R is rotating about itscenter with angular speed 100RPM . What would the hoop’sangular speed become if its radius suddenly doubled withoutchanging its mass? Hint: The moment of inertia for a hoop isI = MR2.
R
ω(a) 25RPM
(b) 50RPM
(c) 100RPM
(d) 200RPM
(e) 400RPM
Torque April 17, 2013 - p. 8/8
Angular Momentum Exercise
An isolated hoop of mass M and radius R is rotating about itscenter with angular speed 100RPM . What would the hoop’sangular speed become if its radius suddenly doubled withoutchanging its mass? Hint: The moment of inertia for a hoop isI = MR2.
R
ω(a) 25RPM
(b) 50RPM
(c) 100RPM
(d) 200RPM
(e) 400RPM
Doubling radius⇒ If = 4Ii
⇒ ωf =1
4ωi
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