Physics 1A, Lecture 15: Torque and Angular Momentum

Physics 1A, Lecture 15: Torque and Angular Momentum Summer Session 1, 2011 Turn in your Homework! The quiz will commence at 9:33 AM. Key QuesFons: (Discuss with neighbors before quiz) 1) What is torque? 2) What is a moment arm? 3) What are the condiFons for translaFonal and rotaFonal equilibrium? 4) What is the version of Newton's second law for rotaFonal moFon? 5) What is angular momentum? Reading Quiz 13-­‐1 •  What is torque? A)  the net force on an object that rotates B)  the resistance of an object to change its rotaFon C)  the product of the mass of an object and its angular acceleraFon D)  the force that provides the centripetal force E)  the tendency of a force to cause an angular acceleraFon Reading Quiz 13-­‐2 •  What is a moment arm? A) The distance between the line of acFon of a force and an object’s center of mass. B) The perpendicular distance between the line of acFon of a force and an object’s moment of inerFa. C) The distance between the line of acFon of a force and an object’s moment of inerFa. D) The perpendicular distance between the line of acFon of a force and an object’s axis of rotaFon E) The distance between the line of acFon of a force and an object’s axis of rotaFon. Reading Quiz 13-­‐3 •  What are the condiFons for both translaFonal and rotaFonal equilibrium? A) net force is zero. B) net force and net torque are zero. C) no translaFonal moFon. D) no rotaFon. E) no translaFonal moFon and no rotaFon. Reading Quiz 13-­‐4 •  What is the version of Newton's second law for rotaFonal moFon? A) Στ = I ω B) Στ = I α C) Στ = m α D) Στ = ma E) Στ = ½ I ω2 Reading Quiz 13-­‐5 •  What is angular momentum? A) L = I ω B) L = I α C) L = m α D) L = ma E) L = ½ I ω2 Announcements •  No reading quiz tomorrow •  No clicker points, but we might have clicker review quesFons. •  Homework due at 4pm in Evan’s office hours –  My office hour at noon, –  Evan’s office hours 2-­‐4 pm, –  Evan’s review session 5-­‐6 pm CalculaFng your grade •  Points already earned: –  0.35 x (Final quiz grade) –  0.2 x (Homework grade) (all of them are out of 3) ≈ 20% –  0.1 x (Reading quiz grade) –  0.05 x (Clicker grade) •  Final Grade cutoffs (that might be lowered): – 
– 
– 
– 
A … above 88% B … above 75% C … above 63% D … above 50% •  Take the cutoff that you want to reach, subtract points already earned, divide by 0.35 … That is what you need to get on the Final to guarantee that grade. Anonymous poll How much of the homework assignment have you been gelng correct? A)  close to 100% B)  close to 75% C)  close to 50% D)  close to 25% E)  close to 0% Anonymous poll How much of the extra homework assignment have you been trying? A)  close to 100% B)  close to 75% C)  close to 50% D)  close to 25% E)  close to 0% Moments of InerFa •  In general: I=
�
mi ri 2
i
•  Changes when you you consider different axes of rotaFon. m
m
r r 2r m
m
•  Pre-­‐calculated for common 3D objectsà Clicker quesFon There are two baseball bats of the same size and weight. Bat A has its mass concentrated in the handle, while Bat B has its mass concentrated near the other end. Which bat will have the larger moment of inerFa? A)  Bat A B)  Bat B C)  they will have the same moment of inerFa D)  not enough informaFon Clicker QuesFon What is the moment of inerFa? m
2m L L/3 A) 5/4 mL2 2m m
B) 9/4 mL2 L/2 C) 9/2 mL2 D) 5/2 mL2 �
E) 3/4 mL2 2
I
=
m
r
i
i
i
RotaFonal KineFc Energy •  TranslaFonal KineFc Energy: 1
2
KE = mv
2
•  RotaFonal KineFc Energy: 1 2
KER = Iω
2
Rolling without slipping •  TranslaFonal velocity is equal to tangenFal velocity! Ihoop = M R2
vT
v
ω=
=
R
R
� v �2
1
1
KER = Ihoop ω 2 = (M R2 )
2
2
R
KEtot = KE + KER = M v 2
Rolling without slipping •  TranslaFonal velocity is equal to tangenFal velocity! Icylinder
1
= M R2
2
vT
v
ω=
=
R
R
1
1
2
KER = Icylinder ω =
2
2
KEtot
�
1
M R2
2
��
3
= KE + KER = M v 2
4
v �2
R
Races If the hoop and the sphere have the same mass and the same radius, which one will have a larger final velocity? Clicker quesFon •  A solid cylinder and a hollow cylinder of the same mass and radius roll down an incline. Which one will reach the bopom first? A) the solid one B) the hollow one C) both will reach the bopom at the same Fme Rolling down a ramp sliding (no fricFon) rolling (cylinder) rolling (hoop) m m
m m
m m
P Eg
P Eg
P Eg KE
P Eg
KE
P Eg
KE KE
R
KE KER
P Eg
KE KER
KE KER
Torque Axis of rotaFon aka pivot point: •  A point that is not moving relaFve to the rotaFon Moment arm: •  A vector that extends from the pivot to the place where the force is being exerted. In order for a force to exert a torque on an object: •  Force must have a component that is perpendicular to the moment arm. �r
F�
�r
F�
�r
F�
F�⊥
Math review: Cross product of two vectors •  Two vectors in any direcFon (place tail to tail) •  Find component of A
�
in the direcFon perpendicular to B
�
and mulFply by |B|
. •  Tells you a measure of how much two vectors lie perpendicular to each other •  Input: Two vectors •  Output: One vector �
�
�� ��
�A × B � = AB sin θ
Max Torque F�⊥
�r
•  F�
θ
F�
τ = −F r
F�
τ =0
�r
�r
F� 90o
Zero Torque �r
θ
τ = +F r sin θ
τ = +F⊥ r
�r
F�
τ =0
F�
nd
Newton’s 2 law for rotaFon �
•  τ = Iα = −ArA +BrB sin(180 − θ)
�rB θ
�rA
�
A
�
B
RotaFonal equilibrium •  When pivot point is not fixed, even if net torque is zero, you can get translaFonal moFon when there is a net force. So you need to check both. �
Fx = 0
�
τ =0
�
Fy = 0
Plasorm clicker The forces are all of equal magnitude. Which of these rods are in equilibrium? A)  rod 1 B)  rod 2 C)  rod 3 D)  none of them Think about it… •  What looks wrong with this photo? Ladder problem •  Find the minimum angle that the ladder can make with the floor so that the ladder doesn’t slip, given m, L, and μs between the ladder and the ground. �
�
Nw
Nf
θ
mg
fs
Fx = N w − f s = 0
Fy = Nf − mg = 0
→�
Nw = µs mg
�
�
L
τ = +mg
cos θ
2
−µs mgL sin θ = 0 �
→ θ = tan
−1
1
2µ
�
Think about it… Which hula hoop should be easier to use? The heavy one or the light one? •  KineFc fricFon (in rotaFon) happens when hula hoop slips against my clothes. This means that the hoop looses rotaFonal energy over Fme. •  I can move my hips to give the hoop small amounts of torque. •  The heavy hoop has more rotaFonal energy, so it will be easier to keep moving. •  The faster the hoop goes à the more centripetal force which pushes it up against my body à the more normal force it feels from my body à the more staFc fricFon is keeping it from sliding down Angular momentum •  RotaFonal analogy with the impulse equaFon: ∆L = τ ∆t
•  Angular momentum L=
�
τ dt =
�
Iαdt →
L = Iω
•  If no external torques on a rotaFonal “collision”: ∆L = 0
Angular momentum conservaFon L = Iω
∆L = 0
If angular momentum is conserved and I
changes à angular velocity must change! Example A merry-­‐go-­‐round (m = 100kg, r = 2.00m) spins with an angular velocity of 2.50(rad/s). A monkey (m = 25.0kg) hanging from a nearby tree, drops straight down onto the merry-­‐go-­‐round at a point 0.500m from the edge. What is the new angular velocity of the merry-­‐go-­‐
round? Announcements •  No reading quiz tomorrow •  No clicker points, but we might have clicker review quesFons. •  Homework due at 4pm in Evan’s office hours –  My office hour at noon, –  Evan’s office hours 2-­‐4 pm, –  Evan’s review session 5-­‐6 pm