IF YOU’RE NEW THIS WEEK
Read last week’s lecture on http://www.physics.rutgers.edu/~jn511/PHY124H.html
Make sure you have homework (Wiley+) set up ([email protected])
AFTER reading the website/Lecture 1, come see me in office hours if you have further questions
Make sure I have anything needed from ODS.
Gradebook is usual Rutgers credentials (NETID and password)
WHICH REACHES THE BOTTOM FIRST?
A.) Hollow sphere of Mass M
B.) Solid Sphere of Mass M
C.) Not enough info
D.) Solid disk of Mass M
E.) Hoop of Mass M
C.) Not enough info
WHICH REACHES THE BOTTOM FIRST?
Both a hoop and solid cylinder have the same
I =12
MR2
K =12
Iω2 = const If they are all the same mass only!
What would happen if we doubled the mass of one of them?
KINETIC ENERGY OF ROTATION
K =12
Icomω2 +12
Mv2comKinetic energy of rolling wheel
Where does this come from?
K =12
Ipw2Kinetic Energy at point p
New perspective, wheel is rotating around point P
KINETIC ENERGY OF ROTATION
K =12
Icomω2 +12
Mv2comKinetic energy of rolling wheel
K =12
Ipω2 But we want the motion at the com
Parallel-axis theorem Ip = Icom + MR2
K =12
Icomω2 +12
MR2ω2
KINETIC ENERGY OF ROTATION
K =12
Icomω2 +12
Mv2comKinetic energy of rolling wheel
What is missing? vcom = ωr
K =12
Icomω2 +12
MR2ω2
K =12
Icomω2 +12
Mv2com
Iclicker #1. A circular hoop rolls without slipping on a flat horizontal surface. Which one of the following is necessarily true?
a) All points on the rim of the hoop have the same speed.
b) All points on the rim of the hoop have the same linear velocity.
c) Every point on the rim of the wheel has a different linear velocity.
d) All points on the rim of the hoop have acceleration vectors that are tangent to the hoop.
e) All points on the rim of the hoop have acceleration vectors that point toward the center of the hoop.
c) Every point on the rim of the wheel has a different linear velocity.
A bicycle wheel of radius 0.70 m is turning at an angular speed of 6.3 rad/s as it rolls on a horizontal surface without slipping. What is the linear speed of the wheel?
a) 1.4 m/s
b) 28 m/s
c) 0.11 m/s
d) 4.4 m/s
e) 9.1 m/s
d) 4.4 m/s
BICYCLE LINEAR VELOCITY
v = ωr
v = 6.3 rad/s ⋅ 0.70 m
v = 4.4 m/s
Radians can be thought of as an “invisible” unit i.e. if we used
.
rads
⋅ ft =fts
rads
⋅ m =ms
A solid cylinder is rotating about an axis that passes through the center of both ends. The radius of the cylinder is r. At what angular speed must this cylinder rotate to have the same total kinetic energy that it would have if it were moving horizontally with a speed v without rotation?
A.)
B.)
C.)
D.)
E.)
ω
ω =v2
2rω =
vr
2
ω =vr
ω =v2r
ω =v2
r2
ω =vr
2
r
Versus
r
Rotating around an axis Moving without rotating
v
ωI =
12
Mr2
Krot =12
Iω2 Kslide =12
Mv2
ROLLING+FRICTION
No sliding? Only static friction fs
Wheels need friction to accelerate , which leads to a com
But, wheels also tend to slip, then there is kinetic friction
α
P
a com
fs
Smooth rollingfk
This is called smooth rolling.
acom = αR
ROLLING DOWN A HILL+FRICTION
https://www.youtube.com/watch?v=XjMDvptIbP0
Start at 46 seconds. What happens at ~56 seconds?
ROLLING DOWN A RAMP
Newton’s 2nd Law F = Ma
fs − Mg sin θ = Macom,x
θ
R
F N
F gF g cos θ
F g sin θθ
f sx+
y+
ROLLING DOWN A RAMP
Newton’s 2nd Law (angular) τnet = Icomα
Torque τ = r⊥F
Rfs = Icomα
θ
R
F N
F gF g cos θ
F g sin θθ
f sx+
y+
DIRECTION OF ACCELERATION
Is the wheel rolling clockwise or counterclockwise?
With smooth rolling, wheel goes counter clockwise, so α > 0But, we have acom < 0 so acom = − α/R
θ
R
F N
F gF g cos θ
F g sin θθ
f sx+
y+
Demo
ROLLING DOWN RAMP+FRICTION
Rfs = Icomα
Collecting the equations
fs = Icomα/R
fs = − Icomacom,x /R2
acom,x = −g sin θ
1 + Icom/MR2
Recall
fs − Mg sin θ = Macom,x
TORQUE AS A VECTOR (DEMO)
τ = r × F
c = a × bRecall where |c | = |b | |c |sin ϕ
Clockwise rotation points into the page
Counter clockwise points out of the page
⊗
⊙
ANGULAR MOMENTUM L
l = r × p Recall p = m v
l = m ( r × v )Before, we thought of p when doing collisions.
Where does a baseball batter what to hit a ball?
or
ANGULAR MOMENTUM AS A VECTOR
l = m ( r × v ) Magnitude
Direction
| l | = m | r | | v |sin ϕ
Right-hand ruleSmallest angle between
r and p.
RELATING TORQUE AND ANGULAR MOMENTUM
Newton’s second law F net =d pdt
How do you think angular momentum will look like?
τnet =d ldt
WHAT ABOUT MANY PARTICLES?
12
3 L = l1 + l2 + … + ln =n
∑i
l i
n
∑i
τnet,i =n
∑i
d l i
dt=
d Ldt
This would include internal and external torques. But what does Newton’s third imply for internal torques?
For every action there is an equal and opposite reaction, so we only need to consider external torques.
τnet =d Ldt
CONSERVATION OF ANGULAR MOMENTUM For an isolated system (no external forces)
L = const
L i = L f
Demo (chair)
EXAMPLE
1
2
3
Find the moment of Inertia
I = ∑i
mir2i
dd
Give the stick that is rotating around point 1 where all three balls are of mass, m, and the stick’s
mass is negligible.
I = m ⋅ 0 + m ⋅ d2 + m ⋅ (2d)2 = 5md2
EXAMPLE
1
2
3
Find the angular momentum of particle 2.d
dGive the stick that is rotating around point 1
where all three balls are of mass, m, and the stick’s mass is negligible.
L2 = I2ωω
I2 = md2
L2 = md2ω
EXAMPLE
1
2
3
What is the total angular momentum?d
dGive the stick that is rotating around point 1
where all three balls are of mass, m, and the stick’s mass is negligible.
L2 = I2ωω
L = 0 + md2ω + 4md2ω = 5md2ω
EXAMPLE
1
2
3
If the ball 3 falls off, how fast with the stick rotate?d
dGive the stick that is rotating around point 1
where all three balls are of mass, m, and the stick’s mass is negligible.
ω Li = Lf
Li = 5md2ωi Lf = Ifωf
If = md2ωf = 5md2ωi/If
ωf = 5ωi
GYROSCOPE (DROP A SPINNING TOP)
Why doesn’t it fall over?
τ =d Ldt
τ = Mgr sin 90∘ = Mgr
Rapidly spinning, begins to rotate horizontally about its vertical axis
L i ≠ 0 Drop doesn’t change angular momentum
GYROSCOPE CONTINUED
L = IωWe know that
andτ =
d Ldt
d L = τdt
For the rapidly spinning gyroscope, is fixed, so only direction changes.
| L |
d L = τdt τ = Mgr
d L = Mg rdt
GYROSCOPE CONTINUED
d L = Mg rdtTo understand the change in angle, we use Ldϕ = dL
dϕ =dLL
=Mg rdt
L=
Mg rdtIω
Ω ≡dϕdt
=Mg rIω
Precession rate
Ω ⇑ when ω ⇓
A hollow cylinder of mass M and radius R rolls down an inclined plane. A block of mass M slides down an identical inclined plane. Complete the following statement: If both objects are released at the same time,
a) the cylinder will reach the bottom first.
b) the block will reach the bottom first.
c) the block will reach the bottom with the greater kinetic energy.
d) the cylinder will reach the bottom with the greater kinetic energy.
e) both the block and the cylinder will reach the bottom at the same time.
b) the block will reach the bottom first.
WHAT IS THE DIRECTION OF THE EARTH’S ORBITAL ANGULAR MOMENTUM AS IT SPINS ABOUT ITS AXIS?
a) north
b) south
c) east
d) west
e) radially inward
a) north