Transcript Lecture 11.1

Welcome back to Physics 215
Today’s agenda:
• Moment of Inertia
• Rotational Dynamics
• Angular Momentum
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Lecture 11-1
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Current homework assignment
• HW8:
– Knight Textbook Ch.12: 8, 26, 54, 58, 60, 62, 64
– Due Monday, Nov. 5th in recitation
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Discussion
Dw/Dt) S miri2 = tnet
a - angular
acceleration
Moment of inertia, I
Ia = tnet
compare this with Newton’s 2nd law
Ma = F
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Moment of Inertia
N
I = m r + m2 r2 + … + m N rN = å mi ri
2
11
2
2
i=1
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* I must be defined with
respect to a particular axis
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Moment of Inertia of
Continuous Body
Dm a 0
å
N
Þ
I = å mi ri
i=1
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ò
r
Dm
Þ I = ò r dm
2
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Tabulated Results for Moments of Inertia of
some rigid, uniform objects
Physics 215 – Fall 2014
Lecture 11-1
(from p.299 of University Physics, Young & Freedman)
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Parallel-Axis Theorem
D
I = ICM + M D
CM
2
D
*Smallest I will always be along axis passing through CM
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Practical Comments on Calculation of
Moment of Inertia for Complex Object
1.
To find I for a complex object, split it into simple
geometrical shapes that can be found in Table 9.2
2.
Use Table 9.2 to get ICM for each part about the axis
parallel to the axis of rotation and going through the
center-of-mass
3.
If needed use parallel-axis theorem to get I for each
part about the axis of rotation
4.
Add up moments of inertia of all parts
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Beam resting on pivot
N
r
CM of beam
r
rm
x
M=?
Vertical equilibrium?
Mb = 2m
SF =
Rotational equilibrium?
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m
St =
M=
N=
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Suppose M replaced by M/2 ?
• vertical equilibrium?
• rotational dynamics?
• net torque?
SF =
St =
• which way rotates?
• initial angular acceleration?
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Moment of Inertia?
I = Smiri2
* depends on pivot position!
I=
* Hence a = tI =
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Rotational Kinetic Energy
K = Si(12mivi2 = (1/2)w2Simiri2
• Hence
K = (1/2)Iw2
• This is the energy that a rigid body
possesses by virtue of rotation
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Spinning a cylinder
2R
F
Cable wrapped around
cylinder. Pull off with
constant force F. Suppose
unwind a distance d of cable
• What is final angular speed of cylinder?
• Use work-KE theorem
W = Fd = Kf = (1/2)Iw2
• Mom. of inertia of cyl.? -- from table: (1/2)mR2
– from table: (1/2)mR2
w = [2Fd/(mR2/2)]1/2 = [4Fd/(mR2)]1/2
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cylinder+cable problem -constant acceleration method
N
F
extended free body diagram
* no torque due to N or FW
* why direction of N ?
FW
radius R
* torque due to t = FR
* hence a = FR/[(1/2)MR2]
= 2F/(MR)
D = (1/2)at2 = d/R; t = [(MR/F)(d/R)]1/2
 w = at = [4Fd/(MR2)] 1/2
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Angular Momentum
• can define rotational analog of linear
momentum called angular momentum
• in absence of external torque it will be
conserved in time
• True even in situations where Newton’s
laws fail ….
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Definition of Angular Momentum
* Back to slide on rotational dynamics:
miri2Dw/Dt = ti
* Rewrite, using li =
Dli/Dt = ti
miri2w

w
pivot
ri
Fi
mi
* Summing over all particles in body:
DL/Dt = text
L = angular momentum = Iw
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An ice skater spins about a vertical axis through
her body with her arms held out. As she draws
her arms in, her angular velocity
•
•
•
•
1. increases
2. decreases
3. remains the same
4. need more information
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Angular Momentum 1.
r

p
O
Point particle:
|L| = |r||p|sin() = m|r||v| sin()
vector form  L = r x p
– direction of L given by right hand rule
(into paper here)
L = mvr if v is at 900 to r for single particle
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Angular Momentum 2.
w
o
rigid body:
* |L| = I w fixed axis of rotation)
* direction – along axis – into paper here
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Rotational Dynamics
t = Ia
DL/Dt = t
 These are equivalent statements
 If no net external torque: t = 0 
* L is constant in time
* Conservation of Angular Momentum
* Internal forces/torques do not contribute
to external torque.
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Bicycle wheel demo
• Spin wheel, then step onto platform
• Apply force to tilt axle of wheel
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Linear and rotational motion
• Force
• Torque
• Acceleration
• Angular acceleration
Fnet = å F = ma
• Momentum
• Kinetic energy
K = mv
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• Angular momentum
L = Iw
p = mv
1
2
t net = åt = Ia
• Kinetic energy
2
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A hammer is held horizontally and then
released. Which way will it fall?
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General motion of extended objects
• Net force  acceleration of CM
• Net torque about CM  angular acceleration
(rotation) about CM
• Resultant motion is superposition of these
two motions
• Total kinetic energy K = KCM + Krot
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Three identical rectangular blocks are at rest on a
flat, frictionless table. The same force is exerted
on each of the three blocks for a very short time
interval. The force is exerted at a different point
on each block, as shown.
After the force has stopped acting on each block,
which block will spin the fastest?
1.
2.
3.
4.
A.
B.
C.
A and C.
Physics 215 – Fall 2014
Top-view diagram
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Three identical rectangular blocks are at rest on a
flat, frictionless table. The same force is exerted
on each of the three blocks for a very short time
interval. The force is exerted at a different point on
each block, as shown.
After each force has stopped acting, which block’s
center of mass will have the greatest speed?
1.
2.
3.
4.
A.
B.
C.
Top-view diagram
A, B, and C have the same C.O.M. speed.
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Reading assignment
• Rotational dynamics, rolling
• Remainder of Ch. 12 in textbook
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