Download Chapter 10: Dynamics of rotational motion

Chapter 10: Dynamics of rotational motion
•  Torque: Is it a force?
•  torques  rotational motion
(just as forces  linear accelerations)
•  combination of translation and rotation: rolling objects
•  Calculation of work done by a torque
•  Angular momentum conservation
•  rotational dynamics and angular momentum: they are related
The race of objects with different moments
•  This problem is a “classic” and appears on most
professors’ exams on this material. (example 10.5)
•  Which objects reaches the bottom first?
Torque: How to rotate an object more effectively
•  A force applied at a
right angle to a
lever will generate a
torque.
•  The distance from
the pivot to the
point of force
application will be
linearly
proportional to the
torque produced.
Lines of force and calculations of torques
•  The directions of
torques: use the righthand rule (RHR).
  
τ = r × F = rF sin φ = Ftan r
There are many ways to understand
the above formula; What would you
choose? (Level arm)
φ: angle between the two vectors r and F
How to turn an object more effectively?
Larger force!
Longer distance from the rotating axis
Apply that force perpendicular to the radius!
Calculate an applied torque: Loosening a pipe fitting
•  To be a good plumber (they make more money than
you do!), what do you need?
•  Strength; Knowledge; and good tools.
A plumber pushes straight
down on the end of a long
wrench as shown. What is the
magnitude of the torque he
applies about the pipe at lower
right?
A. (0.80 m)(900 N)sin 19°
B. (0.80 m)(900 N)cos 19°
C. (0.80 m)(900 N)tan 19°
D. none of the above
Calculate an applied torque: Loosening a pipe fitting
•  To be a good plumber (they make more money than
you do!), what do you need?
•  Strength; Knowledge; and good tools.
A plumber pushes straight
down on the end of a long
wrench as shown. What is the
magnitude of the torque he
applies about the pipe at lower
right?
A. (0.80 m)(900 N)sin 19°
B. (0.80 m)(900 N)cos 19°
C. (0.80 m)(900 N)tan 19°
D. none of the above
Q10.1
The four forces shown all have the
same magnitude: F1 = F2 = F3 = F4.
Which force produces the greatest
torque about the point O (marked by
the blue dot)?
F1
F3
O
F2
F4
A. F1
B. F2
C. F3
D. F4
E. not enough information given to decide
Q10.2
Which of the four forces shown here
produces a torque about O that is
directed out of the plane of the
drawing?
A. F1
B. F2
C. F3
D. F4
E. more than one of these
F1
F3
O
F2
F4
Q10.4
A force
acts an object at a point
located at the position
What is the torque that this force applies about the origin?
A. zero
B. C. D. E. Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley
Newton’s laws still valid: Τ = Iα is just like F = ma
•  Refer to Figures 10.6 and 10.8 while considering
the application of Figure 10.7.
What will the newton’s laws about motions
become in rotational dynamics?
Unwinding cable: Example 10.3
•  Example 9.9 used energy
methods.
What is speed of the block when it
hits the floor?
vy
1
1 1
mgh = mv y2 + ( MR 2 )ω 2 ,ω =
2
2 2
R
v y2
2gh
g
⇒ vy =
⇒ ay =
=
M
2h 1+ M
1+
2m
2m
•  Use rotational dynamics:
∑τ = RT = Iα = (0.5MR )α
∑ F = mg + (−T) = ma
2
z
€
y
z
y
•  Block and pulley’s
accelerations are related
v y = Rω ⇒
ay = Rα z
€
dv y
dω
=R
⇒
dt
dt
z
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A rigid body in motion about a moving axis
•  What is the total kinetic energy of a rigid body rotating about a moving axis
1
1
2
K = Mv cm + Icmω 2
2
2
Translational
rotational
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How to derive this:
1


K i = mi (v cm + v i' ) 2
2
1
1


2
K = ∑ K i = ∑ ( miv cm
) +∑ (miv cm • v i' ) +∑ ( miv 'i2 )
2
2
Translational
0
rotational


(∑ miv i' ) • v cm ) = 0
Rolling with and without slipping
•  Rolling with slipping may be calculated. Slipping
makes things worse (for driving and calculations).
•  What does “without slipping mean”?
The race of objects with different moments
•  This problem is a “classic” and appears on most
professors’ exams on this material. (example 10.5)
•  Which objects reaches the bottom first?
Icm = cMR 2
1
1
2
Mgh = Mv cm
+ Icmω 2
2
2
1
1
v
2
= Mv cm
+ (cMR 2 )( cm ) 2
2
2
R
⇒
v cm =
€
2gh
1+ c
What does c=0 mean?
Compact objects wins the race!
Consider the acceleration of a rolling sphere
•  Refer to Example 10.7 and Figure 10.19.
Consider the acceleration of a rolling sphere II
•  Figure 10.20 will introduce an effect of friction.
Work and power in rotational motion
•  Figure 10.21 is a game I loved as a child.
•  Examples 10.8 and 10.9 allow us to follow calculations
of power and torque.
The race of objects with different moments
•  We learned that the most compact object rolls down fastest:
smallest moment of inertia: “I”
•  What caused these objects to roll in the first place?
Q10.6
A lightweight string is wrapped several
times around the rim of a small hoop. If
the free end of the string is held in place
and the hoop is released from rest, the
string unwinds and the hoop descends.
How does the tension in the string (T)
compare to the weight of the hoop (w)?
A. T = w B. T > w C. T < w D. not enough information given to decide
Copyright © 2008 Pearson Education Inc., publishing as Pearson Addison-Wesley
Consider the acceleration of a rolling sphere
•  Friction is the reason the ball rolls without slipping
•  Friction also is the reason the ball rotates!
•  Does the friction do work?
∑F
x
= Mgsin β + (− f ) = Macm −x
2
=
Rf
=
I
α
=
(
MR 2 )α z
z
z
5
acm −x = Rα z
∑τ
⇒
acm −x = ?g
f = ? Mg
µs−min = ?
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Work and angular momentum in rotational motion
•  We have already learnt “work” in linear motion.
•  Just change the symbols!
W = τ (θ 2 − θ1 )
W = F(x 2 − x1 )
W =
∫
x2
x1
1
1
F(x)dx = mv 22 − mv12
2
2
p = mv


dp
Ftot =
dt
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€
θ2
1
1
W = ∫θ τ (θ )dθ = Iω 22 − Iω12
1
2
2

L = Iω

New concepts

dL
τ tot =
dt
Spinning professor: Not me (Example 10.11)
I1prof = 3.0kg • m 2
ω1 = 0.5rev /s
I2prof = 2.2kg • m 2
r2dumbbell = 0.20m
ω 2 = ?rev /s
r1dumbbell = 1.0m
mdumbbell = 5.0kg
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total
total
Angular momentum conservation
I1 ω1 = I2 ω 2
(Is the kinetic energy conserved? Decreased? Or increase?
Remember the pair skaters pushing each other?:
€
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m1 v1 = m2v 2
How does a car’s clutch works?
•  Engine flywheel (A)
will run against the
clutch plate until they
have the same angular
velocities
•  Different clutch plates
have different
moments of inertia
•  Is the total kinetic
energy conserved?
Assume Ftot-external=0
Gyroscopic precession
•  The precession of a
gyroscope shows up
in many “common”
situations.
If we release the flywheel axis
without the wheel spinning,
the system will drop.
What will happen if the flywheel
was spinning fast at the beginning?
Consider a similar situation:
Gyroscope: A device to
measure/maintain orientation
Precession:Change of the rotation axis
v0=0
v0 not 0
F not 0
F not 0
A rotating flywheel


dL  
τ tot =
=r×F
dt
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•  The shifting self-rotation axis
of the earth is a special case of
precession! (The reason?)
Since the torque τ is always
perpendicular to the angular
Momentum L,
|L| remains the same
The only thing changing is the
L direction, hence the rotation axis
Summary
  
τ = r × F = rF sin φ = Ftan r
• Torque is not a force;
Cross products
• Torque changes the angular momentum/angular velocity

dω
 dL
τ=
= Iα = I
dt
dt
Newton’s 2nd law
• Work-energy theorem in rotational dynamics
W =
• Angular momentum
∫
θ2
θ1
1
1
τdθ = Iω 22 − Iω12
2
2
   

L = r × p = r × mv


L = Iω
Particle
Rigid body
• Angular momentum is conserved when total external torque is 0
• Total kinetic energy of a rotating rigid body
K=
1
1
2
Mv cm
+ Icmω 2
2
2
• Code word rolling without slipping: zero velocity at contact point
• Precession: changing rotation axis
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