Download Lecture 21

WileyPLUS Assignment 3 is available
Chapters 6 & 7
Due Wednesday, November 11 at 11 pm
Next Week
Tutorial and Test 3
Chapters 6 & 7
See web page for tutorial questions
Friday, October 30, 2009
1
Equations for rotational motion
Analogous to the equations for linear motion.
! = !0 + "t
1
! = ("0 + ")t
2
1
! = "0t + #t 2
2
v = v0 + at
ω
v
α
a
θ
x
1
x = (v0 + v)t
2
1
x = v0t + at 2
2
v 2 = v02 + 2ax
!2 = !20 + 2"#
arc length, l = rθ
speed, v = rω
π radians = 180o
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2
8.33/27: A child is running around a stationary merry-go-round at
0.25 rad/s. At the moment he sees his favourite horse, one quarter
turn away, the merry-go-round starts to move in the same direction
as the child is running, accelerating at 0.01 rad/s2.
What is the shortest time it takes the child to catch up with the
horse?
Child
!c = 0.25 rad/s
What are the angular positions of child and horse?
Horse
" = 0.01 rad/s2
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3
8.26/21: A spinning top is made to spin by pulling on a string that is
wrapped around it. The length of the string is L = 64 cm and the
radius of the top is r = 2 cm. The string is pulled so that the angular
acceleration of the top is 12 rad/s2.
What is the final angular velocity of the top?
ω 2 = ω02 + 2αθ
(think v2 = vo2 + 2ax)
What is θ?
The number of turns of string around the top is N = L/(2πr),
so θ= N x 2π = (L/2πr) x 2π = L/r = 64/2 = 32 rad
ω = 28 rad/s
Friday, October 30, 2009
4
8.75/14: A golf ball passes through a windmill,
which has 8 blades and rotates at ω = 1.25 rad/s.
The opening between successive blades is equal
to the width of a blade.
A golf ball is of diameter d = 0.045 m. What must
be the minimum speed of the golf ball so that it
passes through an opening between blades?
There are 8 blades and 8 gaps between blades.
The angular width of each gap is 2π/16 rad.
The golf ball must travel at least a distance equal to its
diameter while the blade moves the width of one gap.
That is, t =
d
2π/16
8dω
=
→v=
v
ω
π
v = 0.143 m/s
d
v
v
t = θ/ω
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Chapter 9: Rotational Dynamics
Sections 1, 2, 3, 6 only
• Action of torques
• The two conditions of equilibrium
• Centre of gravity
• Conservation of angular momentum
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Circular Motion
B
r
!
l
r
l = r!
v = r"
v
" = #$/#t
! = #!/#t
A
! in radians: 1800 = % rad.
" in rad/s
! in rad/s2
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Torques
w = width of door
Torque = (magnitude of force) ! (lever arm)
Lever arm, l = w sin 90◦
Torque, ! = Fl = Fw
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Le
ve
ra
rm
Torques
!
Pivot, axis
of rotation
w = width of door
Lever arm: l = w sin !
Torque, ! = Fl = Fw sin "
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Torques
!=0
w = width of door
Torque, ! = Fl = Fw sin " = 0
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9.3/1: The torque
applied to the bolt is:
! = Fl
l = 0.28 sin 50◦ m
lever arm
l
Line of action of
the force
= 45 N
! = (45 N) × (0.28 sin 50◦ m)
= 9.65 N.m
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F = 720 N
of he
e ft
n
Li n o e
c
o
t i or
ac f
Torque generated by Achilles tendon
about ankle joint:
l = (0.036 m) × cos 55◦ = 0.0207 m
! = Fl = (720 N)(0.0207 m) = 14.9 N.m
Sign convention: torque is positive when tending to rotate
counterclockwise about reference point. So, torque = –14.9 N.m.
Friday, October 30, 2009
12
Two ways to calculate torques
P
d
Lever arm, l
Line of action of
the force
F# = component of F
perp. to line to P
F#
900
$
900
$
F
Point of
application of F
Torque about point P:
1) Torque = (force) & (lever arm): τ = F l = F × d sin θ
τ = F⊥ d = F sin θ × d
2) Torque = F# & d :
Choose the method that is the more convenient...
Friday, October 30, 2009
13
9.-/8: One end of a metre stick is pinned to a table, so that the
stick can rotate freely around the surface of the tabletop. Two
forces, both parallel to the tabletop, are applied in such a way that
the net torque is zero. Where along the stick must the 6 N force
be applied?
x
Friday, October 30, 2009
14
9.2/3: The engine applies a torque of τeng = 295 N.m to the wheel of
a car, which does not slip against the road surface because the
static friction force applies a countertorque. The car is travelling at
constant velocity.
What is the static friction force?
As the wheel is turning at constant rate, the
net torque applied to it must be zero.
Therefore, 'eng = Fs & r
Fs = 'eng/r = 295/0.35 = 843 N
O
v
r = 0.35 m
'eng
Fs
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15
The Two Conditions of Equilibrium
Balance is important!
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16
The Two Conditions of Equilibrium
For an object to be in equilibrium:
1) The sum of forces acting on the object must be zero:
!�Fi = �F1 + �F2 + . . . = 0
Then the acceleration is zero by Newton’s first law.
2) Balance: the sum of torques about any point must be zero:
!"i = "1 + "2 + . . . = 0
Then the angular acceleration is zero.
Friday, October 30, 2009
17
Clickers!
Are the three forces shown sufficient
to keep the rod in equilibrium?
A
For the rod to be in equilibrium, the sum of
the forces and of the torques about any point
must be zero.
A) It is not possible to adjust the magnitudes of the forces so that
the net force is zero.
B) It is possible to adjust the magnitudes of the forces so that the
net force is zero, but the net torque cannot be zero.
C) It is possible to adjust the magnitudes of the forces so that the
net force and the net torque are both zero.
Friday, October 30, 2009
18
For the rod to be in equilibrium, the sum of the
forces and of the torques about any point must
be zero.
A
The sum of forces could be made to be zero by
adjusting the magnitudes of the forces.
That is, F1 + F2 + F3 = 0
First condition of equilibrium is OK
Second condition of equilibrium? Calculate torques about A.
F1 and F2 exert no torque about A, but F3 does.
$ the net torque cannot be zero unless F3 = 0, but then the first
condition of equilibrium fails (unless the other forces are zero too!).
$ the rod cannot be in equilibrium if only these forces act.
Friday, October 30, 2009
19
Focus on Concepts, Question 6
Five hockey pucks are sliding across frictionless ice. The drawing
shows a top view of the pucks and the three forces that act on each
one. The forces can have different magnitudes (F, 2F, or 3F), and
can be applied at different points on the puck. Only one of the five
pucks could be in equilibrium. Which one?
A) 1
B) 2
C) 3
D) 4
E) 5
D) 4 The net force is zero and the net torque is zero.
Friday, October 30, 2009
20
Conditions of equilibrium:
Forces acting
on the diving
board
1) Forces acting on the board:
−F1 + F2 −W = 0
2) Torques about the axis:
0 × F1 + 1.4F2 − 3.9W = 0
That is, F2 =
3.9W
= 2.786W
1.4
Friday, October 30, 2009
21
-F1 + F2 - W = 0 (first condition)
F2 = 2.786W
(second condition)
F1 = F2 - W = (2.786 - 1)W = 1.786W
If the weight of the diver is W = 530 N, then
F1 = 947 N,
F2 = 2.786W = 1480 N
Friday, October 30, 2009
22
What weight can the muscle lift?
Maximum tension in
muscle = 1840 N
Friday, October 30, 2009
23
(Wa acting at centre
of gravity of arm)
Forces on arm
from shoulder
joint
Calculate torques about the shoulder joint to
eliminate Sx, Sy from the torque equations.
Friday, October 30, 2009
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= 1840 N
lM = 0.15 sin 13◦ = 0.0337 m
0.15 m
= 31 N
= 0.28 m
= 0.62 m
Torques about shoulder joint (axis):
1840 × 0.0337 − 31 × 0.28 −Wd × 0.62 = 0
Wd = 86 N
Friday, October 30, 2009
25
= 1840 N
= 31 N
= 86 N
Forces:
x : Sx − M cos 13◦ = 0
Sx = 1840 cos 13◦ = 1793 N
y : M sin 13◦ + Sy −Wa −Wd = 0
Sy = −1840 sin 13◦ + 31 + 86 = −297 N
(shoulder joint exerts a downward force on the arm)
Friday, October 30, 2009
26
Forces acting on ladder
50◦
WF = 875 N
WL = 355 N
A
A
Ladder is 8 m long. No friction at wall.
Friday, October 30, 2009
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