Transcript Ch 12 Rotation of a Rigid Body

Chapter 12
Rotation of a Rigid Body
Vector (or “cross”) Product
Cross Product is a vector perpendicular to the plane of vectors A and B
A  B  A B sin 
Cross Product of Vectors
A

B
A sin 
B sin 

B
A  B  A B sin 
Position  Force  A B sin 
A
Right hand rule: Curl your right hand around the center of rotation with the
fingers going from the first vector to the second vector and the thumb will
be pointing in the torque direction
A ×B ≠ B× A
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Cross Product Problem
F in d : E = D × C if C = 2 N a n d D = 1 m
Csin(110)
C
110°
D
AP Physics C
E = C D s in θ
E =  2   1  s in  1 1 0  = 1 .8 8 N m
4
Test your Understanding
Which way will it rotate once the
support is removed?
1.
2.
3.
4.
Clockwise.
Counter-clockwise.
Not at all.
Not sure what will happen.
Torque
If the forces are equal, which will open the heavy door more easily?
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Interpretation of torque
Measures tendency of any force to cause rotation
Torque is defined with respect to some origin –
must talk about “torque of force about point X”,
etc.
Torques can cause clockwise (+) or anticlockwise
rotation (-) about pivot point
Torque con’t
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Torque con’t
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Definition of Torque:
  r  F

where r is the vector from the reference point (generally either the
pivot point or the center of mass) to the point of application of the

force F . If r and F are not perpendicular then:
|  | r F sin 
where  is the angle between the vectors

r and
.
Definition of Torque:
|  | r F sin 
Torque Problem
Adrienne (50 kg) and Bo (90kg) are playing on a 100 kg
rigid plank resting on the supports seen below. If
Adrienne stands on the left end, can Bo walk all the way
to the right end with out the plank tipping over? If not,
How far can he get past the support on the right?
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Torque Problem con’t
N1
2m
N2
3m
4m
x
50kg
100kg
90kg
τ N et = 90x - 100( .5)
- 50( 5)= 0
x = 3 .3
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Moments
M1
M2
d1
d2
Suppose we have masses m1 and m2 on the seesaw at
distances d1 and d2, respectively, from the fulcrum, when
does the seesaw balance?
By Archimedes’ Law of the lever, this occurs when
m1d1 + m2d2 = 0
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Moments con’t
M1
M2
x1
x2
If we place a coordinate system so that 0 is at the
fulcrum and if we let xi be the coordinate at which is
placed then:
m1x1 + m2x2 = m1d1 + m2d2 = 0
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Moments con’t
M1
M2
M3
M4
M5
x1
x2
x3
x4
x5
More generally, if we place masses m1, m2, …, mr at
points x1, x2, …. , xr, respectively, then the see saw
balances with the fulcrum at the origin, if and only if
m1x1 + m2x2 + …+ mrxr = 0
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Moments con’t
M1
M2
M3
x1
x2
x3
M4
M5
x5
x4
Now, suppose that we place masses m1, m2, … , mr at
points x1, x2, … xr, respectively, then where should we place
the fulcrum so that the seesaw balances?
The answer is that we place the fulcrum at x-bar where:
m1(x1 - (x-bar) ) + m2(x2 - (x-bar) ) + …+ mr(xr - (x-bar) )= 0
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Moments con’t
M1
M2
M3
x1
x2
x3
M4
M5
x5
x4
m1(x1 - (x-bar) ) + m2(x2 - (x-bar) ) + …+ mr(xr - (x-bar) )
is called the moment about x-bar.
x-bar
Moment is from the Greek word for movement, not time.
If positive, movement is counter-clockwise, negative it is
clockwise.
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Moments con’t
M1
M2
M3
x1
x2
x3
x-bar
M4
M5
x4
x5

0
Suppose that




m 1 x 1  x  m 2 x 2  x  ...  m r x r  x
We want to solve for x

m 1 x 1  m 2 x 2  ...  m r x r  m 1 x  m 2 x  ...  m r x
0
m 1 x 1  m 2 x 2  ...  m r x r   m 1  m 2  ...  m r  x  0
r
x 
AP Physics C
m
i
xi
i 1
r
m
i 1
i
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Center of Mass
M1
M2
M3
x1
x2
x3
x-bar
M4
M5
x4
x5
r
x 
m
i
xi
i 1
r
m

M0
m to t
i
i 1
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Center of Mass con’t
Suppose m1, m2, … , mr are masses located at
points (x1, y1), (x2, y2), … , (xr, yr).
M1
M4
M2
r
The moment about the y-axis is:
y
My 
m
i
xi
m
i
yi
i 1
r
The moment about the x-axis is:
Mx 
i 1
r
M3
x
x 
m
r
i
yi
i 1
r

i 1
mi

My
m to t
y 
m
i
xi
i 1
r

mi

Mx
m to t
i 1
 My
Mx 
,
Center of Mass is

m
m
to
t
to
t


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Center of Mass con’t
Now lets find the center of mass of a thin plate with uniform density, ρ.
b
First we need the mass of the plate: M    A rea     f ( x )  g ( x )d x
a
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Center of Mass con’t
Next we need the moments of the region:
M x 
1
2
y m 
Mx 
1
2
1
2
f ( x )   f ( x ) x  
1
2
g ( x )   g ( x ) x 
b
2
2
   f ( x )  g ( x )  d x
f(x)
a
 M y  x  m  x    f ( x )  g ( x )  x 
g(x)
∆x
b
M y    x f ( x )  g ( x ) d x
a
To find the center of mass we divide by mass:
y 
Mx
m to t

1
2
b
2
2
   f ( x )  g ( x )  d x
a
b


b
a
b
  f ( x )  g ( x )d x
2  f ( x )  g ( x )d x
a
x 
AP Physics C
My
M
f ( x )2  g ( x )2  d x



1
2A

b
a
 f ( x )2  g ( x )2  d x


a

1
A

b
a
x f ( x )  g ( x ) d x
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Center of Mass Problem
Determine the center of mass of the region
bounded by y = 2 sin (2x) and y = 0 on the
interval, [0, π/2]
Given the symmetry of the curve it is obvious
that x-bar is at π/4.
First find the area.
A 


2
0

2 sin (2 x ) d x   co s (2 x ) |0 2  2
Using the table of integrals:
y 
1
2A

b
a
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
1x
1
2

2
 f ( x )2  g ( x )2  d x 


sin
(2
x
)
d
x


sin
4
x







4 0 
4 2
4
0
1

2
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Moment of Inertia
a  r
F
T
T
T
F
 ma
 F r  m ra
2
 mr r  mr 
 I
 ma
AP Physics C


M
r
CR
F
25
Moment of Inertia con’t
M   w ( y )d y
I 
AP Physics C

b
a
y w  y  dy
2
26
Calculating Moment of Inertia
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Calculating Moment of Inertia
dm
dA

M
dm 
A
M
dA
A
d A   2  r  d r an d A   R
dm 
I 
AP Physics C

M
R
2
2  rd r 
r dm 
2
2M
R
2

R
0
2M
R
2
rd r
r dr 
3
2
MR
2
2
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Moment of Inertia con’t
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Parallel Axis Theorem
I ||  I C M  M d
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2
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||-axis Theorem Proof
I 

x dm 
2

 x ' d  d m 

 x '  d m  2d  x ' d m  d
2
I  IC M  M d
AP Physics C
2
 dm
2
31
Rotational Kinetic Energy
K ro t 
1
2
m 1v
K ro t 
1
2
m r  

1
2
2
1
2
1 1

2
1
2
2
2
m 2v
1
2
m r   ...

2
2
  m i ri   
 i

2
2 2
1
2
K ro t 
E m ech 
AP Physics C
 ...
1
2
I
1
2
2
2
I
2
I  M g y cm
2
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Rotational Dynamics
Ft = m a t = m rα
rFt = m r α
2
 = mr α
2
 N et =
mr
2
i i
α
i
 N e t  Iα
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Rotation About a Fixed Axis
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Bucket Problem
A 2.0 kg bucket is attached to a
mass-less string that is wrapped
around a 1.0 kg, 4.0 cm diameter
cylinder, as shown. The cylinder
rotates on an axel through the
center. The bucket is released
from rest 1.0 m above the floor,
How long does it take to reach the
floor
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Bucket Problem con’t
FB = m a y = T - m g
 N et = T R
 net = α I = T R
α=
TR
2
=
2T
.5 M R
MR
2T
2T
a y = -α R = R =MR
M
AP Physics C
I
=
TR
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Bucket Problem Con’t
2T
ay = -
M
may = -
ay = Δy =
AP Physics C
1
2
T =
May
2
2m g
-a y M
2
- mg
a y = -7 .8 4 m s 2
2m + M
ay Δt
2
Δ t = 0 .5 0 s
37
Static Equilibrium
FN e t = 0

AP Physics C
=
0
N et
38
Statics Problem
A 3.0 m ladder leans against a
frictionless wall at an angle of 60°.
What is the minimum value of μs,
that prevents the ladder from
slipping?
F
x
F
y
= n 2 - fs = 0
= n1 - m g = 0
τ N e t = d 1FG - d 2 n 2
=
1
 L c o s 6 0  M g -  L s in 6 0  n
o
2
= .5 ( .7 5 ) M g AP Physics C
o
3
32n
2
2
=0
0
39
Statics Problem
n 2 = fs = μ s n1
fs =
n2 =
n1 = M g
Mg
2 ta n 6 0
o
2 ta n 6 0
o
= μ sn1  μ s M g
μsM g =
μs =
Mg
Mg
2 ta n 6 0
o
1
2 ta n 6 0
o
μ s ≥ 0 .2 9
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Balance and Stability
θc
h
t/2
θc h
t/2
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Rolling Motion
Δ x cm = 2π R = v cm T
AP Physics C
v cm =
2πR
T
= Rv
42
Rolling Motion con’t
ri = rc m + ri,re l
v i = v c m + v i,re l
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Rolling Motion con’t
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Rolling Motion con’t
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Rolling Kinetic Energy
K
ro t,P
=
I ω
1
2 P
IP = I c m + M R
K =
AP Physics C
2
I ω +
1
2 cm
2
1
2
M R ω 
46
2
Great Downhill Race
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Downhill Race con’t
2
I ω +
1
2 cm
1
2
Ic m = c M R
P article c = 0
S phere c = 2 / 5
C ylin d e r c = 1 / 2
H oop c = 1
AP Physics C
1
2
2
Mv
v cm =
= M gh
2
cM R ω +
2
2
cm
2
1
2
 v cm 
M
 = M gh
 R 
2gh
1+ c
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Angular Momentum
p = mv
L = Iω = m rv
L = r × p = r  m v  sin 
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Angular Momentum
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Angular Momentum
L z = m rv t = r × p
dL
dt
=
d
dt
r × p =
dr
dt
p+r
dp
dt
= v × p + r × Fn e t
dL
dt
AP Physics C
=  net
51
Conservation of Angular
Momentum
1
2
2
Lf = Li
m lf ω f =
2
2
m li ω i
1
2
2
i
lf ω f = l ω i
Two equal masses are at the ends of a massless 500 cm long rod. The rod spins at 2.0
Rev/s about an axis through its midpoint. If
the rod lengthens to 160 cm, what is the
angular velocity
2
2
 li 
 50 
ωf    ωi = 
 2 = .2 0
 160 
 lf 
AP Physics C
re v
s
52
Testing Understanding
There is no torque on the buckets so angular momentum is conserved.
Increased mass in buckets increases inertia so angular velocity must
decrease.
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Problem 1
An 18 cm long bicycle crank arm with a pedal at one end
is attached to a 20 cm diameter sprocket, the toothed
disk around which the chain moves. A cyclist riding this
bike increases her pedaling rate from 60 rpm to 90 rpm in
10 s.
a. What is the tangential acceleration of the pedal?
b. Want length of chain passes over the top of the
sprocket during this interval?
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Problem 1 con’t
a. Since at = rα, find α first. With 60 rpm
= 6.28 rad/s and 90 rpm = 9.43 rad/s:
α=
Δω
Δt
=
9 .4 3 ra d - 6 .2 8 ra d / s
10 s
= 0 .3 1 4 ra d / s
2
The angular acceleration of the sprocket
and chain are the same.
a t = rα =  0 .1 8 m   0 .3 1 4 ra d / s
2
 = 0 .0 5 7 m / s
2
b. Since L = r∆θ, find r∆θ.
θ f = θ i + ω iΔ t +
1
2
α Δt 
2
θ f - θ i = Δ θ =  6 .2 8 ra d / s  1 0 s  +
1
 0 .3 1 4 ra d / s  1 0 s 
2
2
2
= 7 8 .5 ra d
The length of chain which has passed over the top of the sprocket is
L e n g th = 0 .1 0 m  7 8 .5 ra d  = 7 .9 m
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Problem 2
A 100 g ball and a 200 g ball are connected by a 30 cm mass-less rigid rod.
The balls rotate about their center of mass at 120 rpm. What is the speed of
the 100g ball?
x cm =
1 0 0 g   0 c m  +  2 0 0 g   3 0 c m 
100 g + 200 g
= 20 cm
 2π rad   m in
v 1 = rω = x cm ω =  0.20 m  120 rev / m in  

 rev   60 s
AP Physics C

 = 2.5 m / s

56
Problem 3
A 300 g ball and a 600 g ball are
connected by a 40 cm mass-less
rigid rod, The structure rotates
about its center of mass at 100 rpm.
What is its rotational kinetic energy?
( 3 0 0 g ) ( 0 c m+)( 6 0 0 g ) ( 4 0 c m )
x cm =
300 g + 600 g
I =( 3 0 0 g ) ( x
2
)+(c 6
00 g) ( 40 cm - x
m
= 2 6 .6 7 c m
)
2
cm
2
2
=( 0 .3 0 0 k g ) ( 0 .2 6 6 7 m )+( 0 .6 0 0 k g ) ( 0 .1 3 3 3
K rot
AP Physics C
1
1
2
= Iω = ( 0 .0 3 2 kg m
2
2
2
m )= 0 .0 3 2 k g m
2
2
 100 × 2π

)
ra d / s  = 1 .7 5 J

60


57
Problem 4
A 25 kg solid door is 220 cm tall, 91 cm wide. What is the door’s moment
of inertia for (a) rotation on its hinges (b) rotation about a vertical axis
inside the door, 15 cm from one edge.
I=
1
3
 2 5 k g   0 .9 1 m 
2
= 6 .9 k g m
2
 0 .9 1 m

d=
- 0 .1 5 m  = 0 .3 0 5 m .
2


2
I = Ic m + M d =
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1
12
 2 5 k g   0 .9 1 m 
2
2
+  2 5 k g   0 .3 0 5 c m  = 4 .1 k g m
2
58
Problem 5
An athlete at the gym holds a 3.0 kg steel ball in his hand. His arm is 70
cm long and has a mass of 4.o kg. What is the magnitude of the torque
about his shoulder if he holds the ball (a) straight out to his side, parallel
to the floor and (b) straight, but 45 degrees below horizontal.
τ = τ b a ll + τ a rm = m b g rb s in 9 0 ° + m a g r a s in 9 0 °
= 3 .0 k g  9 .8 m / s
2
 0 .7 0 m  + 4 .0 k g  9 .8 m / s   0 .3 5 m 
2
= 34 N m
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Problem 5 con’t
τ = τ b a ll + τ a rm = m b g rb s in 4 5 ° + m a g r a s in 4 5 °
= 3 .0 k g  9 .8 m / s
2
  0 .7 0 m   0 .7 0 7  + 4 .0 k g  9 .8 m / s   0 .3 5 m   0 .7 0 7 
2
= 24 N m
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Problem 6
Starting from rest, a 12 cm
diameter compact disk takes
3.0 s to reach its operating
angular velocity of 2000 rpm.
Assume that the angular
acceleration is constant.
The disk’s moment of inertia
is 2.5x10-5 kgm2. (a) How
much torque is applied? (b)
How many revolutions does
it make before reaching full
speed?
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Problem 6 con’t
a. Using the rotational kinematic
equation ω 1 = ω 0 + α ( t1 - t 0 )
 2π 
( 2 0 0 0 rp m )  rad / s = 0 ra d +
 60 
α=
200π
9
ra d / s
2
τ = Iα =( 2 .5 × 1 0
= 1 .7 5 × 1 0
-3
-5
Nm
2  200π
k g m )
ra d / s
9

b. θ 1 = θ 0 + ω 0( t1 - t 0 )+
= 1 0 0 π ra d =
AP Physics C
α  3 .0 s  - 0 s
1
2
100π
2π
2



2
α ( t1 - t0 )= 0 ra d + 0 ra d +
1  200π
2
2 
ra d / s   3 .0 s - 0 s 

2 9

re v o lu tio n s = 5 0 re v
62
Problem 7
The two objects in the figure are
balanced on the pivot. What is the
distance d?
-  FG 1 -  FG  2 + P = 0 N
P =  FG 1 +  FG  2
= 1 .0 k g  9 .8 m / s
2
 + 4 .0 k g  9 .8 m / s 
2
= 49 N
 net = 0 N m
P d - w (1 1 .0 m-) w ( 1 .5
m=) 0 N m
2
4 9 N d - 1 .0 k g  9 .8 m / s
2
 1 .0 m  - 4 .0 k g  9 .8 m / s  1 .5 m  = 0 N
2
d = 1 .4 0 m
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