Transcript Rotational Inertia & Kinetic Energy

Rotational Inertia
&
Kinetic Energy
Linear & Angular
Linear
Angular
Displacement
x
θ
Velocity
v
Acceleration
a
Inertia
m
I
KE
½ mv2
½I
N2
F = ma
=I
Momentum
P = mv
L=I
2
Rolling Motion
If a round object rolls without slipping, there is a fixed relationship
between the translational and rotational speeds:
Rolling Motion
We may also consider rolling motion to be a combination of pure rotational and
pure translational motion:
Rolling Motion
We may also consider rolling motion at any given instant to be a pure rotation
at rate w about the point of contact of the rolling object.
A Rolling Tire
A car with tires of radius 32 cm drives on a
highway at a speed of 55 mph.
(a) What is the angular speed w of the tires?
(b) What is the linear speed vtop of the top to the
tires?
v (55 mph)(0.447 m/s/mph)
w 
 77 rad/s
r
(0.320 m)
(77 rad/s) / (2 rad/rev)  12.25 rev/s
vtop  2v  110 mph
Rotational Kinetic Energy
• Consider a mass M on the end of a
string being spun around in a circle with
radius r and angular frequency w
– Mass has speed v = w r
– Mass has kinetic energy
M
• K = ½ M v2
• K = ½ M w2 r2
• Rotational Kinetic Energy is energy due
to circular motion of object.
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Rotational Inertia I
• Tells how much “work” is required to get
object spinning. Just like mass tells you
how much “work” is required to get
object moving.
– Ktran = ½ m v2 Linear Motion
– Krot = ½ I w2
Rotational Motion
• I = S miri2
(units kg m2)
• Note! Rotational Inertia (or “Moment of
Inertia”) depends on what you are
spinning about (basically the ri in the
equation).
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Inertia Rods
Two batons have equal mass and
length.
Which will be “easier” to spin?
A) Mass on ends
B) Same
C) Mass in center
I = S m r2 Further mass is from axis of rotation,
greater moment of inertia (harder to spin)
A Dumbbell
Use the definition of moment
of inertia to calculate that of a
dumbbell-shaped object with
two point masses m separated
by a distance of 2r and rotating
about a perpendicular axis through
their center of symmetry.
I   mi ri  m r  m r  2mr
2
2
11
2
2 2
2
Nose to the Grindstone
A grindstone of radius r = 0.610 m is being
used to sharpen an axe.
If the linear speed of the stone is 1.50 m/s
and the stone’s kinetic energy is 13.0 J,
what is its moment of inertia I ?
w  v / r  (1.50 m/s) / (0.610 m)  2.46 rad/s
K  Iw
1
2
2
2K
2(13.0 J)
2
 I 2 
 4.30 kg m
2
w
(2.46 rad/s)
Moment of Inertia of a Hoop
I   mi ri   mi R    mi  R  MR
2
2
2
2
All of the mass of a hoop is at the same distance R from the center
of rotation, so its moment of inertia is the same as that of a point
mass rotated at the same distance.
Moments of Inertia
More Moments
I is Axis Dependent
mR  mR  2mR
2
2
2
m  0  m  2 R   4mR 2
2
2
Rotation Plus Translation
vi  vcm  vi,rel
vbottom  0
vaxel  vcm  w R
vtop  2vcm  2wR
Rolling Objects
v  rw
vcm  Rw
acm  R
s  R
2
K  12 mvcm
 12 Icm w2
Like a Rolling Disk
A 1.20 kg disk with a radius 0f 10.0 cm rolls without slipping. The
linear speed of the disk is v = 1.41 m/s.
(a) Find the translational kinetic energy.
(b) Find the rotational kinetic energy.
(c) Find the total kinetic energy.
Kt  12 mv2  12 (1.20 kg)(1.41 m/s)2  1.19 J
Kr  12 Iw2  12 ( 12 mr 2 )(v / r)2  14 (1.20 kg)(1.41 m/s)2  0.595 J
KS  Kt  Kr  (1.19 J)  (0.595 J)  1.79 J
Preflight: Rolling Race
(Hoop vs Cylinder)
A hoop and a cylinder of equal mass roll
down a ramp with height h. Which has
greatest KE at bottom?
A) Hoop
B) Same
C) Cylinder
20%
50%
30%
Preflight: Rolling Race
(Hoop vs Cylinder)
A hoop and a cylinder of equal mass roll
down a ramp with height h. Which has
greatest speed at the bottom of the
ramp?
A) Hoop
B) Same
C) Cylinder
I = ½ MR 48%
22% I = MR
30%
2
2
Rolling Down an Incline
0
0
Ki  Ui  K f  U f
1 2
mgh  mv  I w
2
1
2
2
Hollow Cylinder : I  mr 2
v
mgh  mv   mr   
r
1
2
2
1
2
gh  12 v 2  12 v 2
gh  v
2
v  gh
2
2
Hollow Cylinder : I  mr 2 ; v  gh
Solid Cylinder: I  12 mr 2 ; v 
Solid Sphere: I  52 mr 2 ; v 
4
3
gh
10
7
gh
Compare Heights
A ball is released from rest on a noslip surface, as shown. After reaching
the lowest point, it begins to rise again
on a frictionless surface.
When the ball reaches its maximum
height on the frictionless surface, it is
higher, lower, or the same height as
its release point?
The ball is not spinning when released, but will be spinning when it
reaches maximum height on the other side, so less of its energy will
be in the form of gravitational potential energy. Therefore, it will
reach a lower height.
Spinning Wheel
A block of mass m is attached to a string
that is wrapped around the circumference of
a wheel of radius R and moment of inertia I,
initially rotating with angular velocity w that
causes the block to rise with speed v . The
wheel rotates freely about its axis and the
string does not slip.
To what height h does the block rise?
Ei  E f
E f  mgh
Ei  12 mv2  12 Iw2  12 mv2  12 I (v / R)2  12 mv2 (1 I / mR2 )
 v2  
I 
h
 1 
2 
2
g
mR



A Bowling Ball
A bowling ball that has an 11 cm radius and
a 7.2 kg mass is rolling without slipping at 2.0
m/s on a horizontal ball return. It continues to
roll without slipping up a hill to a height h
before momentarily coming to rest and then
rolling back down the hill.
Model the bowling ball as a uniform sphere
and calculate h.
Wext  Emech  Etherm  0  Emech  0
1
2
2
U f  K f  Ui  Ki  Mgh  0  0  12 Mvcm

I
w
i
2 cm i
Mgh  Mv
1
2
2
cm i

1
2

2
5
MR
2

2
vcm
7
2
i

Mv
cm i
R 2 10
2
2
7vcm
7(2.0
m/s)
i
h

 0.29 m
2
10 g 10(9.8 m/s )
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