Physics for Scientists
and Engineers
Chapter 12
Rotation of a Rigid Body
Spring, 2008
Ho Jung Paik
Rigid Body
„
A rigid object is one that is non-deformable
„
„
„
The relative locations of all particles making up the
object remain constant
All real objects are deformable to some extent
The rigid object model is very useful in many
situations where the deformation is negligible
9-Apr-28
Paik
p. 2
Rotation of a Disk
„
„
„
Axis of rotation is through
the center of the disk, O
Every particle on the disk undergoes the same
circular motion about the origin
Polar coordinates are convenient to use to
represent the position of P
„
„
P is located at (r, θ)
With θ measured in radian, the arc length and r are
related by s = θ r
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Paik
p. 3
Angular Speed and Acceleration
„
Angular speed:
„
Angular acceleration:
„
Angular speed is positive (negative) for counterclockwise (clockwise) rotation
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Paik
p. 4
Rotational Kinematics
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Paik
p. 5
Linear and Angular Quantities
„
The linear velocity is always
tangent to the circular path
„
„
Called the tangential velocity
The tangential acceleration is
the derivative of the tangential velocity
9-Apr-28
Paik
p. 6
Center of Mass
„
There is a special point in a system or object,
called the center of mass (CM), that moves as
if all of its mass (M = Σmi) is concentrated at
that point
„
„
The system will move as if an external force were
applied to a single particle of mass M located at
the CM
A general motion of an extended object can
be represented as the sum of a linear motion
of M at CM plus a rotation about the CM
9-Apr-28
Paik
p. 7
Center of Mass, cont
„
The coordinates of the CM are
„
CM can be located by its position vector, rCM
where ri is the position of the i th particle:
9-Apr-28
Paik
p. 8
Example 1: CM of Two Masses
Two masses, m1 = 1.0 kg and m2
= 2.0 kg are located on the x-axis
at x1 = 1.0 m and x2 = 4.0 m,
respectively.
Find the CM of the system.
r
m1 x1 + m2 x2 ˆ
rcm =
i
m1 + m2
„
(1.0 kg)(1.0 m) + (2.0 kg)(4.0 kg) ˆ
=
i
1.0 kg + 2.0 kg
= 3.0 m iˆ
9-Apr-28
Paik
„
The CM is on the xaxis
The CM is closer to
the particle with the
larger mass
p. 9
CM of Extended Object
„
„
An extended object can be
considered a distribution of
small mass elements, Δm
The coordinates of the CM are
1
1
1
xCM =
xdm, yCM =
ydm, zCM =
zdm
∫
∫
∫
M
M
M
„
„
Vector position of the CM:
The CM of any symmetrical object lies on an axis of
symmetry and on any plane of symmetry
9-Apr-28
Paik
p. 10
Notes on Various Densities
„
„
„
Volume mass density, mass per unit volume:
ρ = m /V, dm=ρdV
Surface mass density, mass per unit area of a
sheet of uniform thickness t :
σ = m /A = ρt, dm = σdA
Line mass density, mass per unit length of a
rod of uniform cross-sectional area A:
λ = m /L = ρΑ, dm = λdx
9-Apr-28
Paik
p. 11
Example 2: CM of a Rod
Find the center of mass of a
rod of mass M and length L.
Let λ be the linear density.
Then,
dm = λdx
M = ∫ dm = λL
The location of the CM is on the x - axis :
yCM = zCM = 0,
1
1
λL2
xCM =
xdm =
λxdx =
∫
∫
M
M
2M
= L/2
9-Apr-28
Paik
„
The rod is symmetric
with respect to x = L/2
p. 12
Motion of a System of Particles
„
The velocity of the CM of the system of particles is
drCM
1
v CM =
=
∑i mi v i
dt
M
„
The momentum can be expressed as Mv CM = ∑ mi v i = p tot
i
„
The acceleration of the CM can be found as
„
The acceleration can be related to a force
9-Apr-28
Paik
p. 13
Rotational Energy
„
The object’s rotational
kinetic energy is the sum
of kinetic energies of each
of the particles:
K rot
„
1
1
1
1
2
2
2
2
= m1v1 + m2 v2 + ⋅ ⋅ ⋅ = m1r1 ω + m2 r22ω 2 + ⋅ ⋅ ⋅
2
2
2
2
1⎛
1 2
2⎞
2
= ⎜ ∑ mi ri ⎟ω = Iω
⎠
2⎝ i
2
2
m
r
The quantity ∑ i i is called the moment of inertia
i
9-Apr-28
Paik
p. 14
Moment of Inertia
„
Moment of inertia is defined as I = ∑ mi ri
2
i
„
„
Dimensions of I are ML2 and its SI units are kg.m2
For an extended object, we replace mi with small
mass element Δmi and take the limit Δmi →0:
I = lim
Δmi → 0
„
∑r
i
i
2
Δmi = ∫ r dm
2
Expressing the mass element as dm =
ρ dV
I = ∫ ρ r 2 dV
9-Apr-28
Paik
p. 15
Example 3: I of a Rod
Uniform rigid rod
about its center
„
„
9-Apr-28
The shaded area has a mass
dm = λ dx = (M/L) dx
Then the moment of inertia
is
Paik
p. 16
Example 3, cont
Uniform rigid rod
about its end
„
„
The shaded area has a mass
dm = λ dx = (M/L) dx
Then the moment of inertia
is
3
L
I = ∫ r 2 dm = ∫ x 2 λdx = λ
3
0
0
1
2
I = ML
3
L
9-Apr-28
Paik
L
p. 17
Example 4: I of a Slab
Slab about its center
or edge
„
„
The derivation of I for
the rod did not involve
the width or thickness
Therefore, the same I
holds for a rod or slab
of the same length and
mass
9-Apr-28
Paik
p. 18
Example 4, cont
Slab about its
center
„
„
The mass element is
M
dm = σdxdy =
dxdy
ab
The moment of inertia is
I = ∫ r dm
2
I=
+a / 2
r =x +y
2
2
2
+b / 2
∫ dx ∫ dy σ (x + y )
2
−a / 2
2
−b / 2
1
= M (a 2 + b 2 )
12
9-Apr-28
Paik
p. 19
Example 5: I of a Hoop
Uniform thin hoop
about symmetry axis
„
„
9-Apr-28
Since this is a thin hoop,
all mass elements are to
a good approximation the
same distance from the
center
Then the moment of
inertia is
Paik
p. 20
Example 6: I of a Cylinder
Uniform solid cylinder
about symmetry axis
„
The moment of inertia is
R
I = ∫ r 2 dm
dm = ρdV = ρ ⋅ 2πrLdr
0
4
R
I = ∫ r 2 ρ 2πrLdr = 2πLρ ∫ r 3 dr = 2πLρ
4
0
0
1
1
2
2
= (ρπR L )R = MR 2
2
2
R
9-Apr-28
R
Paik
p. 21
Example 6, cont
Cylindrical objects
9-Apr-28
Paik
p. 22
Example 7: I of a Sphere
Spherical objects
9-Apr-28
Paik
p. 23
Parallel-Axis Theorem
„
„
„
In the previous examples, the axis of rotation
coincided with the axis of symmetry of the object
For an arbitrary axis, the parallel-axis theorem
often simplifies calculations
The theorem states
I = ICM + Md 2
where d is the distance from
the CM axis to the rotation axis
9-Apr-28
Paik
p. 24
Parallel-Axis Theorem, cont
„
„
The moment of inertia of
the rod about its center is
D is 1/2 L
9-Apr-28
Paik
p. 25
Vector Products of Vectors
„
The vector product of
vectors A and B, A x B,
is defined as a vector
with the magnitude of
|A x B| = AB sinθ
and the direction given by the right-hand rule
„
„
„
If A is parallel to B, then A x B = 0 ⇒ A x A = 0
If A is perpendicular to B, then |A x B| = AB
The vector product is also called the cross product
9-Apr-28
Paik
p. 26
Properties of Vector Products
„
„
„
The vector product is not commutative:
BxA=−AxB
The vector product obeys the distributive law:
A x (B + C) = A x B + A x C
The derivative of the cross product is
„
It is important to preserve the multiplicative order of A
and B
9-Apr-28
Paik
p. 27
Using Components
„
The cross product can be expressed as
„
„
Expanding the determinants gives
The cross products of the unit vectors:
ˆi × ˆi = ˆj × ˆj = kˆ × kˆ = 0,
ˆi × ˆj = kˆ , ˆj × kˆ = ˆi , kˆ × ˆi = ˆj
9-Apr-28
Paik
p. 28
Torque
„
A torque is defined as
τ=rxF
„
„
„
„
τ = r F sinφ = Fd
The direction of τ is given by the right hand rule
d = r sinφ is the moment arm (or lever arm), the
perpendicular distance from the axis of rotation to a line
drawn along the direction of the force
A torque is the tendency of a force to rotate an
object about some axis
9-Apr-28
Paik
p. 29
Net Torque
„
Force F1 will tend to cause a
counterclockwise rotation
about O
„
„
Force F2 will tend to cause a
clockwise rotation about O
„
„
Positive torque
Negative torque
Στ = τ1 + τ2 = F1d1 – F2d2
9-Apr-28
Paik
p. 30
Torque vs Force
„
Forces can cause a change in linear motion
„
„
Forces can also cause a change in rotational motion
„
„
Described by Newton’s 2nd Law
The effectiveness of this change depends on the force
and the moment arm, thus on the torque
The SI units of torque are N.m
„
„
Although torque is a force multiplied by a distance, it is
very different from work and energy
The units for torque are reported in N.m and not changed
to Joules
9-Apr-28
Paik
p. 31
Torque & Angular Acceleration
„
„
„
The tangential force provides a
tangential acceleration:
Ft = mat = mrα
The torque produced by Ft
about the center is
τ = Ft r = (mrα)r = (mr 2)α
Since mr 2 is the moment of inertia of the particle,
τ = Iα
„
This is analogous to F = ma.
9-Apr-28
Paik
p. 32
Torque & Angular Acc, Extended
„
„
„
From Newton’s 2nd Law,
dFt = (dm)at
The torque associated with the
force is
dτ = r dFt = atr dm = αr 2dm
The net torque is given by
2
τ
=
α
r
∑
∫ dm =Iα
„
„
This is the same relationship that applied to a particle
The result also applies when the forces have radial
components
9-Apr-28
Paik
p. 33
Static Equilibrium
„
The condition for a rigid
body to be in static
equilibrium is ΣFx = 0,
ΣFy = 0 and Στ = 0
„
9-Apr-28
You can choose any pivot
point since the net torque
about any point should be
zero
Paik
p. 34
Example 8: Ladder
A 3.0-m-long ladder leans against
a frictionless wall at angle θ. The
coefficient of static friction with the
ground μs is 0.20.
What is the minimum angle θmin for
which the ladder does not slip?
θ
Choose the bottom corner of the ladder as the pivot.
∑F
x
= n 2 − f s = n 2 − μ s n1 = 0 ,
∑F
y
= n1 − Mg = 0
1
(L cos θ )Mg − (L sin θ )n 2 = 0
2
1
1
Mg
Mg
tan θ ≥
=
=
=
= 2 .5 ⇒ θ min = tan −1 2 .5 = 68 °
2 n 2 2 μ s Mg
2 μ s 0 .40
∑τ =
9-Apr-28
Paik
p. 35
Hinged Rod
Why does the ball not move
with the end of the rod?
For the rod,
I = (1/3)mL2, τ = −mg (L/2) cosθ
τ = Iα
−mg (L/2) cosθ = 1/3 mL2α
α = −(3/2)(g /L) cosθ
at = Lα = −(3/2)g cosθ
ay = at cosθ = −(3/2)g cos2θ
y
For the cup to fall faster than the ball,
⏐ay⏐ ≥ g
¾ cos2θ ≥ 2/3 θ ≤ 35.3°
9-Apr-28
Paik
x
p. 36
Example 9: Wheel
A block of mass m is suspended from a
cable which is wrapped around a
frictionless wheel of mass M and radius R.
What is the acceleration of the block?
„
The wheel is rotating and so we apply
Στ = Ια
„
„
„
The tension supplies the tangential force
Friction between the wheel and string
causes the wheel to rotate
The mass is moving in a straight line,
so apply Newton’s 2nd Law
ΣFy = may = −mg +T
9-Apr-28
Paik
p. 37
Example 9, cont
Wheel: Στ = Iα
+TR sin90° = Ia /R
T = Ia /R2, I = (1/2)MR2
Mass: T − mg = −ma
Solve for a :
a = −T/m +g = −Ia /mR2 +g
a = g /(1 + I /mR2)
a < g for M > 0
T = mg /(1 + mR2/I) = mg /(1 + 2m/M)
As M → 0, T → 0, i.e. m is in free fall
9-Apr-28
Paik
p. 38
Example 10: Pulleys and Masses
Two masses, m1 and m2,
are connected to each
other through two identical
massive pulleys, as shown
in the figure.
Find the acceleration of the
masses.
„
„
„
a
Both masses move in linear directions, so apply Newton’s
2nd Law
Both pulleys rotate, so apply the torque equation
Combine the results
9-Apr-28
Paik
p. 39
Example 10, cont
Newton’s 2nd Law for the masses
T1 − m1g = −m1a, T3 − m2g = m2a
⇒ T3 − T1 = (m1 − m2)g − (m1+m2)a
The torque equation for the pulleys
(T1 − T2)R = Iα, (T2 − T3)R = Iα
⇒ T1 − T3 = 2Iα = 2Ia /R
Combine these results to obtain
(m − m )g
a=
m + m + 2(I R )
1
2
2
1
9-Apr-28
2
Paik
p. 40
Work in Rotational Motion
„
Work done by F on the object as
it rotates through an infinitesimal
distance ds = rdθ is
dW = F.d s = (F sinφ) rdθ
⇒ dW = τdθ
„
The radial component of F does
no work because it is perpendicular
to the displacement
9-Apr-28
Paik
p. 41
Power in Rotational Motion
„
„
The rate at which work is being done in a
time interval dt is
This is analogous to P = F⋅v in a linear
system
9-Apr-28
Paik
p. 42
Work-Kinetic Energy Theorem
„
„
The net work done by external forces in rotating a
rigid object about a fixed axis equals the change in
the object’s rotational kinetic energy:
1 2 1 2
dω
∑ W = ∫ τdθ = ∫ Iαdθ = ∫ I ωdt = Iω f − Iωi
2
2
dt
In general, net work done by external forces on an
object is the change in its total kinetic energy, the
sum of translational and rotational kinetic energies:
⎛1 2 1 2⎞ ⎛1 2 1 2⎞
∑ W = ΔK tot = ⎜ mv f + Iω f ⎟ − ⎜ mvi + Iωi ⎟
2
2
⎝2
⎠ ⎝2
⎠
9-Apr-28
Paik
p. 43
Energy Conservation
„
If no friction is involved, total mechanical
energy is conserved
Kf + Uf = Ki + Ui
„
K includes both rotational and translational kinetic
energies, if appropriate
9-Apr-28
Paik
p. 44
Example 11: Rotating Rod
What is the rod’s angular
speed when it reaches its
lowest position?
Two methods:
1) ∑W = Kf − Ki
MgL/2 = (1/2)Iωf2, I = (1/3)ML2
⇒ ωf = (3g /L)1/2
2) Since there is no friction
mentioned, assume that energy is conserved.
Kf + Uf = Ki + Ui
(1/2)Iωf2 + 0 = 0 + MgL/2 ⇒ ωf = (3g /L)1/2
9-Apr-28
Paik
p. 45
Example 12: Atwood Machine
Two masses, m1 and m2, connected by a
string around a pulley, are released from rest.
Find the linear and the angular velocities as a
function of h.
Assume energy is conserved:
Δ E = ΔK + ΔU = 0
Ki = 0, Kf = ½ (m1 + m2)v 2 + ½ Iω 2
ΔU = Uf
− Ui = −m1gh + m2gh
⇒ ½ (m1 + m2)v 2 + ½ Iω 2 + (m2 − m1)gh = 0
Substituting ω = v / R,
2(m2 − m1 ) gh
v=
≅
2
m1 + m2 + ( I / R )
9-Apr-28
2(m2 − m1 ) gh
m1 + m2 + M
Paik
p. 46
Example 13: Downhill Race
A sphere, a cylinder, and a
circular hoop, all of mass M and
radius R, are placed at height h
on a slope of angle θ.
Which one will win the race to
the bottom of the hill?
For work done by gravitational force: Wg = mgh
W = Kf − Ki = (½Iωf2 + ½mvf2) − (½Iωi2 + ½mvi2)
If no slipping occurs, ω = v /R. Also, ωι = 0.
W = ½ (I /R 2 +m)vf2, vf = (2Wg /(I /R 2 +m)1/2 = 2gh /(I /mR 2+1)1/2
The smaller I, the bigger vf .
I /mR 2 values: Particle: 0, Sphere: 2/5, Cylinder: ½, Hoop: 1
9-Apr-28
Paik
p. 47
Summary of Useful Equations
9-Apr-28
Paik
p. 48
Rolling Cylindrical Object
„
The red curve shows the path moved by a point
on the rim of the object
„
„
This path is called a cycloid
The green line shows the path of the CM of the
object
9-Apr-28
Paik
p. 49
Rolling Object, No Slipping
„
„
„
The velocity of the CM is
The acceleration of the
CM is
Rolling motion can be modeled as a combination of
pure translational motion & pure rotational motion
9-Apr-28
Paik
p. 50
Rolling Motion
„
„
Relationship between rotation and translation
At any instant, the point on
the rim located at point P is
at rest relative to the surface
since no slipping occurs
9-Apr-28
Paik
p. 51
Kinetic Energy of a Rolling Object
„
The total kinetic energy of a rolling object is the
sum of the translational energy of its CM and the
rotational energy about its CM
2
1
1
1
⎛ vCM ⎞ 1
2
2
2
K = I CM ω + MvCM = I CM ⎜
⎟ + MvCM
2
2
2
⎝ R ⎠ 2
1
⎞
2 ⎛ I CM
= MvCM ⎜
+ 1⎟
2
2
⎝ MR
⎠
„
Accelerated rolling motion is possible only if rolling
friction is present
„
The friction produces the torque required for rotation
9-Apr-28
Paik
p. 52
Example 14: Ball in a Sphere
A uniform solid sphere of radius
r is placed on the inside surface
of a hemispherical bowl with
much larger radius R. The
sphere is released from rest at
an angle θ to the vertical and
rolls without slipping.
Determine the speed of the
sphere when it reaches the
bottom of the bowl.
9-Apr-28
Paik
Mechanical energy is
conserved
I = (2/5)mr 2 for a solid
sphere
p. 53
Example 14, cont
Energy is conserved:
Uf + Kf = Ui + Ki
Ui = mg [R − (R −r ) cosθ]
Uf = mgr
Ki = 0
Kf = ½mv 2 + ½Iω 2 = ½mv 2 (1 + I /mr 2)
= ½mv 2 (1 + 2/5) = (7/10)mv 2
Substituting U ’s and K ’s into energy conservation,
mg [R − (R −r ) cosθ] − mgr = (7/10)mv 2
⇒ v = [(10/7)g(R −r )(1 − cosθ)]1/2
9-Apr-28
Paik
p. 54
Example 15: Falling Disk
A string is wound around a uniform disk
of radius R and mass M. The disk is
released from rest with the string vertical
and its top end tied to a fixed bar.
Show that (a) the tension in the string is
one-third the weight of the disk, (b) the
magnitude of the acceleration of the CM
is 2g/3, and (c) the speed of the CM is
(4gh/3)1/2 after the disk has descended
through distance h. (d) Verify your
answer to (c) using the energy approach.
9-Apr-28
Paik
p. 55
Example 15, cont
(a)
∑F
y
= T − Mg = − Ma ⇒ T = M ( g − a)
1
2
⎛a⎞
⎝R⎠
∑τ = −TR = − Iα = − MR 2 ⎜ ⎟ ⇒ a =
Solving these equations for T , T =
(b) a =
2T
M
Mg
3
2T
2 ⎛ Mg ⎞ 2
=
⎟= g
⎜
M M⎝ 3 ⎠ 3
4
⎛2 ⎞
(c) v = v + 2a (x f − xi ) = 0 + 2⎜ g ⎟(h − 0) ⇒ v f =
gh
3
⎝3 ⎠
(d) Energy conservation : U gf + K tra , f + K rot , f = U gi + K tra ,i + K rot ,i
2
f
2
i
1
1⎛1
⎞⎛ v
0 + Mv 2f + ⎜ MR 2 ⎟⎜⎜ f
2
2⎝2
⎠⎝ R
9-Apr-28
2
⎞
4
⎟⎟ = Mgh + 0 + 0 ⇒ v f =
gh
3
⎠
Paik
p. 56
Angular Momentum
„
The instantaneous angular
momentum L of a particle relative
to the origin O is defined as the
cross product of the particle’s
instantaneous position vector r
and its instantaneous linear
momentum p:
L=rxp
„
The SI units of angular momentum are (kg.m2)/s
9-Apr-28
Paik
p. 57
Angular Momentum, cont
„
Both the magnitude and direction of L depend on
the choice of origin
„
„
„
The magnitude of L is mvr sinφ
The direction of L is perpendicular to the plane formed
by r and p
Example: Uniform circular motion
„
„
„
L is pointed out of the diagram
L = mvr sin 90o = mvr
L is constant and is along an axis
through the center of its path
9-Apr-28
Paik
p. 58
Torque and Angular Momentum
„
Consider a particle of mass m located at the
vector position r and moving with linear
momentum p
dr
Note
=v
dt
dr
∴
× p = v × mv = 0
dt
9-Apr-28
Paik
p. 59
Torque & Ang Momentum, cont
„
„
The torque is related to the angular
momentum similar to the way force is related
to linear momentum:
dL
∑τ =
dt
This is another rotational analog of Newton’s
2nd Law
„
„
9-Apr-28
Στ and L must be measured about the same origin
This is valid for any origin fixed in an inertial
frame, i.e. not accelerating
Paik
p. 60
L of a System of Particles
„
„
„
Differentiating Ltot with respect to time,
Any torques associated with the internal forces
acting in a system of particles are zero:
The resultant torque acting on a system about an
axis through the CM equals d Ltot/dt of the system
regardless of the motion of the CM
9-Apr-28
Paik
p. 61
L of a Rotating Rigid Object
„
„
Each particle of the object
rotates in the xy plane about
the z axis with an angular
speed of ω
The angular momentum of an
individual particle is:
|L i|=|ri × (mi v i )|=mi ri 2 ω
„
L and ω are directed along
the z axis
9-Apr-28
Paik
p. 62
L of a Rotating Rigid Object, cont
„
„
„
To find the angular momentum of the entire object,
add the angular momenta of the individual particles
This gives the rotational form of Newton’s 2nd Law
If a symmetrical object rotates about a symmetry
axis, the vector form holds with I as a scalar:
L=Iω
9-Apr-28
Paik
p. 63
Conservation of Ang Momentum
„
The total angular momentum of a system is
constant in both magnitude and direction if
the net external torque acting on the system
is zero
„
„
Στext = 0 means the system is isolated
If the mass of an isolated system undergoes
redistribution, the moment of inertia changes
„
„
9-Apr-28
The conservation of angular momentum requires
a compensating change in the angular velocity
Ii ωi = If ωf
Paik
p. 64
The Merry-Go-Round
„
„
Isystem = Iplatform + Iperson
As the person moves
toward the center of the
rotating platform, the
angular speed will increase
„
„
To keep L constant
As the figure skater retracts
her stretched arms while
spinning, her angular speed
increases
9-Apr-28
Paik
p. 65
Intrinsic Angular Momentum
„
„
„
Angular momentum has been used in the
development of modern theories of atomic,
molecular and nuclear physics
The angular momentum has been found to
be an intrinsic property of these objects
Angular momenta are multiples of a
fundamental unit of angular momentum
„
9-Apr-28
h is called the Planck constant
Paik
p. 66
Example 16: Woman on Turntable
A 60.0-kg woman stands at the rim of a horizontal turntable
having a moment of inertia of 500 kg·m2 and a radius of
2.00 m. The turntable is initially at rest and is free to rotate
about a frictionless, vertical axle through its center. The
woman then starts walking around the rim clockwise (as
viewed from above the system) at a constant speed of 1.50
m/s relative to the Earth.
(a) In what direction and with what angular speed does the
turntable rotate? (b) How much work does the woman do to
set herself and the turntable into motion?
9-Apr-28
Paik
p. 67
Example 16, cont
Initially the turntable is not moving so its angular momentum
is zero. Since there is no external torque on the table, the
angular momentum stays zero when the woman walks. Since
she walks cw, ωw < 0.
v
(a) Lwoman + Lturntable = 0, I wω w + I tωt = 0, I tωt = − mw r 2 (− )
r
mw rv 60.0 kg (2.00 m)(1.50 m/s)
=
= 0.360 rad/s ccw
ωt = +
2
Ii
500 kg m
1
1 2
2
(b) W = ΔK = K f − 0 = mwoman vwoman
+ Iωtable
2
2
1
1
2
W = (60.0 kg)(1.50 m/s) + (500 kg m 2 )(0.360 rad/s) 2 = 99.9 J
2
2
9-Apr-28
Paik
p. 68
Example 17: Bullet and Block
A wooden block of mass M
resting on a frictionless
horizontal surface is attached
to a rigid rod of length l and
of negligible mass. The rod
is pivoted at the other end.
A bullet of mass m traveling
parallel to the horizontal surface and normal to the rod with
speed v hits the block and becomes embedded in it.
(a) What is the angular momentum of the bullet-block system?
(b) What fraction of the original kinetic energy is lost in the
collision?
9-Apr-28
Paik
p. 69
Example 17, cont
Consider the block and bullet as a system. Initially the angular
momentum is all in the bullet. Since there are no external
torques on the system, the angular momentum is constant.
(a) Since ∑ τ ext = 0, L f = Li . Li = mvl, L f = (m + M )v f l = mvl.
1 2
1
(b) K i = mv , K f = ( M + m)v 2f
2
2
1 2⎛ m ⎞
⎛ m ⎞
vf = ⎜
⎟
⎟v, K f = mv ⎜
2
⎝M +m⎠
⎝m+M ⎠
ΔK
Ki
=
Ki − K f
9-Apr-28
Ki
−1
1 2⎛
m ⎞⎛ 1 2 ⎞
M
= mv ⎜1 −
⎟⎜ mv ⎟ =
2
M +m
⎝ M + m ⎠⎝ 2
⎠
Paik
p. 70
Example 18: Space Station
A space station is constructed in the
shape of a hollow ring of mass 5.00 x
104 kg. (Other parts make a negligible
contribution to the total moment of
inertia.) The crew walk on the inner
surface of the outer cylindrical wall of
the ring, with a radius of 100 m. At
rest when constructed, the ring is set
rotating about its axis so that people
inside experience an effective free-fall acceleration equal to g. The
rotation is achieved by firing two small rockets attached tangentially
to opposite points on the outside of the ring.
(a) What angular momentum does the space station acquire?
(b) How long must the rockets fire if each exerts a thrust of 125 N?
9-Apr-28
Paik
p. 71
Example 18, cont
v2
g
9.80 m/s 2
2
=
= 0.313 rad/s
(a) Re quire g = = ω r. ω =
r
r
100 m
I = Mr 2 = 5.00 × 10 4 kg (100 m) 2 = 5.00 ×108 kg m 2
L = Iω = (5 × 108 kg m 2 )(0.313 rad/s ) = 1.57 ×108 kg m 2 /s
(b) τ = 2rF = 2(100 m )(125 N ) = 2.50 × 10 4 N m
dL L f − Li
τ=
=
= 2.50 ×10 4 N m
dt
Δt
1.57 ×108 kg m 2 /s − 0 kg m 2 /s
= 6280 s
Δt =
4
2.50 ×10 Nm
The total torque on the ring, multiplied by the time interval found in part
(b), is equal to the change in angular momentum, found in part (a). This
equality represents the angular impulse-angular momentum theorem.
9-Apr-28
Paik
p. 72