Download University Physics AI No. 8 Spin and Orbital Motion

University Physics AI
No. 8 Spin and Orbital Motion
Class
Number
Name
I. Choose the Correct Answer
v
1. A particle moves with position given by r = 3tiˆ + 4 ˆj , where
v
r is measured in meters when t
is measured in seconds. For each of the following, consider only t > 0. The magnitude of the angular
（ B ）
momentum of this particle about the origin is
(A) increasing in time.
(B) constant in time.
(C) decreasing in time.
(D) undefined
Solution: Assume the mass of the particle is m, so the angular momentum of this particle about the
origin is
v
r v v v
dr
v v
L = r × P = r × mv = r × m
= (3tiˆ + 4 ˆj ) × m(3iˆ) = −12mkˆ
dt
Thus the magnitude of the angular momentum of this particle L = 12m kg ⋅ m /s = constant .
2
r
r
2
momentum about the origin of L = (20kg ⋅ m /s) kˆ when t = 0s. The magnitude of the angular
2. A particle moves with constant momentum p = (10kg ⋅ m/s)iˆ . The particle has an angular
（
momentum of this particle is
(A) decreasing
(B) constant.
(C) increasing.
B
）
(D) possibly but not necessarily constant.
r
v
v
Solution: The angular momentum of this particle about the origin is L = r × P , so the position
v
vector of the particle when t = 0s is r = −2 ˆj .
Assume the mass of the particle is m, the position vector of the particle at any time t is
p
v
r = vtiˆ − 2 ˆj = tiˆ − 2 ˆj
m
So the angular momentum of this particle is
r v v
p
L = r × P = ( tiˆ − 2 ˆj ) × Piˆ = 2 pkˆ = (20kg ⋅ m 2 /s)kˆ
m
Thus the magnitude of the angular momentum of this particle L = 20kg ⋅ m /s = constant .
2
3. A solid object is rotating freely without experiencing any external torques. In this case （ A ）
(A) Both the angular momentum and angular velocity have constant direction.
(B) The direction of angular momentum is constant but the direction of the angular velocity might
not be constant.
(C) The direction of angular velocity is constant but the direction of the angular momentum might
not be constant.
(D) Neither the angular momentum nor the angular velocity necessarily has a constant direction.
Solution: Using conservation of angular momentum and the definition of the angular momentum ,
r
r
L = Iω
r
r
4. A particle is located at r = 0iˆ + 3 ˆj + 0kˆ , in meter. A constant force F = 0iˆ + 0 ˆj + 4kˆ (in
Newton’s) begins to act on the particle. As the particle accelerates under the action of this force，the
（
torque as measured about the origin is
(A) increases.
(B) decreases.
(C) is zero.
D ）
(D) is a nonzero constant.
Solution: The torque as measured about the origin is
v
v
v
τ = r × F = (0iˆ + 3 ˆj + 0kˆ) × (0iˆ + 0 ˆj + 4kˆ) = 12iˆ
So it is a nonzero constant.
II. Filling the Blanks
y(m)
1. The total angular momentum of the system of
particles pictured in Figure 1 about the origin at O
− v0 iˆ
is ( 4.4mv0 kg ⋅ m /s) kˆ
3m
2
-4.00
Solution: Using the definition of the angular
momentum
v v v v
v
L = r × P = r × mv ,
the
2.00 2m
v0iˆ
1.20
1.00
O
-2.00
total
x(m)
v0 ˆj
2m
Fig.1
angular momentum of the system of particles is
v
Ltotal = (−4iˆ + 2 ˆj ) × 3m(−v0 iˆ) + (1iˆ + 2 ˆj ) × 2m(v0 iˆ) + (1.2iˆ − 2 ˆj ) × 2m(v0 ˆj )
= 6mv kˆ − 4mv kˆ + 2.4mv kˆ = 4.4mv kˆ(kg ⋅ m 2 /s)
0
0
0
0
2. A particle of mass 13.7g is moving with a constant velocity of magnitude 380m/s. The particle,
moving in a straight line, passes with the distance 12cm to the origin. The angular momentum of the
particle about the origin is 62.472 kg·m2/s .
Solution: Using the definition of the angular momentum, the angular momentum of the particle
about the origin is
v v
v
L = r × mv = rmv sin θ = dmv = 12 × 10 −3 × 13.7 × 10 −3 × 380 = 6.25 × 10 −2 kg ⋅ m 2 /s
3. The rotor of an electric motor has a rotational inertia Im=2.47×10-3kg⋅m2 about its central axis.
The motor is mounted parallel to the axis of a space probe having a rotational inertia Ip=12.6kg⋅m2
about its axis. The number of revolutions of the motor required to turn the probe through 25.0°
about its axis is 354rev .
Solution: Assume the two axes is coaxial, the angular momentum is conserved, we have
I mωm = ( I m + I p )ω
Integrate both sides of the equation with respect to time, we get
∫
θ1
0
θ2
I mωm dt = ∫ ( I m + I p )ωdt
0
I mθ1 = ( I m + I p )θ 2
( I + I )θ
θ
n = 1 = m o p 2 = 354rev
2π
360 I m
4. Two particles each of mass m and speed v, travel in opposite directions along parallel lines
separated by a distance d. The total angular momentum of the system about any origin is
mvd .
Solution:
y
See figure, select any point O and set up coordinate
system. So the total angular momentum of the system
v0iˆ
1
about the origin is
r1
v
v v
v
v v
v
Ltotal = L1 + L2 = r1 × mv + r2 × mv
= r1 mv sin θ1 + r2 mv sin θ 2
− v0 iˆ 2
= mv(r1 sin θ1 + r2 sin θ 2 ) = mvd
O
θ2
θ1
d
x
r2
r
5. A particle is located at r = (0.54m)iˆ + ( −0.36m) ˆj + (0.85m) kˆ . A constant force of
magnitude 2.6 N acts on the particle. When the force acts in the positive x direction, the components
of the torque about the origin is 2.21 ˆj + 0.936kˆ , and when the force acts in the negative x
direction, the components of the torque about the origin is − 2.21 ˆj − 0.936kˆ .
Solution: According to the definition of the torque
v
v
v
τ =r×F,
2.6 Niˆ , the torque about the origin is
r v
τ = r × F = (0.54iˆ − 0.36 ˆj + 0.85kˆ) × (2.6iˆ) = 2.21 ˆj + 0.936kˆ
when the force is − 2.6 Niˆ , the torque about the origin is
v r v
τ = r × F = (0.54iˆ − 0.36 ˆj + 0.85kˆ) × (−2.6iˆ) = −2.21 ˆj − 0.936kˆ
when the force is
v
6. A cylinder having a mass of 1.92kg rotates about its axis of symmetry.
Forces are applied as shown in Fig.2: F1=5.88N, F2=4.13N, and
F3=2.12N. Also R1=4.93cm and R2=11.8cm. The magnitude and direction
of the angular acceleration of the cylinder are
90.3rad/s2 .
v v
Solution: Using the definition of the torque τ = r × F , suppose the
v
clockwise is the positive. The magnitude of the total torque about its axis
r
F1
R2
r
R1 F
3
Fig.2
of symmetry is
τ = F1R2 − F2 R2 − F3 R1
= 5.88 × 11.8 × 10− 2 − 4.13 × 11.8 × 10− 2 − 2.12 × 4.93 × 10− 2 = 10.2 N ⋅ m
r
F2
30 o
Using the equation
v
v
1
2
1
2
τ total = Iα , and I = MR 2 = × 1.92 × 11.8 × 10− 2 = 0.113 kg ⋅ m 2
Thus the magnitude of the angular acceleration of the cylinder is
α=
τ
I
=
10.2
= 90.3 rad/s 2
0.113
The direction of he angular acceleration is counterclockwise, point out of the page.
7. A disk of mass m and radius R is free to turn about a fixed, horizontal axle. The
disk has an ideal string wrapped around its periphery from which another mass m
(equal to the mass of the disk) is suspended, as indicated in Figure 3. The magnitude
of the acceleration of the falling mass is
2g/3 , the magnitude of the angular
acceleration of the disk is 2g/3R .
Solution: Assume the acceleration of the falling mass is a, and the angular
acceleration of the disk is α . We have
m
R
m
⎧
2g
⎧
⎪a = Rα
⎪⎪a = 3
⎪
⎨mg − T = ma ⇒ ⎨
⎪α = a = 2 g
⎪
2
α
mR
⎪⎩
⎪TR =
R 3R
2
⎩
Fig.3
8. A uniform beam of length l is in a vertical position with its lower end on a rough surface that
prevents this end from slipping. The beam topples. At the instant before impact with the floor, the
angular speed of the beam about its fixed end is
3g
.
l
Solution: Use the CWE theorem. Assume the zero PE is at the point of the end of the beam. So
1
⎫
I CM ω 2 + 0⎪
1
1
1 1 2 22 1
3g
⎪
2
2
⎬ ⇒ I CM ω = mgl ⇒ ⋅ ml ⋅ ω = mgl ⇒ ω =
1
l
2
2
2 3
2
Ei = mgl + 0 ⎪⎪
2
⎭
Ef =
III. Give the Solutions of the Following Problems
1. A pulley having a rotational inertia of 1.14×10-3kg⋅m2 and a radius of 9.88cm is acted on by a
force, applied tangentially at its rim that varies in time as F=At+Bt 2, where A=0.496N/s and
B=0.305N/s2. If the pulley was initially at rest, find its angular speed after 3.60s.
Solution:
Using the equation
v
v
v
v
τ = r × F = Iα , we have
rF 9.88 × 10 −2
αz =
=
(0.496t + 0.305t 2 ) = 42.987t + 26.433t 2
−3
I
1.14 × 10
For
α=
dω
dω
, so α =
= 42.987t + 26.433t 2 ,
dt
dt
If the pulley was initially at rest, thus its angular speed after 3.60s is
3.6
3.6
0
0
ω = ∫ (42.987t + 26.433t 2 )dt = 21.494t 2 + 8.811t 3
= 689.6 rad/s
T2
2. Two identical blocks, each of mass M, are connected by a
I
M
light string over a frictionless pulley of radius R and rotational
inertia I (Fig.4). The string does not slip on the pulley, and it is
T1
not known whether or not there is friction between the plane
and sliding block. When this system is released, it is found
M
that the pulley turns through an angle θ in time t and the
Fig.4
acceleration of the blocks is constant. (a) What is the angular
acceleration of the pulley? (b) What is the acceleration of the
two blocks? (c) What are the tensions in the upper and lower sections of the string? All answer are
to be expressed in terms of M, I, R, θ, g, and t.
Solution: Sketch the forces diagram of the system shown in figure.
N
N
T2
f
T1
T2
Mg
v
a
mg T1
Mg
Apply the Newton’s second law and the counterpart of Newton’s law for a spinning rigid body, we
have
⎧Mg − T1 = Ma
⎪(T − T ) R = Iα
2
⎪⎪ 1
⎨a = α R
⎪
⎪θ = 1 αt 2
⎪⎩
2
Solving them, we get
(a) The angular acceleration of the pulley is
α=
2θ
.
t2
(b) The acceleration of the two blocks is a =
2θR
t2
(c) The tensions in the lower sections of the string is T1 = M ( g − a ) = M ( g −
(d) The tensions in the lower sections of the string is T2 = T1 −
3. A disk with moment of inertia I1 is rotating with
initial angular speed ω0; a second disk with moment of
inertia I2 initially is not rotating (see Figure 5). The
arrangement is much like a LP record ready to drop
onto an unpowered, freely spinning turntable. The
second disk drops onto the first and friction between
them brings them to a common angular speed. Find the
common angular speed ω.
2θR
)
t2
2θR 2θI
Iα
= M (g − 2 ) − 2
R
t
t
I2
r
ω0
Direction of spin
I1
Fig.5
Solution:
Let two disks be a system, the total torque about the axis is zero, applying the conservation of
angular momentum, we have the common angular speed ω.
r
r
L f = Li
r
r
r
⇒ I 1ω + I 2ω = I 1ω 0
r
⇒ ( I 1 + I 2 )ω = I 1ω 0
⇒ ω=
I1
ω0
I1 + I 2
4. Two cylinder having radii R1 and R2 and rotational inertias I1 and
+ω
I2, respectively, are supported by axes perpendicular to the plane of
+
0
Fig.6. The large cylinder is initially rotating with angular velocity ω0. I2
R2
I1
The small cylinder is moved to the fight until it touches the large
R1
cylinder and is caused to rotate by the frictional force between the
two. Eventually, slipping ceases, and the two cylinders rotate at
Fig.6
constant rates in opposite directions. Find the final angular velocity
ω2 of the small cylinder in terms of I1, I2, R1, R2, and ω0. (Hint: Angular momentum is not conserved.
Apply the angular impulse equation to each cylinder.)
Solution: Apply the rotational counterpart of Newton’s second law of motion
we get
v
τ total
v
dL
=
,
dt
⎧ R2 fdt = I 2ω 2 − 0
⎪∫
⎪
⎨− ∫ R1 fdt = I 1ω1 − I 1ω 0
⎪
⎪⎩ω1 R1 = ω 2 R2
Solving the equations, we can get the final angular velocity ω2 of the small cylinder is
I 1ω 0 R1 R2
I 1 R22 + I 2 R12
ω2 =
5. One way to determine the moment of inertia of a disk of radius
Wheel
R and mass M in the laboratory is to suspend a mass m from it to
M
R
Hub
r
give the disk an angular acceleration. The mass m is attached to a
string that is wound around a small hub of radius r, as shown in
Figure 7. The mass and moment of the inertia of the hub can be
m
neglected. The mass m is released and takes a time t to fall a
distance h to the floor. The total torque on the disk is due to the
Fig.7
torque of the tension in the string and the small but unknown
frictional torque
r
τ frictn of the axle on the disk. Once the mass m hits the floor, the disk slowly stops
spinning during an additional time t ′ because of the sole influence of the small frictional torque on
the disk. Find the moment of inertia of the disk in terms of m, r, g, h, t and t ′.
Solution:
Assume the acceleration of the falling mass is a, and the angular acceleration of the disk is α . The
ω = α t while the falling mass reaches the floor. So
angular speed of the disk is
h=
1 2
2h
at ⇒ a = 2
2
t
Using Newton’s second law
⎧mg − T = ma
⎪
⎨Tr − τ fric = Iα
⎪
⎩ a = rα
(1)
Apply the rotational analog of Newton’s second law to the disk
r
τ total
r
r
dLtotal ∆L
=
≈
∆t
dt
During the course of the mass hits the floor
τ fric =
ω −0
t'
I=
αIt
t'
(2)
Using the equation (1) and (2), we get the moment of inertia of the disk
I=
mr ( g − a) 1
⋅
=
t
a
1+
t'
2
gt 2
− 1)
2h
t
1+
t'
mr 2 (