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Chapter 2. Fundamental Laws of Mechanics
Chapter Two
The Fundamental laws of Mechanics
Contents
• The concepts of displacement (位移), velocity, acceleration, angular velocity, angular
acceleration and angular momentum (角动量)
• Newton’s laws of motion, rotational laws,
• Conservational (守恒的) laws of momentum, energy and angular momentum
§2.1 The motion of Point masses
2.1.1 Displacement and equation of motion
When a particle is moving in space, it changes its position with time. Generally, the
motion could be described by a function of time.
For example, a particle moves along x-direction in a constant speed v, its position can
be written as
x = vt + x0  x(t)
(2.1)
x0 can be determined by the initial condition.
If the particle moves in a three dimensional space, its path at time t can be described
by the following equation:
r(t) = x(t) i + y(t) j + z(t) k
(2.2)
r rr(t)
(2.3)

where r denotes vector r and same for i, j, and k. Quite a few books used such bold form
letters as vectors for simplicity. At the time of t+t, r(t) becomes r(t+ t). The change of its
path during the interval t is given by rr = r(t+ t) - r(t).
rr(t) is called the displacement of the particle
r(t)
in t period. It also has three components in x, y and z
r(t)
directions. r(t) is called the equation of motion for the
particle.
2.1.2 Velocity and Acceleration
r(t+ t)
1. Instantaneous velocity and acceleration
Fig. 2.1 The displacements
The average velocity of the particle during t with displacement of rr is rr/t. It is also a
vector, its direction is the same as the increment rr. The instantaneous velocity of the
particle at time t is



r dr
v  lim

(2.4)
t 0 t
dt
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The instantaneous velocity is not necessarily a constant. It can be a function of time. At the

moment t, r (t ) will give the exact position of that particle. Using equation (2.4), we could get
the exact value of velocity at that particular moment. Generally speaking, the magnitude of
the velocity is called speed, denoted by

 dr
ds
v v 

,
dt
dt
where v represents instantaneous speed, ds is the absolute path length displaced at the time
interval dt. In the Cartesian coordinate system, as i, j and k do not change with time, we have
 dx  dy  dz 
v
i
j k.
(2.5)
dt
dt
dt
The three components on the axes of x, y and z are expressed as
dx
dy
dz
vx  , v y 
, vz 
(2.6)
dt
dt
dt
And the magnitude of the velocity is
v  v x2  v y2  v z2
(2.7)
Example 2.1 The position of a particle moving in x-y plane is described by the following
parametric equations given by
x  r  r cos t
y  r sin t
where r  1m,  

(E.1.1)
s 1 . Find
4
(1) the path (or orbit) function f(x,y) = 0;
(2) velocity at any time;

(3) position vectors at t = 0 and t = 6s, the displacement  r and path length s during
this time interval.
Solution
(1) rearrange the equation (E.1) as
x  r  r cos t
(E.1.2)
y  r sin t
Square on both sides and delete t, then we have
(E.1.3)
(x  r) 2  y 2  r 2
Where is the path function of a circle with radius r and its centre locates at (r, 0). i.e. x = r and
y = 0.




 dx  dy 
i
j  r sin ti  r cos tj  r  sin ti  cos tj 
(2) v 
(E.1.4)
dt
dt
and the magnitude of the velocity is

v  v x2  v y2  r  ms 1
(E.1.5)
4
Which means that the motion is a circular motion with constant speed. The angle  between

v and x-direction can be simply found as
vy
tan  
  cot t
(E.1.6)
vx
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Chapter 2. Fundamental Laws of Mechanics
By inspection of the sign vx and vy at a particular time, you can determine which quadrant the
angle is in.
(3) at first, we could get the two position vectors at t = 0 and t = 6s respectively,


r0  2ri
t=0 
(E.1.7)





3
3
r6 s  (r  r cos )i  r sin
j  ri  rj
t = 6s 
(E.1.8)
2
2
the displacement during t  6s is
 
 

r  r6 s  r0  r (i  j )
Its shortcut length is

r  (r ) 2  (r ) 2  2r  2m
While the path length during the time interval should be calculated by the following:
s  the angle moved by radian  radius of the circle orbit
= angular velocity × time interval × r
3
m
4
2
In above calculations, the concept that the arc length is equal to the angle in radian multiplied
by radius was used.


s 1  6s  1m 
2. Acceleration
The path of a particle moving in two or three dimensions is a curve in general, its velocity
changes both in magnitude and in direction. The magnitude of the velocity changes when the
particle speeds up or slows down. The direction of the velocity changes because the velocity
is tangent to the path and the path bends continuously. To describe the average velocity rate
of change in velocity for the time interval t, the average acceleration is defined as

 v
a
.
(2.8)
t
The instantaneous acceleration of a particle is defined in the same way as instantaneous
velocity as



v dv dv x  dv y  dv z 
a  lim


i
j
k
t 0 t
dt
dt
dt
dt
(2.9)

d 2r d 2 x  d 2 y  d 2 z 
 2  2 i  2 j 2 k
dt
dt
dt
dt
Example 2.2 Suppose that the motion of a particle is described by the equation of motion x =
20 + 4t2. Find the speed and the acceleration of the particle at t = 2s.
Solution: (a) find speed
using the differentiation formula
d n
( x )  nx n 1 ,
dx
we have
(1) the speed at any moment t
dx
v(t ) 
 8t ,
dt
(2) the speed at t = 2 seconds
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v(t) |t = 2s = 8 t|t = 2s = 16 (m/s)
dv
a
 8(ms  2 )
(b) Acceleration
dt
Example 2.3 The coordinate of a particle moving in the three dimensional space given as
functions of time by




r (t )  x(t )i  y (t ) j  z (t )k
where
x(t) = 1 + 2t2 (m), y(t) = 2t + t3(m) and z(t) = 3t + 4t2(m).
Find the particle’s vectors and magnitudes of its position, velocity and acceleration at t = 2s.

Solution: (1) position at t = 2s, this is r ( 2)




r (t )  9i  12 j  22k
Its magnitude is
r  9 2  12 2  22 2 (m)
(2) find the velocity at time t



 dx  dy  dz 
v
i
j  k  4ti  (2  3t 2 ) j  (3  8t )k .
dt
dt
dt
At t = 2s, we have




v  8i  14 j  19k ,
the magnitude of the velocity is
v  8 2  14 2  19 2 (ms 1 )
(3) Acceleration



 dv  dv y  dv z 
a xi 
j
k  4i  6tj  8k ,
dt
dt
dt
At t = 2s, the acceleration is




a  4i  12 j  8k ,
The magnitude of the acceleration is
a  4 2  12 2  8 2 (ms 2 )
Example 2.4 A person standing on a cliff pulls a boat by a pulley, as shown in Fig. 2.2.
Suppose that the height of the cliff is h, the rate of the rope pulled is u. Find: (1) the velocity
of the boat; (2) its acceleration.

h

O
x
x
Fig. 2.2 diagram for example 2.4
Solution: As the boat moves on the surface of water, its motion is in one dimension. Set the
x-axis point to the right, choose the origin at the foot point of the pulley, and let  represent
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Chapter 2. Fundamental Laws of Mechanics
the variable length of the rope at any time. So the relation among h,  and x can be found
easily by Pythagorean Theorem (勾股定理)
x2  2  h2 ,
and the equation of motion for the boat is


r  xi ,
Take the time derivative of x, we have
1 1
1 / 2
dx 1 2
d
u
   h 2  2  2 
  2  h 2    (u )  
dt 2
dt
x
d
 u , as the boat moves towards negative direction of x-axis. Therefore, the speed
where
dt
of the boat is
u
v
x2  h2
x
Obviously, the velocity is expressed as a function of coordinate x. The acceleration of the boat
is then
dv
u 2h2
a
 3
dt
x
So as long as we know where it is, i.e. position x, we could simply calculate the acceleration.
The result indicates that acceleration and velocity are in the same direction, so the speed of
the boat will become larger and larger with the value of x decreased.
§2.2 Newton’s Law of Motion
We know from experience that an object at rest never starts to move by itself; in order to
move a body, a push or pull must be exerted on it by some other body. Similarly, a force is
required to slow down or stop a body already in motion, and to make a moving body deviate
from straight line motion requires a sideways force. All these processes (speeding up, slowing
down, or changing direction) involve a change in either the magnitude or direction of the
velocity. Thus in each case the body accelerates, and an external force must act on it to
produce the acceleration. Before discussing Newton’s Law of motion, the definitions of
momentum and force should be stated.
Momentum: The momentum of a body is defined as the product of its mass times its
vector velocity.


p  mv
(2.10)

Force: If a single force F is applied to a body of mass m, the value of the force is
defined as the time rate of change of momentum of the body
 dp
F
(2.11)
dt
for a body of constant mass, (2.11) becomes

 d (mv )
dv

F
m
 ma
(2.12)
dt
dt
2.2.1 Newton’s first law of motion
Newton’s first law describes the motion of an isolated object which, for our purposes,
is an object on which no net force is acting.
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If an isolated object is at rest, it will remain at rest; if it is in motion, it will continue
moving along a straight line at a constant speed.
A body remains in a state of rest or constant velocity (zero acceleration) when left to
itself.
In the most general case, a single force acting on a body produces a change in both its
translational and rotational motion. However, when several forces act on a body
simultaneously, their effects can compensate one another, with the result that there is no
change in either the translational or rotational motion. When this is the case, the body is said
to be in equilibrium. This means (1) that the body as a whole either remains at rest or moves
in a straight line with constant speed, and (2) that the body is either not rotating at all or is
rotating at a constant rate.
Mathematically, this says


a  0,
when Fnet  0
(2.13)

where Fnet is the vector sum of all the forces acting on the body.
Note: Newton’s first law is often called the law of inertia. Inertia is the property of an
object that resists acceleration.
2.2.2 Newton’s second law of motion
The time rate of change of momentum of a body is equal to the net force acting on the body.
For constant mass
Fnet = dP
P/dt or
Fnet = ma
a
(2.11)
That is, if the sum of all forces acting on an object is not zero, then it will be accelerated. The
acceleration depends on the net force and on the mass m of the object as well.
If you think of inertia as the qualitative term for the property of a body that resists
acceleration, then mass (a scalar quantity) is the quantitative measure of inertia. If the mass is
large, the acceleration produced by a given force will be small.
2.2.3 Newton’s third law of motion
Wherever two bodies interact, the force on the first body due to the second is equal and
opposite to the force on the second due to the first.
FA due to B = - FB due to A
(2.12)
Or we can say: if object A exerts (施加) a force on object B, then object B exerts an equal and
opposite force on object A.
For every action there is an equal and opposite reaction. This law is true for any type
of force, including frictional, gravitational, electrical, and magnetic forces.
2.2.4 Inertial frame of reference (inertial system)
A frame of reference (参考系) in which bodies move in straight lines with constant
speeds unless acted upon by external forces, i.e. A frame of reference in which free bodies are
not accelerated.
Newton’s laws of motion are valid in an inertial system but not in a system that is itself
accelerated with respect to such a frame.
§2.3 Work and Energy
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Chapter 2. Fundamental Laws of Mechanics
2.3.1 Work
In everyday life, the term work is applied to any activity requiring the exertion of
muscular or mental effort. In physics, the term is used in a much more specific sense, which
involves the application of a force to a body and displacement of that body.
If a displacement drr of a body is caused by a force F, the work (dA) done by the force
is expressed as
dA = F·dr = F·dr Cos
(2.13)
 is the angle between the F and dr. Work is a scalar, its dimension ( 量 纲 ) is
“Newton·Meter” and its unit is “joule” (J).
If a body moves along a curve from a to B, the work done by the force F is
AAB = dA = F
F·dr
(2.14)
“” is the symbol of integral, finding the result of a continuous summation. The formula of
integration we used here is
x n 1
n
(C is a constant)
(2.15)
 x dx  n  1  C
If this integration is from x = A to x =B, the result should be
B
B n 1  A n 1
n
A x dx  n  1
2.3.2 Energy
1. Kinetic Energy
If a body is drawn by a force on a horizontal frictionless surface, the work done by the force
will increase the speed of the body. So the work can be described by the kinetic energy. It can
be evaluated as
B 
B
B

AAB   F  dr   F cos dr   Fdr (  0)
A
A
A
(2.16)
1
 dv 
   m   vdt   m  vdv  m v A2  v B2
dt 
2
tA 
vA
When the length of the curve AB is s, the projection value of F is F (resultant force), the work
done by F from A to B should be F·s, therefore
1
F  s  mv A2  v B2 
2
Using Newton’s second law of motion, we have
(2.17)
2as  v A2  v B2
2
So this shows that the integration is correct. The quantity 1/2mv is called kinetic energy.
Therefore we could conclude that “The work done by the resultant external force on a body is
equal to the change in kinetic energy of the body.”
tB
vB


2. Potential Energy
Potential energy can be thought of as energy stored up for future potential use. In many cases,
it can be converted into other useful forms of energy.
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3 Conservative force If the system is conservative, the total work done on a particle when it
moves around any closed curve is zero:
 
(2.18)
 F  ds  0
Now consider two points A and B, and any two paths connecting them. We can combine these
two paths to make a closed curve, to which (2.18) applies. This means that
B 
A 


 F  ds (lower _ path)   F  ds (higher _ path)  0
A
B
B 
 

F

d
s
(
lower
_
path
)


 F  ds (higher _ path)  0
B
A
B

A
B



 F  ds (lower _ path)   F  ds (higher _ path)
A
A
So that the work done on the particle by conservative force when it moves from A to B is
independent of the path from A to B. In this case we choose a reference point C, and define
the potential energy at point A by
B
 
U ( A)    F  ds
A
A
C
Then we can write
A 
B 
  C   B  


 F  ds   F  ds   F  ds   F  ds  ( F  ds )  U ( A)  U ( B)
B
A
A
C
C
C
In such a force field, there is a special function which is determined by the relative positions
of the particle, and it is called potential Energy function.
Example 2.5 Gravitational potential energy function is mgh. Mg is gravitational force.
0
U (h)    mgds  mgh
h
That means that from a height h, the conservative force (gravity) does a work of mgh.
Therefore, at that height h, it has potential energy of mgh.
Example 2.6 Elastic potential energy is 1/2 kx2 (k is elastic constant of a spring). The elastic
force is also a conservative force and usually is expressed as –kx, choose a proper point as the
zero point of the elastic potential energy, it could be easily derived.
In such a system, potential energy and kinetic energy can be converted each other. It can be
shown that the total kinetic energy plus potential energy is a constant at any time. Therefore if
a system has a conservative force only, it has the property of conservation of energy.
E = K + U = Constant
(2.19)
(hints of proof:
B
 
dv
1
U ( A)  U ( B)   F  ds   (m )  (vdt)   mvdv  m v B2  v A2
dt
2
A
A
vA
B
v
B


This shows that all the potential energy between A and B is transformed into kinetic energy.)
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Chapter 2. Fundamental Laws of Mechanics
§2.4 Rotational Motion
2.4.1 Rotational kinematics
Displacement x ←→ 
angular displacement
Velocity
dx/dt ←→ d/dt
angular velocity
d/dt
(2.20)
For motion a circle, there is a simple relation between w and the velocity v along the
circumference. See fig. 4.1
s=R
(2.21)
Now differentiating the both sides with respect to t, we have
ds/dt = R d/dt
v=R
(2.22)
The velocity v is the distance traveled in one second. So
v = 2R f
f is the number of revolution
2R f = R 
Therefore we have the very important relation between the frequency and the angular velocity
 = 2f
or
f=

(2.23)
The letter f stands for frequency in revolution per second and  is radius per second.
Displacement
x ←→
 (angular)
Velocity
dx/dt ←→ d/dt (angular)
Acceleration a = d2x/dt2 ←→  = d2/dt2 (angular)
dv/dt = R d/dt
a = R
(2.24)
Obviously, the acceleration a is the tangential acceleration. This relation is not only valid for
the circular motion but also for any kind of curve motion. The only difference is the radius R
is changeable at different moment.
2.4.2 Angular momentum and its conservation
1. Definition of angular momentum
A single particle can have angular momentum even if it moves in a straight line. It is
defined as
  
Lrp
(2.25)
where p is the particle’s linear momentum and r is the position vector from the origin to the
particle. According to the definition of cross product of two vectors, we have
L  rp sin   mrv sin 
Where  is the angle between r and p and the direction of angular momentum is determined
by the right-handed rule. See the fig. 4.1 L is in k (z) direction.
Where
r = |rr| = (x2 + y2)1/2
v = |v| = (vx2 +vy2)1/2
for r = x i + y j and
for v = vx i + vy j
2. Conservation of Angular momentum
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We consider a single particle moves in a central force field which is directed from or
away from the origin. This means that the central force has the same direction with r . Taking
the gravity as an example, the gravity points to the center of the earth.
The conservation of a physical quantity is defined that it is not changed with time.
Therefore if the angular momentum is conserved, the dL
L/dt should be equal to zero. Let’s
calculate it for the central force field.
Product rule of differentiation : d(x·y)/dt = dx/dt · y + x · dy/dt
dL
L/dt = d(rr ×p
p)/dt
= drr/dt ×p
p + r ×dp
p/dt
= v ×mv
v + r ×F
F
= m (v
v ×v
v) + (rr ×F
F)
=0
(2.26)
Thus we have
L = constant
(2.27)
This explains that Angular Momentum in a central force field is a conservative physical
quantity.
2.4.3 Torque (力矩) and Moment of inertia (惯量矩)
A torque is defined as
=r×F
(2.28)
Consider a very small body of mass m attached to the end of a string and accelerating in a
circle of radius R as a result of an applied force F acting tangentially to the circle. If Newton’s
second law of motion is applied to this situation, we have
F = ma,
Where the vector notation is dropped and we recognize the acceleration is tangentially to the
circle because the applied force is in tangent direction.
It is known that from (2.24)
a = R 
Furthermore, as r and F are orthogonal (perpendicular) here,

= | r × F | = r F sin90° = r F
∴
 = R (m a) = R m (R )
(using r = R)
2
= m R 
= I 







（







Where I = m R2 is called the moment of inertia and  = I  is the alternative expression of
Newton’s second law. It is also called rotational law.
§2.5 Elasticity of Materials
In this section, the stress, strain and modulus will be discussed. The contents are the
bases of material properties and will be used in the following chapters.
2.5.1 Stress
The definition of the stresses is
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Chapter 2. Fundamental Laws of Mechanics
dF
(2.30)
dA
Whew dF is the element of force suffered by material on dA area. Obviously, stress has
dimensions of pressure (force/area), and we often measure it in pascals, (1 N/m2 = 1
pascal = 1 Pa). Stress can be classified into two different types. One is called normal
stress or stretching stress, the other is called shearing stress. The normal (stretching)
stress is perpendicular to the surface exerted by a force. It is expressed by
dF

(2.31)
dA
Of course, it is equal to F/A if the force is uniform on the area. Shearing stress is parallel
to the acting area, expressed by
dF

(2.32)
dA
Same as above, it is equal to F/S if the area is uniformly exerted by the force.
Stress 
Fig. 2.3 showing different forces acting on an object
2.5.2 strain（应变）
There are three kinds of strains, which are stretching, volume and shearing strains.
The definition of the three strains is given below respectively.
1. Stretching (tensile) strain is defined by
L

(2.33)
L0
where L = L0–L denotes the length change and L0 is the original length of that object.
2. Volume strain, expressed by , is defined by
V
 
V0
(2.34)
where V = V0–V, V0 is the volume before being depressed and V is the volume under
stain. The minus sign means that the bulk of object is always depressed and becomes
smaller.
3. Shearing strain, denoted by , is defined as
x

 tan 
(2.35)
h
where x is the length change on the direction of acting force, h is the height of the
object and  is the related angle deviated from the vertical line (see Fig. 2.4). Shearing
strain is related to the shearing stress, caused by the pair of shearing forces (also see Fig.
2.4)
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Medical Physics
x
F
F

h
Fig. 2.4 Showing the shearing strain related elements
4. Poisson ratio
When an object is elongated, its cross-section will be getting smaller (contractive). The two
sides of the cross-section are originally labeled as a0 and b0 respectively and its length is
labeled as L0. After elongation, there are some changes along the three edges. The Poisson
ratio is defined as
a L b L


(2.36)
a0 L0
b0 L0
If the material is incompressible,
1
 ，
(2.37)
2
and for other materials,  < ½
2.5.3 Elastic moduli （弹性模量）
1. stretch modulus
A straight wire doesn’t stretch as easily as a coiled wire, but its length does increase when
a force is applied. The wire made of different material will have a different increase. How
far the wire stretches depends on the stress and strain and finally depends on the material
properties. In order to describe such a property, the concept of modulus is introduced.
A stretch modulus E of the wire is the ratio of stress to strain when the length of the wire
changes. It is also called the elastic modulus or Young’s modulus and denoted by Y or E

E
(2.38a)

where  denotes normal stress,  is tensile strain. We can also express it as
streching stress dF / dA
(2.38b)
stretch mosulus 

stretching strain L / L0
2. Shear modulus
Under certain conditions, a force applied to a solid object can change the shape of the
object. Fig. 2.4 illustrates the result of applying a large horizontal force to the top of a
rectangular block welded to a horizontal steel plane. A shearing strain is the displacement
divided by height. If the force is not great enough to produce a permanent distortion of the
block, the block will return to its original shape when the force is removed. The shearing
stress Force/Area is the applied force divided by the area. The shear modulus of an object
is the ratio of shearing stress to strain when the shape of the object changes.
shearing stress dF / dA 
(2.39)
shear mosulus G 


shearing strain
x / h 
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Chapter 2. Fundamental Laws of Mechanics
3. Bulk modulus
A solid object, such as a steel cylinder or a copper block, subjected to a high pressure
decreases slightly in volume. For a given substance, the relative change in volume in
proportional to the applied pressure. The bulk modulus K of an object is the ratio of stress
to strain when the volume of the object changes. That is the bulk modulus is equal to the
volume stress divided by volume strain.
P
P
K 
 V0
(2.40)

V
The minus sign indicates that the increase in pressure causes a decrease in volume.
Example 2.7 An aluminum cylinder 25 cm long and 8 cm in diameter is lowered
approximately one kilometer into the ocean where the pressure is 1.0107N/m2 greater than
the atmospheric pressure. The bulk modulus of aluminum is 7.51010 N/m2. Calculate the
decrease in volume produced by this extreme pressure.
Solution: the original volume of the cylinder is
V  r 2h   (0.04m)2 (0.25m)  1.26  103 m3
Now, you can use Eq.(2.40) to find the decrease in volume:
P  V0
(1  107 N / m2 )(1.26  103 m3 )
V  

 1.68  10 7 m3
10
2
K
7.5  10 N / m
2.6 Mechanical properties of bones and muscles
This part is taken as a self-study reading material in the third section of chapter two in the
textbook written in Chinese.
Problems
A. Questions
1. What is the difference between displacement and distance?
2. What is the difference between velocity and speed?
3. Is it possible for a body to move in a constant speed but a variable velocity?
4. When a body has a constant acceleration, does the direction of its velocity change or
not? If yes, give one of your examples.
5. If an object is acted on under several forces, does the object have to have acceleration?
6. Do the direction of the net force acted on a body and the direction of motion of the
body have to be the same?
B. Problems
1. A 3 kg point mass moves parallel to the x axis along the line y =2 m. the speed of the
object is 4m/s. Find (a) the linear momentum and (b) the angular momentum of this
object.
2. An object moves according to the equation of motion




r (t )  (3  4t 2 )i  (6t 2  8t 4 ) j  3k
Find:
(i).
The initial position of the object; (r(0) =3 k)
(ii).
The velocity of the object at t = 2 second and the angle with zaxis also at t =2 s.
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Medical Physics
(iii).
The magnitude of its acceleration at any moment. (hint: find

d 2r
acceleration by 2 and then using | a | a x2  a y2  a z2 )
dt
3. A force acting on an object has an angle of  with the object’s displacement.
3
(i)
Suppose the magnitude of the force is 4 N and the object moves for 2 meters
2 

under the force. Find the work done by the force using work   F  dr .
0
(ii)
4.
5.
6.
7.
8.
Suppose the initial velocity of the object is zero and final speed of the object
is 6 ms-1. Calculate the mass of the object if it moves on a smooth horizontal
plane. (hint: the change of kinetic energy is equal to the work done by the
force and using the result in (i).)
A particle with mass m moves around a circle of radius r. Suppose that its angular
displacement can be described by a function of   3t 3  3t 2  2 . Find
(i)
its angular velocity at t = 5 second and its tangential speed at t = 2 second;
d
(hint:  
, v  r )
dt
(ii)
its angular acceleration at t = 5 second and tangential acceleration at t =2
d 2 d
second; (hint:   2 
a  r )
dt
dt
(iii)
the magnitude of its angular momentum to the center of the circle at t = 6
  

| L | rp sin  ,  should be equal to ninety
second. (hint: L  r  p,
degree in circular motion case)
Prove equation (2.37)
A compressional stress of 5.09105N/m2 is applied to a steel rod 10cm long and
1.0cm in diameter. The elastic (Young’s) modulus of steel is 201010N/m2. What
strain does this stress produce?
The shear modulus of cast-iron (铸铁) is 4.61010 N/m2. A 10-cm cube of iron is
distorted by a horizontal force of 250N. What horizontal displacement will be
produced?
A vertical steel shaft, having a diameter of 1.0 cm, is fastened to a horizontal plate.
When a horizontal force of 150 N is applied to the top of the shaft, 0.5 m above the
plate to which is fastened, what horizontal displacement will be produced? (The shear
modulus of steel is 8.01010 N/m2.)
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