LESSON 1: RECTILINEAR TRANSLATION MOTION

1. PROBLEM:
A projectile is fired upward from a 15.3 m cliff at a speed of 19.6 m/s and allowed to fall
into a valley below. The acceleration g due to Earth’s gravity is about 9.8 m/s2 , or about
32 ft/s 2, downward.

(a) Given that a(t) = –9.8 m/s2 , find v(t) and use it to find the time at which the
projectile reaches its maximum height. Find the maximum height of the projectile
using geometry.

SOLUTION:

If a(t) = –9.8, then v(t) = –9.8t + C for some constant C. Note that v(0) = C in this
problem, so that C is the initial velocity. Therefore, v(t) = –9.8t + 19.6 m/s. The maximum
height occurs when the velocity is zero, so –9.8t + 19.6 = 0
implies that the maximum height occurs at t = 2 seconds.
Although we do not have a position function, we can find
the maximum height using geometry. Since the maximum
height in this problem is simply the displacement over the
first 2 seconds, and the displacement is the net area
bounded by the velocity curve, we see that the maximum
height is the area of the shaded triangle, which is
(1/2)(2)(19.6) = 19.6 meters.

(b) Use geometry to find the displacement and total distance traveled over the
interval [0, 3].

Solution: The displacement is the net area bounded by v(t), and the total
distance traveled is the total area. In this case, we can use the two triangles in
the figure to compute displacement on [0, 3] as (1/2)(2)(19.6) – (1/2)(1)(9.8) =
14.7 meters, and the total distance traveled on [0, 3] as (1/2)(2)(19.6) +
(1/2)(1)(9.8) = 24.5 meters.
 2. PROBLEM:

A man is rotating a sling with a stone in a 2 m long at the rate of 18 revolutions every 9
seconds. Find the rectilinear speed of the stone.

SOLUTION:

Now the distance traveled by the stone in 9 s is d = (18) × 2 π r

= 18 × 2 π × 2m
= 288 π m.
So the Rectilinear speed of the stone is v = d/t
= 288πm/9s

= 32 π m/s.
3. PROBLEM:

A Clown riding a uni-cycle has the wheel base diameter as 60 cm. if the wheel rotates at
the rate of 120 revolutions per minute, then find the Rectilinear speed at which the clown is
traveling.

SOLUTION:

Step 1:
A clown riding a uni-cycle has the wheel base diameter as 60 cm. if the wheel rotates at the
rate of 120 revolutions per minute, then find the Rectilinear speed at which the clown is
traveling.

Step 2:
The Angular speed (ω) of the wheel is (2 π) x 120 = 240 π / min
The wheels have a radius of 30 cm [60m/2 = 30 cm]
The Rectilinear speed = v = r × (angular speed)
v = r ω.
Rectilinear speed v = r ω = 30 x 240 ππ /min = 7200 cm / min.

4. PROBLEM:

Particles A and B are elevated 12 meters high from a given reference base. Particle A is
projected down an incline of length 20 meters at the same time particle B is let to freely fall
vertically. Find the velocity of projection of particle A if both particles strikes the base at the
same time.

SOLUTION:

For particle B
h=12gt2
12=12gt2
t=√24g
 For particle A
vAi−y/ vAi =12/20
vAi−y=0.6v Ai
Apply the formula s = vot + 0.5at2 to the vertical movement of A
s=vot+12at2
h=vAi−yt+12gt2
12=(0.6vAi)24g−−−√+g2(24g−−−√)2
12=0.6vAi24g−−−√+g2(24g)
12=0.6vAi24g−−−√+12
vAi=0 answer

5. PROBLEM:

If the position of a particle along x axis varies in time as:

x = 2t 2 − 3t + 1
Then :
1. What is the velocity at t = 0?
2. When does velocity become zero?
3. What is the velocity at the origin?
4. Plot position time plot. Discuss the plot to support the results obtained for the
questions above.

SOLUTION:

We first need to find out an expression for velocity by differentiating the given function of
position with respect to time as:
v = t 2t 2 − 3t + 1 = 4t – 3

(i) The velocity at t = 0,

v = 4 x 0 − 3 = − 3m/s
(ii) When velocity becomes zero :
For v = 0,
4t − 3 = 0
⇒ t = 3 4 = 0.75 s
(iii) The velocity at the origin :
At origin,
x = 0, x = 2t 2 − 3t + 1 = 0
 ⇒ 2t 2 − 2t − t + 1 = 0
⇒ 2t ( t − 1 ) − ( t − 1 ) = 0
⇒ t = 0.5 s, 1
This means that particle is twice at the origin at t = 0.5 s and t = 1 s. Now,
v ( t = 0.5 s ) = 4t − 3 = 4 x 0.5 − 3 = −1 m/s.
Negative sign indicates that velocity is directed in the negative x direction.
v ( t = 1 s ) = 4t − 3 = 4 x 1 − 3 = 1 m/s.
Position- Time Plot
 LESSON 2: RECTILINEAR MOTION WITH CONSTANT ACCELERATION

1. PROBLEM:

A cricket ball is projected vertically from the top of a building from a position 40.0 m above
the ground below. Consider the three cases: (a) The ball leaves the hand from rest. (b) The
ball is projected vertically downward at 12.5 m.s-1 For cases (a) and (b) (1) What are the
velocities of the ball after it has been falling for 1.23 s? (2) What are the positions of the ball
after it has been falling for 1.23 s? (3) What are the velocities of the ball as it strikes the
ground? (4) What are the times of flights for the ball to reach the ground? For case (c) only
(5) What is the time it takes to reach its maximum height? (6) What is the maximum height
above the ground reached by the ball? (7) What is the time for the ball to return to the point
at which it was thrown? (8) What is the velocity of the ball as it returns to the position at
which it was thrown?

SOLUTION:

1 How-to-approach the problem Identify / Setup Draw a sketch of the situations. Include
motion maps. Show the frame of reference (coordinate axis & origin). State the type
(category) of the problem. Write down all the equations that might be relevant. Write down
all the given and know information including units. Write down all the unknown quantities
including their units. Execute / Evaluate Use the equations to find the unknowns. Check that
your answers are sensible, significant figures, units and that you have documented your
answer with comments and statements of physical principles. The problem type is free fall –
uniformly accelerated motion in the vertical y direction: a = g = -9.81 m.s -2 and the
displacement s corresponds to the changes in the vertical position of the ball. We can solve
the problem using the equations

(1) v= u  at

(2) s=ut+ ½ at2

(3) v2= u2 + 2as (4) v avg = u+v/2

(5) s= v avg t

Initial state: t = 0 s = 0 u = 0 a = g = -9.81 m.s-2

Final state when t = 1.23 s: s = ? m v = ? m.s-1

Eq(1)  v  u  at 0 (-9.81)(1.23)m.s =-12.1 m.s

Eq(2)  s= ut + ½ at2=0(0.5)(-9.81)(1.23)2 m=- 7.42m

Final state when s = -40.0 m (as ball strikes ground): v = ? m.s-1 t = ? s

CASE B

Initial state: t = 0 s = 0 u = -12.5 m.s-1 g = -9.81 m.s-2

Final state when t = 1.23 s s = ? m v = ? m

Eq(1)  v=u+ at=-12.5+(-9.81)(1.23)m.s-1= -24.6 m.s-1

Eq(2)  s= ut+ ½ at2=(-12.5) (1.23)+ (0.5)(-9.81)(1.23)2 m=-22.8 m

Final state when s = -40.0 m (as ball strikes ground): v = ? m.s -1 t = ? s

2. PROBLEM:

Ball tossed with 10 m/s vertical velocity from window 20 m above ground.

Determine:

• velocity and elevation above ground at time t,

• highest elevation reached by ball and corresponding time, and

• time when ball will hit the ground and corresponding velocity.
3. PROBLEM:

Spotting a police car, you brake a Porsche from a speed of 100 km/h to a speed of 80.0
km/h during a displacement of 88.0 m, at a constant acceleration.

(a) What is that acceleration?

(b) How much time is required for the given decrease in speed?

a)

b)

4. PROBLEM:

Brake mechanism used to reduce gun recoil consists of piston attached to barrel moving in
fixed cylinder filled with oil. As barrel recoils with initial velocity v 0, piston moves and oil is
forced through orifices in piston, causing piston and cylinder to decelerate at rate
proportional to their velocity.

Determine v ( t), x ( t), and v (x).

SOLUTION:
5. PROBLEM:
 LESSON 3: RECTILINEAR MOTION WITH VARIABLE ACCELERATION

1. PROBLEM:

Rotation of the arm about O is defined by θ = 0.15 t2 where θ is in radians and t in

seconds. Collar B slides along the arm such that r = 0.9 - 0.12 t 2 where r is in meters.
After the arm has rotated through 30 o ,
Determine (a) the total velocity of the collar,
(b) the total acceleration of the collar, and
(c) the relative acceleration of the collar with respect to the arm.

SOLUTION:
2. PROBLEM:

The acceleration, a , of a particle, at time t seconds is given by:

a=4−t20 ms−2.

This model is valid for 0 t 80. Given that the particle starts at rest, find the distance
travelled by the particle when t= 80 .

SOLUTION:

First integrate the acceleration to obtain the velocity.

v= 4−t20 dt=4t−40t 2+c1

To find the value of the constant c1 , note that the particle is initially at rest, so
that v=0when t=0. Substituting these values shows that c1=0. Hence the velocity is:

v=4t−40t 2 ms-1

The displacement of the particle can be found by integrating the velocity:

s= 4t−40t 2 dt=2t 2−t 3120+c2

To find the constant c2 , assume that the particle starts at the origin, so that s=0 when t=0.
Hence c2=0 and the displacement at time t is given by:

s=2t 2−t 3120 m

To find the distance travelled substitute t=80.

s=2 802−120803=8530 m (to 3sf)

3. PROBLEM:

The acceleration, a ms−2, of the cyclist, at time t seconds is given by: a=1−t15 for0 t 15.

The cyclist travels along a straight road.

a) Given that the cyclist starts at rest, find an expression for the velocity cyclist at time t.
b) What is the speed of the cyclist after 15 seconds?

c) How far does the cyclist travel in the 15 seconds?

SOLUTION:

A)
= 2152−90153 −0
v= adt=t−t230+c
=75 m
t=0 v=0 c=0

v=t−t2/30

B)

v(15)=15−152.30=7.5 ms-1

C)

s(15)= 015vdt = 015(t−t2/30)dt = t2/2−t3/90

015
4. PROBLEM:

A van slows down from a speed of 20 ms−1 to 5 ms−1 in 40 seconds. At time t seconds
the acceleration, a ms−2 is given by a=−kt, where k is a constant.
a) Find k.
b) Find the distance that the car travels during the 40 seconds.

SOLUTION:
a)

v= adt
= −kt dt
=−2kt2+c t=0 v=20 c=20
v=20−2kt2
Using t=40 v=5gives
5=20−5k 402
800k=15
k= 15/800=3/160

a)
Using the expression for v obtained in part a),

s= 040 20−3t/2320 dt = 20t−t3/960 040 = 20 40−403/960 −0

=733 m

5. PROBLEM:

The acceleration, a ms−2, at time t seconds of a particle, which moves along a straight line,
is given by:
a=t25 .
a) Given that the initial velocity of the particle is 2 ms−1, find an expression for the velocity
of the particle.
b) Find the distance that the particle travels in the first 50 seconds of its motion.

SOLUTION:
s(50)= 050vdt = 050 t2/50+2 dt
a)

= t3/150+2t 050
v= adt=t2/50+c t=0
= 150503+2 50 −0
v=2 c=2 v=t2/50+2

b) =933 m

Using the result for the velocity obtained

in part a)
 LESSON 4: KINETICS OF RECTILINEAR TRANSLATION

1. PROBLEM:

The acceleration of the mass of a spring-mass system is given as a function of the

displacement: a (x) = − (k/m)x. If the mass is started from rest at a displacement of – ρ
determine the velocity of the mass as a function of the displacement.

SOLUTION:
If we note that a (x) = d2x/dt2 we can write (d2x / dt2) + (k / m)x = 0
from the given acceleration. This has the general solution
x = A sin ωt +B cos t
where A, B are arbitrary constants, t is the time and ω = √(k/m) is the natural frequency. A
and B are found by noting that at t = 0 the displacement x is – ρ and the velocity (dx / dt) is
zero (at rest). The first condition gives
− B. To apply the second condition we first find the velocity
v = (dx / dt) = A ω cos ωt – B ω sin t
At t = 0 this gives
0=A
Now if we substitute Eqs. (4) and (2) into Eqs. (3) and (1) we find,
after re-arranging them
− [x/ρ] = cos ωt
[v/ρω] = sin
Now, if we square each of these, add them together and observe
that cos2 ωt + sin2 ωt = 1, we get
(x2 / ρ2) = (v2 / ρ2 ω2) = cos2 ωt + sin2 ωt = 1
x2 + (v2 / ω2) = ρ2
(v2 / ω2) = (ρ2 – x2)
v2 = (ρ2 – x2) ω2
v2 = (ρ2 – x2) (k/m)

2. PROBLEM:

The motion of a particle along a straight line is described bythe equation x = t3 − 3t2 − 45t +
50, where x is expressed infeet and t in seconds. Compute a) the time at which v (t) = 0,b)
the position and distance traveled by the particle at thattime, c) the acceleration of the
particle at that time, d) thedistance traveled by the particle from t = 4 sec. to t = 6 sec.

SOLUTION:
We know from basic concepts in mechanics that if the position of aparticle as a function of
time is x = f (t), then the velocity v = (dx / dt) = [{d f (t)}/dt] and the acceleration a = (dv / dt)
= [{d2 f (t)} / dt2]. Therefore x = t3 – 3t2 – 45t + 50 v = 3t2 – 6t – a = 6t 6. These
equations can be applied to solve the above problems.
(a) setting v = 0 in equation (b) we obtain 3t2 – 6t – 45 = 0.
Applying the quadratic formula we obtain the roots t = − 3 s and
t = + 5 s. Since the negative value represents the time before the particle
started moving, it is rejected. The velocity is zero at the end of 5 seconds.
(b) The position at t = 5 s can now be obtained by substituting t = + 5
sec into equation (a) x (5) = (5)3 – 3 (5)2 – 45 (5) + 50 = − 125 ft.
The initial position is x (0) = + 50 ft, also from equation (a). Before
asserting that the distance traveled is x (5) – x (0) = − 125 – 50 = − 175 ft, we must be
certain that the particle did not retrace its path at some time between t = 0 s and t = 5
s. In order to do so, the particle would have to stop (at least momentarily) before
reversing its motion. Since there was no solution for v = 0 between 0 and 5 seconds,
the particle must have moved directly from x (0) to x (5). Thus, equation (d) is the
correct answer to the
distance traveled by the particle.
(c) Substituting t = 5 s into equation (c) yields
a = 6 (5) − 6 = + 24 ft/s2.
(d) To solve this, consider the fact that the velocity equals zero at one
point during the interval t = 4 s to t = 6 s. Then separate the interval into two
parts, t = 4 sec to t = 5 sec and t = 5 sec to t = 6 sec.
From t = 4 sec to t = 5 sec we have
x (4) = (4)3 – 3 (4)2 – 45 (4) + 50 = − 114 ft from equation (a)
and d1 = x (5) – x (4) = − 125 ft – (− 114 ft)
= − 11 ft.
From t = 5 sec to t = 6 sec we have x (6) = (6)3 – 3 (6)2 – 45 (6) +
50 = − 112 ft hence d2 = x (6) – x (5) = − 112 ft – (− 125) ft = + 13 ft. Note
that d2 and d1 are of opposite sign thus proving thatthe
particle did retrace its path.
The total distance traveled is
D = |d1| + |d2| = 11 ft + 13 ft = 24 ft.

3. PROBLEM:
A stone is dropped into a well and the splash is heard two seconds later. If sound travels
1100 ft/sec, what is the depth of the well?
SOLUTION:

The total time to hear the splash is made up of t1, the time required for the stone to hit the
water and t2 the time required for the sound to travel back, t1 satisfies: h = 1/2
gt1 2 where g = 32.2 ft/sec2 is the acceleration due to gravity and there is no initial velocity .
t2 satisfies
h = vs t2 where vs is the velocity of sound. In both cases the
distance travelled is h, thus we may equate the two expressions
1/2 gt1 2 = vs t2. Upon substitution of available numerical values
we find that
t2 = [(1/2 gt1 2) / (vs)]
= [(16.1) / (1100)] t1 2
= 0.0146 t1 2.
The total time required is 2 sec. Thus,
t1 + t 2 = 2
and t1 + .0146 t1 2 = 2
or 0.146 t1 2 + t1 – 2 = 0.
 This equation is solved by means of the quadratic formula, which yields two values for t1
t1, 1 t1, 2 = [(−1 ± √{1 – 4 (.0146) (−2)}) / {2 (.0146)}].
The two values are
t1 = 1.94 sec, − 70.4 sec.
Negative time has no meaning here therefore we reject it and keep the first of the above
values. The depth of the well can now be found from
Eq. (1) to be
h = 1/2 (32.2) (1.944)2
h = 60.6 ft.

4. PROBLEM:

A particle moves from point A to B under the influence of avariable force. It has an
acceleration of 5 ft/s2 for 6 s, and then its acceleration increases to 7 ft/s2 until its has
reached the velocity of 100 ft/s. For sometime its velocity remains constant, then the
particle starts to decelerate and within 10 s stops at B. The total travel time from A to B is
46 s. Draw the a–t, v–t, and x–t curves, and determine the distance between points A and
B.

SOLUTION:

It is known that 0 < t < 6 a = 5 ft/s2, 6 < t < t2

a = 7 ft/s2, t2 < t < t3 a = 0, t3 < t < 46s a is not known. Since v = 100 ft/s at t = t2 it is possible
to solve for t2 using vf = v0 + a (tf –t0),
100 ft/s = v (6) + 7 ft/s2 (t2 – 6s).
Now to solve for v (6), use (a) again
v (t=6) = 0 + 5 ft/s2 ∙ (6s)
v (6) = 30 ft/s
Subtracting (b) into (a) yields
100 ft/s – 30ft/s = 7 ft/s2 (t2 – 6s)
t2 – 6s = (70 / 7) s
t2 = 6s + 10s = 16s.
t3 is easily obtained. Since the total trip takes 46 seconds and deceleration
starts 10 seconds from the end, t3 = 36 s. Now the value for the
deceleration again follows from equation (a),
0 = 100 ft/s + a (46s – 36s)
a = − (100 / 10) (ft / s2)
= − 10 ft/s2.

The curve containing this information is illustrated in Figure 1. Velocity −Time

Curve. Since the acceleration is either constant or zero, arid velocity is the integral of
acceleration, the velocity curve consists of straight lines connecting points whose
abscissa are the times when the acceleration changes (t = 0, 6, 16, 36, 46s) and
whose ordinates are the total area under the a–t curve up to that value of t.
t=0 v=0 (0, 0)
t = 6, v = (5) (6) (6, 30)
t = 16, v = (5) (6) + (7) (10) (16, 100)
t = 36, v = (5) (6) + (7) (10) (36, 100)
t = 46, v = (5) (6) + (7) (10) + (−10) (10) (46, 0)
Position−Time Curve. Position is the integral of velocity, and the points to be
connected on the x−t curve will be determined by the same method as those on the
v−t curve. However, now the velocity is constant only between t = 16s and t = 36s.
The only straight line segment will be present over this interval of the x−t curve. The
other sections will have parabolic curves, turning upward for t < 16s, and downwards
for t > 36s.
The defining points are:
t = 0, x = 0 (0, 0)
t = 6, x = 1/2 (6) (30) (6, 90)
 t = 16, x = 1/2 (6) (30) + 1/2 (10) (30 + 100) (16, 740)
t = 36, x = 740 + (20) (100) (36, 2740)
t = 46, x = 3740 + 1/2 (10) (100) (46, 3240)

5. PROBLEM:

A braking device in a rifle is designed to reduce recoil. It consists of a piston

attached to the barrel. The piston moves in a cylinder filled with oil. When the shot is
fired the barre land the piston recoil at an initial velocity V0. The oil under pressure
passes through tiny holes in the piston causing the piston and the barrel to
decelerate at a rate proportional to their velocity a = −kv. Express (a) v in terms of t,
(b) x in terms of t, (c) v in terms of x. Draw the corresponding motion curves.

SOLUTION:

(a) The basic definition of acceleration is

a = (dv / dt). Subtracting the expression given for the acceleration yields
– kv = (dv / dt). Rearranging to separate variable yields
− k dt = (dv / v).
Now it is possible to integrate equations (c) to get
− k t∫0 dt = v∫(v)0 (dv / v)
− kt = In (v / v0).
If antilogs of equation (d) are taken, we arrive at the desired form
v = v0 e−kt
This curve is shown in Figure 2

(b) Again, starting with a basic definition, this time for v, we have
v = (dx / dt)
Subtracting of the expression (e) into (f) yields
v0 e− kt = (dx / dt),
v0 e− kt dt = dx.
Integrating (g)
v0 t∫0 e− kt dt = x ∫ 0 dx,
(− v0 / k) e− kt | t 0 = x,
(− v0 / k) (e− kt – e0) = x,
x = (v0 / k) (1 − e− kt).
This curve is shown in Figure 3.
 (c) Part c may be done in two ways: Using the chain rule for differentials,
equation (a) may be rewritten a = (dv/dt) = (dx/dt) (dv/dx) = v (dv/dx). Subtracting a = − kv
in equation (i), we get
− kv = v (dv/dx)
or (dv/dx) = − k,
dv = − k dx.
Integrating to solve yields
v – v0 = − kx
or v = v0 – kx.
As an alternate method, equations (e) and (h) can be combined. From (e) we know e−
kt = (v / v0). Substituting for the exponential in (h) yields
x = (v0 / k) [1 – (v / v0)]
or kx = v0 – v
or v = v0 – kx.
This curve is illustrated in Figure 4.

LESSON 6: CURVILINEAR TRANSLATION MOTION IN TWO DIMENSION

1. PROBLEM:

A slotted link is rotating about fixed pivot O with a counterclockwise angular velocity
of 3 rad/s, and a clockwise angular acceleration of 2 rad/s2. The movement of the
link is causing a rod to slide along the curved channel, as shown. The radius of the
channel as a function of θ is given by, R = 0.7θ (with R in meters and θ in radians).
Determine the velocity and acceleration components of the rod at θ = 45°

SOLUTION:
The angle θ = 45° is equal to π/4 radians. In the equations, counterclockwise angular
velocity is positive, and clockwise angular acceleration is negative (since it acts to “slow
down” the rotational speed of the link).

The radial velocity of the rod is given by equation (1):

The circumferential velocity of the rod is given by equation (3):

The radial acceleration of the rod is given by equation (2):

The circumferential acceleration of the rod is given by equation (4):

2. PROBLEM:
 The position of a particle as a function of time is given by the following expression (SI
units):

Find the expression for the radius of curvature of the trajectory as a function of time
and compute the radius of curvature at the two instants t = 0 and t = 1.

SOLUTION:

Those two expressions can be obtained from the velocity and acceleration vectors.
In Maxima those vectors can be obtained in the following way

(%i1) vector_r: [5*t, 3*t^2/2, 2*(1-t^2)]$

(%i2) vector_v: diff (vector_r, t);
(%o2) [5,3t,-4t]
(%i3) vector_a: diff (vector_v, t);
(%o3) [0,3,-4]

The magnitudes of v and a can be found from the square root of the product of each of
these vectors with itself. The scalar product of two vectors is computed in Maxima writing a
dot between the two lists that represent the vectors:

(%i4) v: sqrt (vector_v.vector_v);

(%o4) /25t2 + 25
(%i5) a: sqrt (vector_a.vector_a);
(%o5) 5

Note that the acceleration is constant, which implies a trajectory either linear or parabolic.
To find the normal component of the acceleration vector, the tangential component can be
found first through the derivative v

(%i6) at: diff (v, t);

(%o6) 25t/ 25t2 + 25

and the normal component is found from equation 3.9:

(%i7) an: ratsimp (sqrt (a^2 - at^2));

(%o7) 5/ t2+1

The tangential and normal components of the acceleration both depend on time, in spite of
the acceleration being constant; that suggests that the radius of curvature does not remain
constant and, hence, the trajectory is parabolic. The expression for the radius of curvature
can be obtained from equation 3.11

(%i8) R: ratsimp (v^2/an);

(%o8) t2+1 (5t2+5)

When t = 0 and t = 1 the radius of curvature is then

(%i9) subst (t=0, R);

(%o9) 5
(%i10) float (subst (t=1, R));
(%o10) 14.14

3. PROBLEM:
 Starting from rest, a motorboat travels around a circular path of ρ = 50 m at a speed
that increases with time, v = (0.2 t2) m/s.

Find: The magnitudes of the boat’s velocity and acceleration at the instant t = 3 s.

SOLUTION:

1) The velocity vector is v = v ut , where the magnitude is given by

v = (0.2t2) m/s. At t = 3s:

v = 0.2t2 = 0.2(3)2 = 1.8 m/s

2) The acceleration vector is a = atut + anun = vut + (v2/ρ)un. .

Tangential component: at = v = d(.2t2)/dt = 0.4t m/s2 At t = 3s: at = 0.4t = 0.4(3)

= 1.2 m/s2 .

Normal component: an = v2/ρ = (0.2t2)2/(ρ) m/s2 At t = 3s: an = [(0.2)(32)]2/(50)

= 0.0648 m/s2

The magnitude of the acceleration is

a = [(at)2 + (an)2]0.5 = [(1.2)2 + (0.0648)2]0.5 = 1.20 m/s2

4. PROBLEM:

Kilauea in Hawaii is the world’s most continuously active volcano. Very active volcanoes
characteristically eject red-hot rocks and lava rather than smoke and ash. Suppose a large
rock is ejected from the volcano with a speed of 25.0 m/s and at an angle above the
horizontal, The rock strikes the side of the volcano at an altitude 20.0 m lower than its
starting point. (a) Calculate the time it takes the rock to follow this path

SOLUTION:

While the rock is in the air, it rises and then falls to a final position 20.0 m lower than its
starting altitude. We can find the time for this by using
y= y0+v0yt- 1/2gt2
If we take the initial position to be zero, then the final position is y= -20.0m. Now the
initial vertical velocity is the vertical component of the initial velocity, found from
v0y=v0 sin0o= (25.0 m/s) sin(35.0) = 14.3 m/s. Substituting known values yields
-20.0 m = (14.3 m/s) t-(4.90 m/s2) t2.
Rearranging terms gives a quadratic equation in :
(4.90 m/s2) t2- (14.3 m/s) t- (20.0m)=0
This expression is a quadratic equation of the form
at2+bt+c=0, where the constants are
A= 4.90,
B= -14.3, and
C =-20.0.
This equation yields two solutions:
t= 3.96 and
t=1.03. (It is left as an exercise for the reader to verify these solutions.) The time is t =3.96
s or-1.03. The negative value of time implies an event before the start of motion, and so we
discard it. Thus,
t=3.96 s
5. PROBLEM:

The motion of two particles (A and B) is described by the position vectors rA = [3t i + 9t(2 –
t) j] m rB = [3(t2 –2t +2) i + 3(t – 2) j] m

Find: The point at which the particles collide and their speeds just before the collision.

SOLUTION:

The point of collision requires that rA = rB, so xA = xB and yA = yB .

Solution: x-components: 3t = 3(t2 – 2t + 2)

Simplifying: t2 – 3t + 2 = 0

Solving: t = {3 ± [32 – 4(1)(2)]0.5}/2(1)

=> t = 2 or 1 s

y-components: 9t(2 – t) = 3(t – 2)

Simplifying: 3t2 – 5t – 2 = 0

Solving: t = {5 ± [52 – 4(3)(–2)]0.5}/2(3)

=> t = 2 or – 1/3 s

So, the particles collide when t = 2 s. Substituting this value into rA or rB yields xA = xB = 6
m and yA = yB = 0

2.) Differentiate rA and rB to get the velocity vectors.

vA = drA/dt = = [3i + (18 – 18t)j] m/s

At t = 2 s: vA = [3i – 18j] m/s
vB = drB/dt = xBi + yBj = [(6t – 6)i + 3j] m/s
At t = 2 s: vB = [6i + 3j] m/s

Speed is the magnitude of the velocity vector.

vA = (32 + 182) 0.5 = 18.2 m/s
vB = (62 + 32) 0.5 = 6.71 m/s
 TABLE OF CONTENTS
 LESSON 1: RECTILINEAR TRANSLATION MOTION
1.1. Problem no.1
1.2. Problem no.2
1.3. Problem no.3
1.4. Problem no.4
1.5. Problem no.5

 LESSON 2: RECTILINEAR MOTION WITH CONSTANT

ACCELERATION
2.1 Problem no.1
2.2 Problem no.2
2.3 Problem no.3
2.4 Problem no.4
2.5 Problem no.5

 LESSON 3: RECTILINEAR MOTION WITH VARIABLE

ACCELERATION
3.1 Problem no.1
3.2 Problem no.2
3.3 Problem no.3
3.4 Problem no.4
3.5 Problem no.5

 LESSON 4: KINETICS OF RECTILINEAR TRANSLATION

4.1 Problem no.1
4.2 Problem no.2
4.3 Problem no.3
4.4 Problem no.4
4.5 Problem no.5

 LESSON 5: DYNAMICS OF EQUILIBRIUM

5.1 Problem no.1
5.2 Problem no.2
5.3 Problem no.3
5.4 Problem no.4
5.5 Problem no.5

 LESSON 6: CURVILINEAR TRANSLATION MOTION IN TWO

DIMENSION
6.1 Problem no.1
6.2 Problem no.2
6.3 Problem no.3
6.4 Problem no.4
6.5 Problem no.5
 LESSON 5: DYNAMICS OF EQUILIBRIUM

1. PROBLEM:
Determine the force in each bar of the truss shown in Fig. P-405 caused by lifting the
120 kN load at a constant velocity of 8 m per sec. What change in these forces, if
any, results from placing the roller support at D and the hinge support at A?
SOLUTION:

The load is lifted at constant velocity (in dynamic equilibrium), thus, the forces involve is similar to
forces under static equilibrium.
ΣFV=0ΣFV=0
CV=120+120(35)CV=120+120(35)
CV=192kNCV=192kN
ΣFH=0ΣFH=0
CH=120(45)CH=120(45)
CH=96kNCH=96kN
By symmetry of vertical forces
RA=DV=12(192) RA=DV=12(192)
RA=DV=96kNRA=DV=96kN
ΣFH=0ΣFH=0
DH=96kNDH=96kN
At Joint A ΣFV=0ΣFV=0
FAB(35)=96FAB(35)=96
FAB=160kNFAB=160kN
ΣFH=0ΣFH=0
FAC=45FAB=45(160)FAC=45FAB=45(160)
FAC=128kNFAC=128kN
At Joint C ΣFV=0ΣFV=0
FBC=192kNFBC=192kN
ΣFH=0ΣFH=0
FCD+96=128FCD+96=128
FCD=32kNFCD=32kN
At Joint B ΣFH=0ΣFH=0
FBD(45)=160(45)FBD(45)=160(45)
FBD=160kNFBD=160kN
ΣFV=0ΣFV=0
FBD(35)+160(35)=192FBD(35)+160(35)=192
160(35)+160(35)=192160(35)+160(35)=192
192=192192=192 Check!
At Joint D
ΣFV=0ΣFV=0
96=160(35)96=160(35)
96=9696=96 Check!
ΣFH=0ΣFH=0
32+96=160(45)32+96=160(45)
128=128128=128 Check!
Summary
AB = 160 kN compression
AC = 128 kN tension
BC = 192 kN tension
CD = 32 kN tension
BD = 160 kN compression
With the roller support at D and the hinge support at A
 2. PROBLEM:

If Markov processes are scalar diffusion processes with drift and diffusion
coefficients −x and 2, respectively, then the corresponding invariant densities
are ρi(t,x)=(2π)−1/2exp{−[x−eoexp(−t)]2/2} where eo is an arbitrary real number. Among
these densities, there is only one which is stationary, namely ρ−(x)=(2π)−1/2exp(−x2/2).
This density is also the only DE density for the above processes.

SOLUTION:

Of considerable interest are generalizations of the notion of a DE solution of an ODE in

Euclidean space to ODEs in function Banach spaces. For instance, the generalization to
diffusion stochastic processes with nonlinear coefficients is presented in. The results
therein on DE (generally, nonstationary) processes of this type represent a direct extension
of the present work. A generalization of the DE-solution notion useful in a number of
applications would be that to advanced kinetic equations. The latter is also discussed
below.

Dynamic-equilibrium solutions of ODEs play an important role in mathematical models for

living systems. For instance, in the case when the DE solutions are uniformly bounded in
time, these solutions are necessary parts of the descriptions for homeorhesis, one of the
features which distinguishes living from nonliving matter. In the field of living systems, the
ODE-based models are the simplest and most transparent ones but not the most
comprehensive ones. A more adequate alternative is the models based on the active-
particle generalized kinetic theory (APGKT). They allow us to take the living-matter features
into account much better. Subsequently, the focus is on the capabilities of the APGKT
equations in describing processes in a living system. These equations are formulated in
terms of the generalized distribution functions (GDFs). Determination of the parameters and
other input characteristics of the equations is included in the related research activities. In
connection with this, we note the following.

Firstly, the DE GDF would provide the behavior of a modelled living system which is free
from the specificity of any initial condition. Loosely speaking, it can show the core evolution
of the system. For this reason, it may also serve as the first tool facilitating the validation of
the model.

Secondly, the above input data, in particular, constitute the initial GDF. This GDF is difficult
to obtain experimentally and to estimate theoretically. This problem would be eliminated by
the availability of the DE GDF.

More generally, the DE-solution techniques can noticeably contribute to the mathematical
tools necessary for analysis, in-depth understanding, and prediction for living systems and
other complicated structures in science and engineering.

Summing up the present work, we note the following results.

•The notion of a DE solution of an ODE is introduced and discussed in connection with

ODEs in Euclidean and function Banach spaces.

•The formulation of the DE solutions in terms of the general solution of the ODE is
presented (in Theorem 1).
•The significance of the initial condition in the ODE-based models is distinguished for the
ODEs of the two types, namely the ones which have DE solutions and the ones which do
not have them (see Remark 2).

•Theorem 2 indicates the sufficient condition for the DE solutions to be globally attracting.
This is the very feature that determines the importance of the DE-solution notion.

•The work include a few examples illustrating different aspects of qualitative properties of
the solutions introduced.

•The role of the DE solutions in some applied problems is discussed, also in connection to
the related results published before. The work analyzes in more detail the advantages of
the aforementioned solutions for the advanced models for living systems.

3. PROBLEM:
Three masses are attached to a uniform meter stick, as shown in Figure 12.9. The mass
of the meter stick is 150.0 g and the masses to the left of the fulcrum are m 1 = 50.0 g and
m2 = 75.0 g. Find the mass m3 that balances the system when it is attached at the right
end of the stick, and the normal reaction force at the fulcrum when the system is
balanced.

For the arrangement shown in the figure, we identify the following five forces acting on the
meter stick:

1. w1 = m1g is the weight of mass m1;

2. w2 = m2g is the weight of mass m2;
3. w = mg is the weight of the entire meter stick;
4. w3 = m3g is the weight of unknown mass m3;
5. FS is the normal reaction force at the support point S

SOLUTION:

With Figure 12.9 and Figure 12.10 for reference, we begin by finding the lever arms
of the five forces acting on the stick:
r1r2rrSr3=30.0cm+40.0cm=70.0cm=40.0cm=50.0cm−30.0cm=20.0cm=0.0cm(becau
seFSisattachedatthepivot)=30.0cm.r1=30.0cm+40.0cm=70.0cmr2=40.0cmr=50.0cm
−30.0cm=20.0cmrS=0.0cm(becauseFSisattachedatthepivot)r3=30.0cm.
Now we can find the five torques with respect to the chosen pivot:
τ1τ2ττSτ3=+r1w1sin90o=+r1m1g(counterclockwiserotation,positivesense)=+r2w2sin
90o=+r2m2g(counterclockwiserotation,positivesense)=+rwsin90o=+rmg(gravitational
torque)=rSFSsinθS=0(becauserS=0cm)=−r3w3sin90o=−r3m3g(counterclockwiserota
tion,negativesense)τ1=+r1w1sin⁡90o=+r1m1g(counterclockwiserotation,positivesen
se)τ2=+r2w2sin⁡90o=+r2m2g(counterclockwiserotation,positivesense)τ=+rwsin⁡90o
=+rmg(gravitationaltorque)τS=rSFSsin⁡θS=0(becauserS=0cm)τ3=−r3w3sin⁡90o=−r
3m3g(counterclockwiserotation,negativesense)
The second equilibrium condition (equation for the torques) for the meter stick is
τ1+τ2+τ+τS+τ3=0.(12.2.1)(12.2.1)τ1+τ2+τ+τS+τ3=0.

When substituting torque values into this equation, we can omit the torques giving
zero contributions. In this way the second equilibrium condition is
+r1m1g+r2m2g+rmg−r3m3g=0.(12.17)(12.17)+r1m1g+r2m2g+rmg−r3m3g=0.
Selecting the +y-direction to be parallel to F⃗ SF→S, the first equilibrium condition for
the stick is−w1−w2−w+FS−w3=0.(12.2.2)(12.2.2)−w1−w2−w+FS−w3=0.
 Substituting the forces, the first equilibrium condition becomes
−m1g−m2g−mg+FS−m3g=0.(12.18)(12.18)−m1g−m2g−mg+FS−m3g=0.
We solve these equations simultaneously for the unknown values m 3 and FS. In
Equation 12.17, we cancel the g factor and rearrange the terms to obtain
r3m3=r1m1+r2m2+rm.(12.2.3)(12.2.3)r3m3=r1m1+r2m2+rm.
To obtain m3 we divide both sides by r3, so we have
m3=r1r3m1+r2r3m2+rr3m=7030(50.0g)+4030(75.0g)+2030(150.0g)=315.0(23)g≃31
7g.(12.19)(12.19)m3=r1r3m1+r2r3m2+rr3m=7030(50.0g)+4030(75.0g)+2030(150.0g
)=315.0(23)g≃317g.

To find the normal reaction force, we rearrange the terms in Equation 12.18,
converting grams to kilograms:
FS=(m1+m2+m+m3)g=(50.0+75.0+150.0+316.7)×(10−3kg)×(9.8m/s2)=5.8N.(12.20)

4. PROBLEM:

A weightlifter is holding a 50.0-lb weight (equivalent to 222.4 N) with his forearm, as shown
in Figure 12.11. His forearm is positioned at ββ = 60° with respect to his upper arm. The
forearm is supported by a contraction of the biceps muscle, which causes a torque around
the elbow. Assuming that the tension in the biceps acts along the vertical direction given by
gravity, what tension must the muscle exert to hold the forearm at the position shown?
What is the force on the elbow joint? Assume that the forearm’s weight is negligible. Give
your final answers in SI units.

SOLUTION:

We see from the free-body diagram that the x-component of the net force satisfies the
equation

+Fx+Tx−wx=0(12.21)(12.21)+Fx+Tx−wx=0

and the y-component of the net force satisfies

+Fy+Ty−wy=0.(12.22)(12.22)+Fy+Ty−wy=0.

Equation 12.21 and Equation 12.22 are two equations of the first equilibrium condition (for
forces). Next, we read from the free-body diagram that the net torque along the axis of
rotation is

+rTTy−rwwy=0.(12.23)(12.23)+rTTy−rwwy=0.

Equation 12.23 is the second equilibrium condition (for torques) for the forearm. The free-body
diagram shows that the lever arms are rT = 1.5 in. and rw = 13.0 in. At this point, we do not need to
convert inches into SI units, because as long as these units are consistent in Equation 12.23, they
cancel out. Using the free-body diagram again, we find the magnitudes of the component forces:
FxTxwxFyTywy=Fcosβ=Fcos60o=F2=Tcosβ=Tcos60o=T2=wcosβ=wcos60o=w2=Fsinβ=Fsin60o=F
3–√2=Tsinβ=Tsin60o=T3–√2=wsinβ=wsin60o=w3–
√2.Fx=Fcos⁡β=Fcos⁡60o=F2Tx=Tcos⁡β=Tcos⁡60o=T2wx=wcos⁡β=wcos⁡60o=w2Fy=Fsin⁡β=F
sin⁡60o=F32Ty=Tsin⁡β=Tsin⁡60o=T32wy=wsin⁡β=wsin⁡60o=w32. We substitute these
magnitudes into Equation 12.21, Equation 12.22, and Equation 12.23 to obtain, respectively,

F2+T2−w2F3–√2+T3–√2−w3–√2rTT3–√2−rww3–
√2=0=0=0.F2+T2−w2=0F32+T32−w32=0rTT32−rww32=0.
When we simplify these equations, we see that we are left with only two independent
equations for the two unknown force magnitudes, F and T, because Equation 12.21 for the
x-component is equivalent to Equation 12.22 for the y-component. In this way, we obtain
the first equilibrium condition for forces

F+T−w=0(12.24)(12.24)F+T−w=0
and the second equilibrium condition for torques
rTT−rww=0.(12.25)(12.25)rTT−rww=0.
The magnitude of tension in the muscle is obtained by solving Equation 12.25:
T=rwrTw=frac13.01.5(50lb)=43313lb≃433.3lb.(12.2.4)(12.2.4)T=rwrTw=frac13.01.5(50lb)=4
3313lb≃433.3lb. The force at the elbow is obtained by solving Equation 12.24:
F=w−T=50.0lb−433.3lb=−383.3lb.(12.2.5)(12.2.5)F=w−T=50.0lb−433.3lb=−383.3lb. The
negative sign in the equation tells us that the actual force at the elbow is antiparallel to the
working direction adopted for drawing the free-body diagram. In the final answer, we
convert the forces into SI units of force. The answer is
F=383.3lb=383.3(4.448N)=1705Ndownward(12.2.6)(12.2.6)
F=383.3lb=383.3(4.448N)=1705Ndownward
T=433.3lb=433.3(4.448N)=1927N upward.
5. PROBLEM:

A uniform ladder is L = 5.0 m long and weighs 400.0 N. The ladder rests against a
slippery vertical wall, as shown in Figure 12.14. The inclination angle between the
ladder and the rough floor is ββ = 53°. Find the reaction forces from the floor and
from the wall on the ladder and the coefficient of static friction μsμs at the interface of
the ladder with the floor that prevents the ladder from slipping.

SOLUTION:
From the free-body diagram, the net force in the x-direction is
+f−F=0(12.28)(12.28)+f−F=0
the net force in the y-direction is
+N−w=0(12.29)(12.29)+N−w=0
and the net torque along the rotation axis at the pivot point is
τw+τF=0.(12.30)(12.30)τw+τF=0.
where τwτw is the torque of the weight w and τFτF is the torque of the reaction F.
From the free-body diagram, we identify that the lever arm of the reaction at the wall is r F =
L = 5.0 m and the lever arm of the weight is rw = L2L2 = 2.5 m. With the help of the free-
body diagram, we identify the angles to be used in Equation 12.10 for torques: θFθF = 180°
− ββ for the torque from the reaction force with the wall, and θwθw = 180° + (90° − ββ) for
the torque due to the weight. Now we are ready to use Equation 12.10 to compute torques:
τw=rwwsinθw=rwwsin(180o+90o−β)=−L2wsin(90o−β)=−L2wcosβ(12.2.11)(12.2.11)τw=rww
sin⁡θw=rwwsin⁡(180o+90o−β)=−L2wsin⁡(90o−β)=−L2wcos⁡β

tauF=rFFsinθF=rFFsin(180o−β)=LFsinβ.(12.2.12)(12.2.12)tauF=rFFsin⁡θF=rFFsin⁡(180o−
β)=LFsin⁡β.

We substitute the torques into Equation 12.30 and solve for F :

−L2wcosβ+LFsinβ=0(12.31)(12.31)−L2wcos⁡β+LFsin⁡β=0

F=w2cotβ=400.0N2cot53o=150.7N(12.2.13)(12.2.13)F=w2cot⁡β=400.0N2cot⁡53o=150.7N
We obtain the normal reaction force with the floor by solving Equation 12.29: N = w = 400.0
N. The magnitude of friction is obtained by solving Equation 12.28: f = F = 150.7 N. The
coefficient of static friction is μsμs = fNfN = 150.7400.0150.7400.0 = 0.377.

The net force on the ladder at the contact point with the floor is the vector sum of the
normal reaction from the floor and the static friction forces:

F⃗ floor=f⃗ +N⃗ =(150.7N)(−i^)+(400.0N)(+j^)=(−150.7i^+400.0j^)N.(12.2.14)(12.2.14)F→floo

r=f→+N→=(150.7N)(−i^)+(400.0N)(+j^)=(−150.7i^+400.0j^)N.

Its magnitude is

Ffloor=f2+N2−−−−−−−√=150.72+400.02−−−−−−−−−−−−√N=427.4N(12.2.15)(12.2.15)Ffloor
=f2+N2=150.72+400.02N=427.4N and its direction is

φ=tan−1(Nf)=tan−1(400.0150.7)=69.3oabovethefloor.(12.2.16)(12.2.16)φ=tan−1⁡(Nf)=tan−
1⁡(400.0150.7)=69.3oabovethefloor.

We should emphasize here two general observations of practical use. First, notice that
when we choose a pivot point, there is no expectation that the system will actually pivot
around the chosen point. The ladder in this example is not rotating at all but firmly stands
on the floor; nonetheless, its contact point with the floor is a good choice for the pivot.
Second, notice when we use Equation 12.10 for the computation of individual torques, we
do not need to resolve the forces into their normal and parallel components with respect to
the direction of the lever arm, and we do not need to consider a sense of the torque. As
long as the angle in Equation 12.10 is correctly identified—with the help of a free-body
diagram—as the angle measured counterclockwise from the direction of the lever arm to
the direction of the force vector, Equation 12.10 gives both the magnitude and the sense of
the torque. This is because torque is the vector product of the lever-arm vector crossed with
the force vector, and Equation 12.10 expresses the rectangular component of this vector
product along the axis of rotation.
 LESSON 1:
RECTILINEAR TRANSLATION MOTION
 LESSON 2:
RECTILINEAR MOTION WITH
CONSTANT ACCELERATION
 LESSON 3:
RECTILINEAR MOTION WITH
VARIABLE ACCELERATION
 LESSON 4:
KINETICS OF RECTILINEAR
TRANSLATION
 LESSON 6:
CURVILINEAR TRANSLATION
MOTION IN TWO DIMENSION
 LESSON 5:
DYNAMICS OF EQUILIBRIUM
 Republic of the Philippines
Eulogio “Amang” Rodriguez Institute of Science and Technology
Nagtahan, Sampaloc, Manila

College of Engineering

DYNAMICS OF
RIGID BODIES
(MCH 322)

Submitted by: Submitted to:

Student: Professor:
Frial, Romel C. Engr. Bernardito L. Fabro

152-0003
BSME 4B
 Republic of the Philippines
Eulogio “Amang” Rodriguez Institute of Science and Technology
Nagtahan, Sampaloc, Manila

College of Engineering

DYNAMICS OF
RIGID BODIES
(MCH 322)

Submitted by: Submitted to:

Student: Professor:
Frial, Romel C. Engr. Bernardito L. Fabro

152-0003