Wednesday, May 7, 2008

Concept Review - Chapter 3 Rest and Motion: Kinematics

Rest, Motion and Reference Frame

Motion is a combined property of the object under study and the observer. There is no meaning of rest or motion without the viewer.

For example, a book placed on the table remains on the table and we say that it is not moving, it is at rest. However, if we station ourselves on the moon, the whole earth is changing its position, and os the room, the table and the book are all continuously changing their positions.

To locate the position of a particle we need a frame of reference. We can fix up three mutually perpendicular axes, name them X,Y and Z and then specify the position of the particle with respect to that frame. Then if the coordinates of the particle change with respect to time, we can say that the body is moving with respect to this frame.

Many times the choice of frame is clear from the context.

The magnitude of the displacement is the length of the straight line joining the initial and final position.

Displacement has magnitude as well as direction (initial position and final position).
It is a vector quantity.

Displacements add according to triangle rule of vectors.

For example, if a book kept on table is displaced and the table is also displaced. The net displacement of the book is obtained by the vector addition of the two displacements.


Average speed

Instantaneous speed

Average velocity

Average speed and average velocity of a body over a specified time interval may not turnout to be same.

Example See the worked out example 2 of HC Verma's book.
The teacher made 10 rounds back and forth in the room and the total distance moved is 800 feet (10 rounds back and forth of 40 ft room). As the time taken is 50 minutes, average speed is 800/50 = 16ft/min.
But because he went out of the same door that he has entered, displacement is zero and hence average velocity is zero.

Instantaneous velocity

Average acceleration

Instantaneous acceleration

Motion in a straight line

Choose the line as the X-axis

Position of the particle at time t is given by x.

Velocity is v = dx/dt

acceleration is a = dv/dt = d²x/dt²

If accelaration is constant dv/dt = a (constant)
initial velocity = u (at time t =0)
final velocity = v (at time t)

The v = u+at

x = distance moved in time t = ut+½at²


v² = u²+2ax

Motion in a plane

Motion in plane is described by x coordinate and y coordinate.
The x-coordinate, the x component of velocity, and the x component of acceleration are related by equations of straight line motion along X axis.
Similarly y components.


Projectile motion is an important example of motion in a plane.

Vertical motion of the projectile is the motion along Y axis and horizontal motion is motion along X axis.

Terms used in describing projectile motion

Point of projection
Angle of projection
Horizontal range
Time of flight
Maximum height reached

The motion of projectile can be discussed separately for the horizontal and vertical parts.

The origin is taken as the point of projection.
The instant the particle is projected is taken as t = 0.
X-Y plane is the plane of motion.
The horizontal line OX is taken as the X axis.
Vertical line OY is the Y axis.
Vertically upward direction is taken as positive direction of Y

Initial velocity of the particle = u
Angle between the velocity and horizontal axis = θ

ux – x-component of velocity = u cos θ
ax – x component of acceleration = 0

uy – y component of velocity = u sin θ
ay = y component of acceleration = -g

Horizontal motion – Equations of motion
ux = u cos θ
ax = 0
vx = ux +axt = ux = u cos θ (as ax = 0)
Hence x component of the velocity remains constant.
Displacement in horizontal direction = x = uxt+1/2ax t²
As ax = 0, x = ux t = ut cos θ

Vertical motion – Equations of motion
uy = u sin θ
ay = -g
vy = uy – gt
Displacement in y direction = y = uyt – ½ gt²
vy² = uy² - 2gy

Time of flight of projectile

A projectile is projected from the ground at point O, and after some travel in the space, it reaches the ground at point B. The time taken for this travel is called time of flight of the projectile. Over this time, the displacement in y direction becomes zero.

Hence we can write y = uyt – ½ gt² = u sin θ*t - ½ gt²

solving we get T = (2u sin θ)/g
Time of flight of the projectile = (2u sin θ)/g

Range of the projectile:

The distance OB travelled by the projetile in the horizontal direction is the range.

OB = (u²sin 2θ)/g

Maximum height reached

When the projectile reaches the highest level in vertical direction, the vertical component of the velocity becomes zero.

The time taken for the vertical component of velocity to become zero is
vy = 0 = uy - gt = u sin θ - gt
So t = (u sin θ)/g
So time taken for vertical component of velocity to become zero is (u sin θ)/g.
Note that this time is half of Time of flight.

so maximum height reached in t = (u sin θ)/g = (u² sin²θ)/2g

Expressing velocity w.r.t. one Frame w.r.t. to a different frame

If XOY is one frame called S and X'O'Y' is another frame called S' we can express velocity of a body w.r.t. S as a combination of velocity of body w.r.t. to S' and velocity of S' w.r.t to S.

V(B,S) = V(B,S')+V(S',S)

V(B,S) = velocity of body wrt to S)
V(B,S') = velocity of body wrt to S')
V(S',S) = velocity of S' wrt to S)

we can rewrite above equation as

V(B,S') = V(B,S)- V(S',S)

We can interpret the above equation in terms of two bodies. Assume S', and B are two bodies. If we know velocities of two bodies with respect to a common frame (in this case S)we can find the velocity of one body with respect to the other body (V(B,S')

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