Question 1 What is meant by uniform circular motion?

Question 2 A body goes around the sun with constant speed in a circular orbit.Is the motion uniform or accelerated?

Question 3 A satellite goes around the earth in a circular orbit with constant speed.Is the motion uniform or accelerated?

Question 4 Give examples of circular motion?

Question 5 What is a centripetal force.Give example?

Question 6 A cyclist goes around a circular track once every 5 minute.If the radius of the circular track is 110 meters,calculate his speed?

Question 7 Derive first equation of motion by graphical method?

Question 8 Derive second equation of motion by graphical method?

Question 9 Derive third equation of motion by graphical method?

__Uniform circular motion__

When a body moves in a circle,it is called **circular motion.**

When a body moves along a circular path,then its direction of motion keeps on changing continuously.

Sine the velocity changes(due to continuos change in direction)therefore the motion along a circular path is said to be accelerated.

When a body moves in a circular path with uniform speed its motion is called **uniform circular motion**.

The velocity of the body moving in a circle with uniform speed is not uniform because the direction of motion is constantly changing.

For Ex:A stone tied to a thread is rotated in circular path with uniform speed in clockwise direction.

A ———>Speed is directed towards east.

B ———>Speed is directed towards south.

Since there is change in direction of speed,the velocity is not uniform.

The motion in a circle with constant speed is an example of accelerated motion.

The force which is needed to make an object travel in circular path is called **centripetal force**.

For Ex: 1)Movement of artificial satellite around earth .

2)Motion of moon around earth.

3)Motion of earth around sun.

4)Tip of seconds hand of a watch.

5)Athlete moving on a circular path.

The speed of a body moving along a circular path is given by:

2 π r

v=———

t

**Equation of motion by graphical method**.

**1)Derivation of v=u +at**

Initial velocity u at A =OA

Velocity changes from A To B in time t(uniform acceleration a)

Final Velocity v=BC

BC=BD+DC

v=BD+Ao

v=BD+u

Slope of velocity time graph is equal to acceleration a.

BD

a = ——–

AD

BD

a = ——-

t

BD = at

v=u+at

1

2)Derivation of S = ut + —- a t^{2}

2

The distance travelled by the body is given by area of the space between velocity time graph AB and time axis OC , which is equal to area of figure OABC.

Distance travelled=Area of figure OABC

=Area of rectangle OADC+Area of triangle ABD

1

=(OA *OC) + (—- *AD * BD)

2

1

=(u*t) + (—- * t * at)

2

1

S = ut + —- a t^{2}

2

3)Derivation of v^{2 } = u^{2 }+ 2as

The distance travelled by body in time t is given by area of figure OABC(which is a trapezium)

s = Area of trapezium OABCSum of parallel sides * height

s=———————————————————————-

2

(OA+OB) *OC

s=——————–

2

(u+v) * t

s=—————–

2

v=u + at

at = v – u

v – u

t = ————

a

(u+v) * (v- u)

s=———————

2a

2as = v^{2 } – u^{2 }

^{
}v^{2 } = u^{2 }+ 2as

Easy