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?

Contents

**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.

Since the velocity changes (due to continuous 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 Example:
**

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 Example :**

(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:

** v=2nr/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.

a= BD

a=BD/AD

a=BD/t

BD = at

v=u+at

**(2) Derivation of S = ut +1/2 x at**^{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

=(OA x OC) + 1/2 x AD x BD)

=(u x t) + (1/2 x t x at)

S = ut + 1/2 x at^{2}

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

^{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 OABC

s= Sum of parallel sides x height / 2

s= (OA+OB) x OC / 2

s=(u+v) x t /2

v=u + at

at = v – u

t= v-u /a

s= (u+v) x (v- u)/2a

2as = v^{2 } – u^{2 }

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

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