1.1. To consider speed & velocity around a circleTo consider speed & velocity around a circle

2.2. To consider acceleration as a change in To consider acceleration as a change in velocityvelocity

3.3. To define an equation for acceleration when To define an equation for acceleration when an object moves in a circular path an object moves in a circular path

4.4. To define an equation for resultant force To define an equation for resultant force when an object moves in a circular pathwhen an object moves in a circular path

If an object is moving in a circle If an object is moving in a circle with a constant linear speed, its with a constant linear speed, its velocity is constantly changing....velocity is constantly changing....

Because the direction is Because the direction is constantly changing....constantly changing....

If the velocity is constantly If the velocity is constantly changing then by definition the changing then by definition the object is acceleratingobject is accelerating

If the object is accelerating, then If the object is accelerating, then an unbalanced force must existan unbalanced force must exist

Velocity v

acceleration

Velocity vB

Velocity vA

Consider an object moving Consider an object moving in circular motion with a in circular motion with a speed speed vv which moves from which moves from point A to point B in point A to point B in tt secondsseconds

(From speed=distance / time), the (From speed=distance / time), the distance moved along the arc AB, distance moved along the arc AB, ss is is vvtt

Velocity vB

Velocity vA

v

C A

B

The vector diagram shows The vector diagram shows the change in velocity the change in velocity vv : :

(v(vBB – v – vAA) )

Velocity vB

Velocity vA

The triangles ABC & the The triangles ABC & the vector diagram are similarvector diagram are similar

If If is small, then is small, then vv / / vv = = ss / / rr

Velocity vB

Velocity vA

v

C A

B

Substituting for Substituting for ss = = vvtt

vv / / vv = = vvtt / / rr

((aa = change in velocity / time) = change in velocity / time)

aa = = vv / / tt = = vv2 2 / / rr

We can substitute for angular velocity....We can substitute for angular velocity....

aa = = vv2 2 / / rr

From the last lesson we saw that:From the last lesson we saw that:

vv = = rr

aa = ( = (rr))2 2 / / r r (substituting for (substituting for vv into above) into above)

aa = = rr22

Velocity v

acceleration

We have seen already that We have seen already that anyany object travelling in a object travelling in a circular path is accelerating circular path is accelerating towards the centre of this towards the centre of this circular path.circular path.

This means that the This means that the resultant force is also resultant force is also pointing to the centre!pointing to the centre!

((ΣF = ma)ΣF = ma)

Velocity v

acceleration

But we know more….But we know more….

We have learnt two things We have learnt two things about the accelerationabout the acceleration

aa = = vv2 2 / / rr (1)(1)

andand

aa = = rr22

(2)(2)

YOUR TASK: Substitute the two equations (1) and (2) YOUR TASK: Substitute the two equations (1) and (2) in Newton’s second law (ΣF = ma) and find the in Newton’s second law (ΣF = ma) and find the magnitude magnitude

ΣFΣF = = mvmv2 2 / / rr

oror

ΣFΣF = = mm22rr

Velocity v

acceleration

You should have found out You should have found out that the that the magnitudemagnitude of the of the resultant force is:resultant force is:

So, for So, for anyany object of object of

•mass mass mm that travels at that travels at

•linearlinear speed speed v, v, moving in a circle of moving in a circle of

•Radius Radius rr, ,

We know the following about the We know the following about the resultant resultant force force ΣFΣF acting on it: acting on it:

•Direction: Direction: pointing towards the centrepointing towards the centre

•Magnitude: Magnitude: ΣFΣF = = mvmv2 2 / / r r = = mm22rr

Planet

satellite

Gravity

Str

ing

The point of support

∑F = FT

∑F= Ffriction

The wheel of the London Eye has a diameter of The wheel of the London Eye has a diameter of 130 m and takes 30 min for 1 revolution. 130 m and takes 30 min for 1 revolution. Calculate:Calculate:

a.a. The linear speed of the capsuleThe linear speed of the capsule

b.b. The linear accelerationThe linear acceleration

c.c. The resultant force acting on a person with a The resultant force acting on a person with a mass of 65 kgmass of 65 kg

The linear speed of the capsule :The linear speed of the capsule :

Using v = rUsing v = r

we know that we do a full revolution (2we know that we do a full revolution (2 rad) rad) in 30mins (1800s)in 30mins (1800s)

v = (130/2) x (2v = (130/2) x (2 / 1800) / 1800)

v = 0.23 msv = 0.23 ms-1-1

The linear acceleration:The linear acceleration:

Using a = vUsing a = v2 2 / r/ r

a = (0.23)a = (0.23)22 / (130/2) / (130/2)

a = 7.92 x 10a = 7.92 x 10-4-4 ms ms-2-2

The resultant force:The resultant force:

Using Using ΣΣF = maF = ma

ΣΣF = 65 x 7.92 x 10F = 65 x 7.92 x 10-4-4

ΣF = 0.051 NΣF = 0.051 N

An object of mass 0.150 kg moves around a An object of mass 0.150 kg moves around a circular path which has a radius of 0.420 m once circular path which has a radius of 0.420 m once every 5.00 s at a steady rate. Calculate:every 5.00 s at a steady rate. Calculate:

a.a. The speed and acceleration of the objectThe speed and acceleration of the object

b.b. The resultant force on the objectThe resultant force on the object

[.528 ms[.528 ms-1-1, 0.663ms, 0.663ms-2-2, 0.100N], 0.100N]