## Episode 225: Quantitative circular motion

Lesson for 16-19

- Activity time 150 minutes
- Level Advanced

This episode discusses: linear and angular velocity; degrees and radians; and angular acceleration. And is accompanied by worked examples, student question sets, a student experiment and a demonstration.

Lesson Summary

- Discussion: Linear and angular velocity (10 minutes)
- Worked example: Calculating
*ω*(10 minutes) - Discussion: Degrees and radians (5 minutes)
- Student questions: Calculating
*v*and*ω*(20 minutes) - Discussion: Angular acceleration (10 minutes)
- Worked example: Centripetal force (5 minutes)
- Student questions: Calculations on centripetal force (10 minutes)
- Student experiment: Verification of the equation for centripetal force (40 minutes)
- Demonstration: Alternative method of verifying the equation for centripetal force (40 minutes)

#### Discussion: Linear and angular velocity

Explain the difference between linear and angular velocity.

The instantaneous linear velocity at a point in the circle is usually given the letter *v* and measured in metres per second (m s^{-1}).

The angular velocity is the angle through which the radius to this point on the circle turns in one second. This is usually given the letter *ω* (Greek omega) and is measured in radians per second (rad s^{-1}) (See below)

Time period for one rotation:

*T* = distancevelocity

*T* = 2 π *r**v*

*T* = 2 π *ω*

Therefore linear and angular velocity are related by the formula:

Linear velocity = radius of circle × angular velocity,

*v* = *r* × * ω *

#### Worked examples: Calculating *ω*

A stone on a string: the stone moves round at a constant speed of 3 m s^{-1} on a string of length 0.75 m.

Linear velocity of stone at any point on the circle is 3 m s^{-1} directed along a tangent to the point.

Note that although the magnitude of the linear velocity (i.e. the speed) is constant its direction is constantly changing as the stone moves round the circle.

Angular velocity of stone at any point on the circle = 3 m s^{-1}0.75 m

*ω* = 3 rad s^{-1}

#### Discussion: Degrees and radians

You will have to explain the relationship between degrees and radians. The radian is a more natural

unit for measuring angles.

One radian (or rad for short) is defined as the angle subtended at the centre of a circle radius *r* by an arc of length *r*.

Thus the complete circumference 2 π *r* subtends an angle of 2 π *r**r* radians

Thus in a complete circle of 360 degrees there are 2 π radians.

Therefore 1 radian = 360 ° 2 π

1 radian = 57.3 °

#### Student questions: Calculating *v* and *ω*

Some radian ideas and practice calculations of *v*, *ω* .

Episode 225-1: Radians and angular speed (Word, 45 KB)

#### Discussion: Angular acceleration

If an object is moving in a circle at a constant speed, its *direction* of motion is constantly changing. This means that its linear velocity is changing and so it has a linear acceleration. The existence of an acceleration means that there must also be an unbalanced force acting on the rotating object.

Derive the formula for centripetal acceleration (
α = *v*^{ 2}*r*,
α = *r* *ω* ,
α = *r* *ω* ^{2}):

Consider an object of mass *m* moving with constant angular velocity ( *ω* ) and constant speed (*v* ) in a circle of radius *r* with centre O.

It moves from P to Q in a time *t* .

The change in velocity Δ *v* is parallel to PO and
Δ *v* = *v* sin( θ )

When θ becomes small (that is when Q is very close to P) sin( θ ) is close to θ in radians.

So
Δ *v* = *v* θ

Dividing both sides by *t* gives:

Δ *v**t* = *v* θ *t*

Since Δ *v**t* = acceleration
and
θ *t* = *ω* ,
we have

* α * =

*v*×

*ω*Since we also have

*v* = * ω * ×

*r*,

this can be written as

*α* = *v*^{ 2}*r*

* α * =

*v*×

*ω** α * =

*×*

*ω*^{ 2}

*r*Applying Newton's Second Law ( *F* = *m* × *a*) gives:

*F* = *m**v*^{ 2}*r*

*F* = *m**r** ω *

^{ 2}

This is the equation for centripetal force; students should learn to identify the appropriate form for use in any given situation.

#### Worked examples: Centripetal force

A stone of mass 0.5 kg is swung round in a horizontal circle (on a frictionless surface) of radius 0.75 m with a steady speed of 4 m s^{-1} .

#### Calculate:

(a) the centripetal acceleration of the stone

acceleration = *v*^{ 2}*r*

acceleration = (4 m s^{-1})^{2}0.75 m

acceleration = 21.4 m s^{-2}

(b) the centripetal force acting on the stone.

*F* = *m* × *a*

*F* = 0.5 kg × 21.4 m s^{-2}

*F* = 10.7 N

Notice that this is a linear acceleration and not an angular acceleration. The angular velocity of the stone is constant and so there is no angular acceleration.

#### Student questions: Calculations on centripetal force

Episode 225-2: Centripetal force calculations (Word, 26 KB)

#### Student experiment: Verification of the equation for centripetal force using the whirling bung

Episode 225-3: Verification of the equation for centripetal force (Word, 28 KB)

A Java applet version of this experiment is available on the National Taiwan Normal University website.

#### Demonstration: Alternative method of verifying the equation for centripetal force

This demonstration is an alternative method of verifying the equation for centripetal force.

Episode 225-4: Verifying the equation for centripetal force (Word, 44 KB)