Circular Motion

Angles in Radians and Angular Speed versus Linear Speed

We usually measure angles in degrees.

360° = 1 rotation

But it's not the most convenient way to measure angles in circular motion.

Here's an alternative: Radians. The radius of a circle and its circumference are related by the equation...

Circumference = 2πr

So the factor that allows you to convert from circumference (distance travelled around the arc of the circle) to radius is 2π.

So in this way, 2π describes a whole circle, just as 360° describes a whole circle.

360° ≡ 2π radians and

180° ≡ π radians

In fact, as long as you use angles in radians you can write this general equation:

s = rθ

Where:

Copyright S-cool

s = arc length covered

r = radius of the circle

θ = angle in radians

Example:

Convert the following angles from degrees into radians.

90°, 135°, 330°

Answer:

180° ≡ π radians.

So multiply any angle in degrees byCopyright S-coolto find the same angle in radians.

Copyright S-cool

Example:

What angle in degrees has a car travelled around a circular track if the track has a radius of 100 m and the distance covered by the car is 470 m?

Answer:

Copyright S-cool

Convert to degrees:

Copyright S-cool

Note: We had to turn the conversion factor upside down to convert from radians to degrees.

Question:

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In linear or straight-line motion, we measure speed by looking at how much distance is covered each second. You can do that in circular motion too, but it's often better to use angular speed, ω.

Angular speed measures the angle of a complete circle (measured in radians) covered per second.

For instance,

Copyright S-cool

Where:

θ = angle of rotation in radians

t = time taken in seconds.

If you consider that the time taken for a complete rotation is the period, T, then

Copyright S-cool

because 2π is the angle covered (in radians) when you do a complete circle.

Remembering thatCopyright S-coolyou can also write this as

ω = 2πf

Example:

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An old record player spins records at 45rpm (revolutions per minute). For a point on the circumference (radius = 10cm) calculate the angular speed in rad s-1.

Answer:

45rpm = 45/60 = 0.75 revolutions per second = f

Angular speed = ω = 2πf x 0.75 = 4.7 rad s-1

Question:

The wheel of a car rotates at 10 revolutions per second as the car travels along. The radius of the rubber on the tyre is 20cm.

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If you are going round in a circle of radius, r, and you are travelling at a linear speed, v ms-1:

The distance covered in 1 rotation = 2πr

The time for one rotation = T, the period.

Copyright S-cool

These equations allow you to relate angular and linear speed.

Copyright S-cool

Question:

A shot put is swung round at 1 revolution per second. The athlete's arm is 60cm long.

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Multiple Choice Questions

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Exam-style Questions

 

  1. This question is about the planet Jupiter and one of the moons that orbits it, called lo.

    Copyright S-cool

     

    lo orbits Jupiter at a speed, v of 1.7 x 104 m s-1 at an orbital radius, r, of 4.2 x 108m.

    a) Show that lo takes approximately 43 hours to orbit Jupiter once.

    (2 Marks)

    b) lo is held in its orbit by a centripetal force,Copyright S-cool, where m is the mass of lo. This force is the gravitational attraction between lo and Jupiter.

    i) Show thatCopyright S-coolwhere M is the mass of Jupiter.

    ii) Show that the mass of Jupiter is about 2.0 x 1027 kg.

    (4 Marks)

    c) Show that the gravitational potential at the top of Jupiter's atmosphere, 7.1 x 107 m from the centre of the planet, is about -2 x 109 Jkg-1.

    Assume that Jupiter is a sphere.

    (2 Marks)

    d) In July 1994, comet Shoemaker-Levy 9 crashed into Jupiter causing dramatic heating of the planet's atmosphere. During the approach to the planet, the comet broke up. One piece that struck the planet had a mass of about 4 x 1012 kg.

    This fragment crossed the orbit of lo heading directly towards Jupiter with a velocity of 10 km s-1.

    i) Show that kinetic energy of the fragment at that moment is 2 x 1020 J.

    (1 Mark)

    ii) Explain why the fragment will enter the atmosphere of Jupiter with a velocity greater than 10km s-1.

    (2 Marks)

    (Marks available: 11)

  2. a) A cyclist goes round a circular track, of radius 10 m, at a constant speed of 8.0 m s-1.

    What is the acceleration of the cyclist and what is its direction?

    (3 Marks)

    b) What is the resultant force on the cycle and the rider if together they have a mass of 90 kg?

    (2 Marks)

    (Marks available: 5)

Answer

 

Answer outline and marking scheme for question:

 

  1. a) Time = r/1.7 x 104

    = 155230s

    = 43.1 hours

    (2 Marks)

    b) i) -GMm/r2

    = -mv2/r

    So: Gm/r2 = v2/r

    Therefore M = v2r/G

    ii) M = (1.7 x 104 )2 x 4.2 x 108/ 6.67 x 10-11

    = 1.82 x 1027 kg

    (4 Marks)

    c) Vg = -GM/r = -6.67 x 10-11 x 1.9 x 1027/7.1 x 107

    = - 1.79 x 109 K kg-1

    (2 Marks)

    d) i) ½ mv2 = ½ x 4 x 1012 x 100002 = 2 x 1020 J

    (1 Mark)

    ii) The fragment will have gained kinetic energy (1 Mark) as it lost gravitational potential energy during the approach to the planet (1 Mark).

    (Or force argument: Attracted by gravity (1 Mark) causes it to accelerate (1 Mark)

    (2 Marks)

    (Marks available: 11)

  2. a)

    a = v2 = 82 = 6.4 m s-2
    r 10

     

    Direction: towards the centre of the track.

    (3 Marks)

    b) F = ma (1 Mark) = 90 x 6.4 = 576 N (1 Mark)

    (2 Marks)

    (Marks available: 5)

Angular Acceleration and Centripetal Force

Note: Put your calculator into radians mode before using circular motion equations!

Remember Newton's First law?

"If an object continues in a straight line at constant velocity, all forces acting on the object are balanced."

Or another way of putting it...

"An object at rest tends to stay at rest and an object in motion tends to stay in motion with the same speed and in the same direction unless acted upon by an unbalanced force."

Objects moving in circular motion clearly aren't going in a straight line so the forces can't be balanced.

There is a resultant force. This is called the centripetal force.

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The centripetal force is always directed towards the centre of the circle (along the radius of the circle).

(There is no such thing as centrifugal force, so don't mention it in your exams!)

If an object is moving with constant speed in circular motion, it is not going at constant velocity. That's because velocity is a vector. Although its magnitude remains the same, its direction varies continuously.

Because its velocity is changing, we say it is accelerating. That sounds odd! You've got something that's accelerating but not speeding up (or slowing down)! But it is experiencing an unbalanced (or resultant) force. Thats what makes it change direction.

This resultant force, the centripetal force, causes the centripetal acceleration. It is always at 90° to the direction of movement of the object - and that's why the object doesn't speed up!

If you think about a satellite orbiting the Earth, the force of gravity provides the centripetal force and acceleration. The satellite is basically falling towards the earth but does so at the same rate as the Earth curves away from it!

Centripetal acceleration can be calculated using:

Copyright S-cool

Where:

a = centripetal acceleration (m/s2)

v = velocity (m/s)

r = radius of the circle (m)

And from Newton's Second Law:

F = ma, so

Copyright S-cool

This is an equation for centripetal force.

Example:

A car of mass 750kg moves around a circular track of radius 50m with a speed of 10ms-1. What's the centripetal force on the car and which direction does it act in?

Answer:

Copyright S-cool

The direction of all centripetal forces is always towards the centre of the circle.

When an object is moving in circles that are vertical, the weight of the object has to be taken into consideration.

Example:

An object of mass 2kg is attached to the end of a string length 1m and whirled in a vertical circle at a constant speed of 4ms-1. Find the tension in the string at the top and the bottom of the circle. Assume gravity = 10 m/s2

Answer:

The centripetal force is the resultant of all the other forces acting on the object. The only forces acting are the tension in the string, T, and weight, W, due to gravity.

Copyright S-cool

At the top:

The tension and weight are both acting down. Hence,

Centripetal force = tension + weight

Fc = T + W

So,

T = Fc - W = Copyright S-cool

Copyright S-cool

At the bottom:

The tension acts up and the weight acts down. Hence,

Centripetal force = tension - weight

Fc = T - W

So,

T = Fc + W = Copyright S-cool

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Note: If you are using v = ωr in your syllabus, you can substitute this into the equations for centripetal force and acceleration to find values using angular velocity.

S-Cool Revision Summary

Note: Put your calculator into radians mode before using circular motion equations!

Remember Newton's First law?

"If an object continues in a straight line at constant velocity, all forces acting on the object are balanced."

Or another way of putting it...

"An object at rest tends to stay at rest and an object in motion tends to stay in motion with the same speed and in the same direction unless acted upon by an unbalanced force."

Objects moving in circular motion clearly aren't going in a straight line so the forces can't be balanced.

There is a resultant force. This is called the centripetal force.

Install Flash

The centripetal force is always directed towards the centre of the circle (along the radius of the circle).

If an object is moving with constant speed in circular motion, it is not going at constant velocity. That's because velocity is a vector. Although its magnitude remains the same, its direction varies continuously.

This resultant force, the centripetal force, causes the centripetal acceleration. It is always at 90ο to the direction of movement of the object - and that's why the object doesn't speed up!

Centripetal acceleration can be calculated using:

Copyright S-cool

Where:

v = velocity (m/s)

a = centripetal acceleration (m/s2)

r = radius of the circle (m)

And from Newton's Second Law:

F = ma, so

Copyright S-cool

This is an equation for centripetal force.

When an object is moving in circles that are vertical, its weight has to be taken into consideration.

Note: If you are using v = ω r in your syllabus, you can substitute this into the equations for centripetal force and acceleration to find values using angular velocity.

The radius of a circle and its circumference are related by the equation...

Circumference = 2πr

As long as you use angles in radians you can write this general equation:

s = rθ

Where:

s = arc length covered

r = radius of circle

θ = angle in radians

Install Flash

In linear or straight-line motion, we measure speed by looking at how much distance is covered each second. You can do that in circular motion too, but it's often better to use angular speed, ω.

Angular speed measures the angle of a complete circle (measured in radians) you cover per second.

For instance,

Copyright S-cool

Where:

θ = angle in radians

t = time taken in seconds.

If you consider that the time taken for a complete rotation is the period, T, then

Copyright S-cool

because 2 is the angle covered (in radians) when you do a complete circle.

Remembering that Copyright S-cool you can also write this as

ω = 2πf

If you are going round in a circle of radius, r, and you are travelling at a linear speed, v ms-1:

The distance covered in 1 rotation = 2πr

The time for one rotation = T, the period.

Linear speed Copyright S-cool So, v = 2&rf and w = 2pf or Copyright S-cool = w and v = wr

So we can relate angular and linear speed.

ω = 2πf

Copyright S-cool

v = rω

Copyright S-cool

ω = angular speed, rad s-1

f = frequency, Hz (No. of rotations per second)

T = the period of rotation, s

v = linear speed, ms-1

r = radius of rotation, m

a = centripetal acceleration, ms-2

F = centripetal force, N

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