Module 5: C16 - Circular Motion

Question 1

What is the angle in radians subtended by any arc?

Answer 1

Angle in Radians = Arc Length / Radius

Question 2

What is angular velocity, ω, of an object

Answer 2

The angular velocity, ω, of an object moving in a circular path is defined as the rate of change of angle.

Therefore,
ω = θ/t

Question 3

What is the equation for angular momentum, ω

Answer 3

ω = 2πf

Question 4

Equation for Velocity of an object travelling in Circular Motion

Answer 4

V = s/t = C / T

=> V = 2πr / T

Question 5

Example Question:

The tyre of a car, radius 52cm, rotates with a frequency of 30Hz. Calculate (a) the period of rotation and (b) the linear speed at the tyres edge.

Answer 5

a)
f = 30Hz
T = 1/f => 1/30
T = 0.033

b)
r = 0.52m
V = 2π x 0.52 / 0.033
V = 98ms^-1

Question 6

The radius of the earth is 6.4x10^6m and it orbits the sun at a distance of 1.5x10^11m. Given that the period of rotation of the earth is 1 day and the period of orbit is 1 year, calculate:

i) The speed of someone standing on the equator as the earth spins on its own axis in ms^-1

ii) The speed of the earth as it travels round the sun in ms^-1

Answer 6

T = 1 day = 86400 s
r = 6.4x10^6
v = 2πr/T
v = 2π x 6.4x10^6/86400 = 465.42ms^-1

= 470ms^-1

r = 1.5x10^11
T = 1 year = 31536000s
v = 2πr/T
v = 2π x 1.5x10^11/31536000
v = 29886

v = 30,000ms^-1

Question 7

Example Question:

The tyre of a car, radius 52cm, rotates with a frequency of 30 Hz. Calculate (a) the period of rotation and (b) the linear speed at the tyres edge.

Answer 7

a) f = 30Hz
T = 1/f

T = 1/30
T = 0.33s

b) r = 52cm = 0.52m
f = 30Hz

v = 2πrf
v = 2πx0.52x30
v = 98ms-1

Question 8

Example Question:

The radius of the earth is 6.4 x 10^6 m and it orbits the sun at a distance of 1.5 x 10^11 m. Given that the period of rotation of the earth is 1 day and the period of orbit is 1 year, calculate:

i.The speed of someone standing on the equator as the earth spins on its own axis in kmh^-1

ii. i. The speed of the earth as it travels round the sun in kmh^-1

Answer 8

i)
r = 6.4x10^6 m = 6400km
T = 1 day = 24 hours

v = 2𝜋𝑟 / T
v = 2𝜋 × 6400 / 24
v = 1680kmh^-1

ii)
r = 1.5x10^11m = 150x10^6
T = 1 year = 8760 hours

v = 2𝜋𝑟 / T
v = 2𝜋 × 150 × 10^6 / 8760
v = 108000kmh^-1

Question 9

For the International Space Station in orbit about the Earth (ISS) Calculate:

(a) the centripetal acceleration and
(b) linear speed

Data:
T = 90 minutes
h orbit = 400km
r Earth = 6400km

Answer 9

T = 90 minutes = 5400 seconds
h orbit = 400km = 400,000m
r Earth = 6400km = 6,400,000m

a)
a = rω^2
a = r (2π/T)^2
a = r (2π/5400)^2
a = (6,400,000 + 400,000)(2π/5400)^2
a = 9.21ms^-2

b)
a = v^2 / r
v = √ar
v = √9.21x (6,400,000 + 400,000)
v = 7913.8 ms^-1
v = 7910 ms^-1

Question 10

What is the Centripetal Force and how does it work?

Answer 10

Any accelerating object requires a resultant force to be acting on it (Newton’s First Law). Any force that keeps a body moving with a uniform speed along a circular path is called a centripetal force. In the case of a rollercoaster, the source of the centripetal force comes from the way your weight and the normal contact force from your seat interact with each other. It is the changes in this interaction that make you feel weightless or pushed down into your seat.

A centripetal force is always perpendicular to the velocity of the object. This means that this force has no component in the direction of motion and so no work is done on the object. As a result it’s speed remains constant.

Question 11

How do we calculate the speed or an object moving in a circle at a constant speed?

Answer 11

For an object moving in a circle at a constant speed, we can calculate its speed using the equation:

Speed = Distance Travelled/Time Taken

Question 12

What is the equation for velocity (when something is moving in a circle)

• Involving the radius and time period
• Involving the radius and angular velocity

Answer 12

v = 2πr / T

and since angular velocity ω = 2π/T we can express the speed as v = rω

Question 13

For objects with the same angular velocity, what is the relationship between linear velocity and the radius

Answer 13

For objects with the same angular velocity, the linear velocity at any instant is directly proportional to the radius. Double the radius and the linear velocity will also double.

Question 14

What is Centripetal Acceleration

Answer 14

The acceleration of any object travelling in a circular path at constant speed is called the centripetal acceleration and this, like the centripetal force, always acts towards the centre of the circle.

Question 15

What is the equation (and alternative equation) for Centripetal Acceleration (and what does centripetal acceleration depend on)

Answer 15

The centripetal acceleration a depends on the speed of the object, v, and the radius r of the circular path. The centripetal acceleration is given by the equation:

a = v^2 / r

Combing this with v = rω gives:
a = ω^2 r

Question 16

Worked example: Spinning on the Equator

Calculate the centripetal acceleration of a person standing on the equator as the Earth rotates. The radius of the Earth = 6370km. The period of the rotation is 24 hours = 86400 seconds.

Answer 16

Using a = ω^2 r

We can determine angular speed using ω = 2π / T

ω = 2π / 86400 = 7.27x10^-5 rads^-1

Substituting this into a = ω^2 r gives:

a = (7.27x10^-5)^2 x 6.370x10^6
a = 3.37x10^-2 ms^-2

Question 17

What is Angular Displacement

Answer 17

Angular displacement, θ is equal to the angle swept out at the centre of the circular path.

An object completing a complete circle will therefore undergo an angular displacement of 360°.

Question 18

Equation for Angular Speed involving Angular Displacement and Time

Answer 18

Angular Speed (rads^-1) = Angular Displacement (rad) / Time (s)

𝝎 = 𝚫𝜽 / 𝚫𝒕
𝝎 = 𝜽 / 𝒕

Question 19

What can angular speed also be measured in

Answer 19

Angular speed can also be measured in revolutions per second (rev s-1) or revolutions per minute (r.p.m.)

Question 20

Example Question:

Calculate the angular speed in rad s-1 of an old vinyl record player set at 78 r.p.m.

Answer 20

𝝎 = 78 / 60 revolutions per second

𝝎 = 1.3 rev s-1 but 1 rev = 2π rad

𝝎 = 1.3 x 2π

𝝎 = 8.2 rad s-1

Question 21

What is angular frequency (ω) the same as

Answer 21

Angular frequency is the same as angular speed.

Question 22

What equations do you get when combining

V = rω
and
a = v^2/r

Answer 22

a = = rω^2

and

a = vω

Question 23

What is Newton’s First Law

Answer 23

An object will remain at rest or continue at constantly velocity unless a resultant force acts upon it

Question 24

What is Newton’s Second Law

Answer 24

Force is equal to the rate of change in momentum (the direction of the resultant force and the acceleration must be the same)

Question 25

What Happens when Centripetal Force is Removed?

Answer 25

When the centripetal force is removed, the object will move along a straight line tangentially to the circular path.

Question 26

Calculate the centripetal tension force in a string used to whirl a mass of 200g around a horizontal circle of radius 70cm at 4.0ms^-1

Answer 26

r = 70cm = 0.7m
v = 4.0ms^-1
m = 200g = 0.2kg

F = mv^2 / r
F = (0.2 x 4^2) / 0.7
F = 4.571 N
Centripetal Force = 4.6N

Question 27

Calculate the maximum speed that a car of mass 800kg can go around a curve of radius 40m if the maximum frictional force available is 8kN.

Answer 27

m = 800kg
r = 40m
F = 8kN = 8000N

F = mv^2 / r
v = √Fr/m
v = √8000x40 / 800
v = 20ms^-1

Question 28

A mass of 300g is whirled around a very old circle using a piece of string of length 20cm at 3.0 revolutions per second.

Calculate the tension in the string at positions:

a) A - Top
b) B - Bottom
c) C - String Horizontal

Answer 28

m = 300g = 0.3kg
r = 20cm = 0.2m
ω = 3.0 revs^-1 = 18.8 rads^-1

F = mrω^2
= 0.3 x 0.2 x 18.8^2
= 21. 2064N
= 21.2 N

a) 21.2 - 0.3g = 18.3N
b) 21.2 + 0.3g = 24.1N
c) 21.2N

Question 29

What does Centripetal Force require?

Answer 29

Therefore centripetal acceleration requires a resultant force directed towards the centre of the circular path (this is Centripetal Force)

Tension provides the centripetal force required.

Question 30

What equation do you get for Centripetal Force, combining F=ma and s = v^2/r

Answer 30

F = mv^2 / r

Question 31

How does F relate to v^2 when mass and radius are constant

Answer 31

For constant mass, m and radius, r, the centripetal force F is directly proportional to v^2. That is F∝v^2

Question 32

How can you write to centripetal force equation F = mv^2/r in terms of angular velocity ω

Answer 32

Since v = ωr,

F = m(ωr)^2/r
F = m ω^2r^2 / r

F = mω^2r

Question 33

How can you investigate circular motion using:

• A mass
• Glass Tube
• Paper Clip
• String
• Weight (mg)

Answer 33

As the mass is swung in a horizontal circle, the suspended weight remains stationary as long as the force it provides (Mg) is equal to the centripetal force required to make the bung travel in the circular path. If the centripetal force required is greater than the weight then the weight moves upwards. The paper clip acts as a marker to make this movement clearer. The weight and thus the centripetal force required for different masses, radii, and speeds (calculate from angular velocity and radius) can then be investigated.

Question 34

Worked Example: Taking a bend at top speed

A racing car of mass 1200kg travels around a bend with a radius of 140m. The maximum frictional force between the wheels and the track is 17kN. Calculate the maximum speed at which the car can travel around the bend.

Answer 34

Step 1: Select the correct equation and rearrange to make r the subject.

F = mv^2 / r
so
v = √Fr/m

Step 2: Substitute in the correct values using SI units, and calculate the maximum speed.

v = √1.7x10^4 x 140 / 1200 = 45ms^-1

Question 35

Why do things travel faster on banked surfaces

Answer 35

On the banked part of a track, a horizontal component of the normal contact force, together with frictional force from the tyres, provides the centripetal force required to follow the circular path at such a high speed. The friction between the tyres and track would not be sufficient to allow the cyclist to travel at that speed on a flat surface.

Question 36

What is the force of friction, F0 (where the centripetal force required to keep the car moving in a circle is provided by the friction between the road and tyres)

Answer 36

F0 = mv^2 / r

Question 37

On a banked surface, assuming friction is negligible the centripetal force is provided by the horizontal and vertical components of the normal reaction force. How do you find the horizontal and vertical components.

Answer 37

Horizontal Component:
Nsin (θ) = mv^2 / r

Vertical Component:
Ncos (θ) = mg

Question 38

How do you find the resultant force at the top of a humped bridge

Answer 38

F = mg - S

mg - S = mv^2/r

Question 39

At the top of a humped bridge of the speed of the car increases, there will eventually be a speed v0, where the car will leave the ground (where the support force S is 0)

Give the equation of v0

Answer 39

mg = mv0^2 / r

V0 = √gr

Question 40

The maximum speed without skidding for a car with mass 750kg on a roundabout of radius 20.0m is 9.0ms^-1. Calculate

a) The centripetal acceleration of the car on the roundabout
b) The centripetal force at this speed

Answer 40

m = 750kg
v = 9.0ms^-1
r = 20.0m

a)
a = v^2 / r
a = 9^2 /20
a = 4.05ms^-2
a = 4.1ms^-2

b)
F = mv^2 / r
F = 3037.5 N
F = 3040N

Question 41

A car is racing on a track banked at 25° to the horizontal on a bend with radius curvature of 350m.

a) Show the maximum speed at which the car can take the bend without sideways friction is 40ms^-1.

b) Explain what will happen if the car takes the bend at ever increasing speeds.

Answer 41

a)
Nsin (θ) = mv^2 / r
Ncos (θ) = mg => N = mg / cos(θ)

mg/cos(θ) x sin(θ) = mv^2/r
gsin(θ) / cos(θ) = v^2/r
gtan(θ) = v^2/r
v = √gr tan(θ)

v = √9.81 x 350 x tan(25)
v = 40.0ms^-1

b) If the car takes the bend at ever increasing speeds, then it will travel further up the track.