M5, C2 Circular Motion and Oscillations

Question 1

define angular velocity

Answer 1

the angle an object rotates through per second

Question 2

what are the units of angular velocity

Answer 2

rads^-1

Question 3

what is the linear speed

Answer 3

distance / time

where distance is the arc length the object moves through in its circular motion

Question 4

what does the equation v = ωr mean

Answer 4

linear speed = angular speed X radius of circle of rotation

Question 5

derive the equation v=ωr

Answer 5

ω = θ / t

for one complete revolution θ = 2π and t = T (period) so
ω = 2π / T

T = 1 / f so ω = 2πf

linear speed = distance travelled / time
for one revolution the distance = 2πr and t = T
so
v = 2πr / T
but you worked out that ω = 2π / T so when you sub it in you get

v=ωr

Question 6

what can be said about the velocity and acceleration of an object moving in circular motion

Answer 6

The object is always accelerating.
Because velocity is a vector quantity. The magnitude of the velocity might be constant but the direction is continually changing.
A changing velocity tells us that the object is accelerating.

Question 7

define centripetal acceleration

Answer 7

an object moving in a circular motion is always accelerating.
this acceleration is directed towards the centre of the circle and is called centripetal acceleration.

Question 8

what does this equation mean

a = v^2 / r

Answer 8

centripetal acceleration = linear speed^2 / radius

Question 9

what does the equation a = ω^2r mean

Answer 9

centripetal acceleration = angular speed^2 X radius

Question 10

Relate Newton’s first law the centripetal acceleration

Answer 10

Netwon’s first law states that an object’s velocity will stay the same unless there’s a force acting on it.
Since an object travelling in a circle has a centripetal acceleration, there must be a force causing this acceleration.
Circular motion is caused by a constant net force perpendicular to the velocity called centripetal force.

Question 11

Why does the kinetic energy of an object travelling in circular motion stay constant

Answer 11

The object isn’t moving towards or away from the centre of the circle so there is no motion in the direction of the force.
Hence, no work is done on the object.

Question 12

Derive the equation F = mv^2 / r
and
F = mω^2r

Answer 12

Using Newton’s 2nd law F=ma
sub into a = v^2 / r

F = mv^2 / r

And sub into a = ω^2r

F = mω^2r

Question 13

What is the centripetal force for the motion of a horizontal circle

Answer 13

The centripetal force is provided by the tension of the spring.
Weight is ignored (it acts vertically, not horizontally)

so T = mv^2 / r or T = mω^2r

Question 14

What is the centripetal force for the motion of a vertical circle

Answer 14

The tension provides the centripetal force. But weight has the be considered.
At A and C (the points at quarter to and quarter past) the weight doesn’t need to be considered so F = mv^2 / r or F = mω^2r

At B (at top of circle) tension and weight act on object so resultant force = tension + weight
T + mg = mω^2r or T + mg = mv^2/r

At D (at bottom of circle) tension and weight act in opposite directions so
T - mg = mv^2/r or T - mg = mω^2r

Question 15

In a verticle circle, where is tension maximum and where is it minimum

Answer 15

It’s maximum at the bottom of the circle. The tension of the string must provide the centripetal force to maintain the motion and it must also support the objects weight.

It’s minimum at the top of the circle because the objects weight is directed towards the circles centre and so it provides some of the centripetal force making the tension small.

Question 16

Derive an equation for a string in circular motion going slack.

Answer 16

Goes slack at the top of a circle when T = mv^2/r - mg
T would equal 0 so
mv^2/r = mg
v^2r = g
v = √rg

Question 17

Design an experiment to investigate circular motion

Answer 17

• Measure the mass of the bung and the mass of the washers, then attach the bung to the string. Thread the string through the plastic tube, and weigh down the free end using the washers.
• Make a reference mark on the string, then measure the distance from the mark to the bung to get the value for the radius.
• Line the reference mark up with the top of the tube, then begin to spin the bung in a horizontal circle. Keep your hand still.
• Measure the time taken for the bung to make one complete circle. This is the period. (or time 10 circles and then divide by 10).
• Use the formula ω = 2π / T to find angular velocity. Then sub this into F = mω^2r
• The centripetal force should be equal to the weight of the washers
• Repeat for different values of radius - you should find the period gets longer with a larger radius but the centripetal force remains the same

Question 18

define displacement in terms of oscillations in harmonic motion

Answer 18

the distance of the object from the equilibrium

Question 19

what are free vibrations

Answer 19

involve no transfer of energy to or from the surroundings
it oscillates at its resonant frequency
it will keep oscillating with the same amplitude forever

Question 20

what are forced vibrations

Answer 20

when there’s an external driving force causing oscillations

Question 21

what are damping forces

Answer 21

an oscillating system loses energy to its surroundings because of frictional forces (damping forces)

systems are often deliberately damped to stop them oscillating

they can vary from light to heavy damping

Question 22

what’s the difference between heavy and light damping

Answer 22

lightly damped systems take a long time to stop oscillating and their amplitude only reduces a small amount each period.
heavily damped systems take less time to stop oscillating and their amplitude gets much smaller each period.

Question 23

give an example of something which is lightly damped and critically damped

Answer 23

lightly damped - pendulum

critically damped - car suspension system

Question 24

what is critical damping

Answer 24

this reduces the amplitude of an oscillation in the shortest possible time

Question 25

what is overdamping

give an example

Answer 25

systems which have even heavier damping
they take longer to return to equilibrium than a critically damped system

eg. heavy doors so they don’t close too quickly

Question 26

what does this equation ω = 2π / T stand for?

2 answers

Answer 26

angular velocity = 2π / period
OR
angular frequency = 2π / period

Question 27

how would you investigate simple harmonic motion using springs, masses and data loggers

Answer 27

• Connect a spring to a clamp stand with some mass on the end. Place a position sensor underneath the spring which is attached to a data logger.
• Lift the mass slightly and release it (it will start oscillating with SHM)
• Place a ruler behind the spring to measure how far you raise the mass.
• As the mass oscillates, the position sensor will measure the displacement of the mass over time.
• Let experiment run until at least you have 10 oscillations
• Generate a displacement-time graph

You should find the amplitude of the oscillations gets smaller over time but the period and frequency remain constant.

You can also complete experiment by changing the weight of the mass or stiffness of the spring.

Question 28

when you hang mass from a spring, why does the amplitude of the oscillations get smaller over time

Answer 28

energy is lost to overcoming air resistance as the mass moves up and down

Question 29

how could you investigate simple harmonic motion using a pendulum

Answer 29

• measure the weight of the mass and use a ruler to measure the string’s length.
• move the mass to the side keeping the string taut. measure the angle between the string and the vertical using the protractor (make sure it’s less than 10 degrees)
• release the mass. position your eye level with the reference mark on the card and start the stopwatch when the mass passes in front of it.
• record the time when the mass passes the mark again moving from the same direction.
• keep recording this at regular intervals as the motion dies away.
• calculate the frequency (1/T)

Question 30

How will increasing the length of string, increasing weight of mass and increasing the angle it was released affect the period of a pendulum?

Answer 30

increasing length of the string will increase the period

the weight of the mass and changing the angle will have no effect on the period

Question 31

define simple harmonic motion

Answer 31

It’s acceleration is proportional to its displacement from the equilibrium position

It’s acceleration is directed towards the equilibrium position

Question 32

what does the equation a = -ω^2x

Answer 32

for simple harmonic motion

acceleration = -(angular frequency)^2 X displacement from equilibrium

Question 33

draw a graph for simple harmonic motion
y axis = acceleration
x axis = displacement

Answer 33

a α -x

so negative gradient through origin

Question 34

for simple harmonic motion:

• when is acceleration at it’s greatest
• when is acceleration at 0
• what happens to speed when you increase amplitude
• what can be said about frequency and period

Answer 34

Acceleration has maximum value when the object is at maximum displacement.
Acceleration is 0 at equilibrium.
Frequency and period are constant.
Increased amplitude = increased average speed.

Question 35

what is restoring force

Answer 35

If an objectis undergoing accelerated motion, Newton’s 2nd law tells us that an unbalanced or resultant force must be acting on it.

Question 36

derive an equation for restoring force

Answer 36

newton’s 2nd law: F=ma
a = -(ω^2)x
so
F = -m(ω^2)x

Question 37

Draw a graph of force against displacement for simple harmonic motion

Answer 37

F α -x

so negative gradient through origin

Question 38

derive an equation for maximum acceleration in simple harmonic motion

Answer 38

max acceleration = (ω^2)A

because at max acceleration it will be max displacement which will be the amplitude

Question 39

define SHM (simple harmonic motion)

Answer 39

an oscillation in which the acceleration of an object is directly proportional to its displacement from the midpoint, and is directed towards the midpoint

Question 40

draw 3 graphs for simple harmonic motion when displacement = max when t = 0

1) displacement-time
2) velocity-time
3) acceleration-time

State what the maximum values equal for the graphs.

Answer 40

displacement-time:
cosine graph
max value = max displacement = amplitude

velocity-time:
minus sine graph
max value = ωA

acceleration-time:
minus cosine graph
max value = (ω^2)A

Question 41

draw 3 graphs for simple harmonic motion when displacement = 0 when t = 0

1) displacement-time
2) velocity-time
3) acceleration-time

Answer 41

displacement-time:
sine graph

velocity-time:
cosine graph

acceleration-time:
-sine graph

Question 42

what does this equation mean

x = Acos(ωt)

Answer 42

WHEN OBJECT IS AT AMPLITUDE (max displacement) AT t = 0

displacement = amplitude(cos(angular frequency X time))

Question 43

what does this equation mean

x = Asin(ωt)

Answer 43

WHEN OBJECT IS AT EQUILIBRIUM (0 displacement) AT t = 0

displacement = amplitude(sin(angular frequency X time))

Question 44

what does this equation mean

v = ±ω√(A^2 - x^2)

Answer 44

v = velocity
ω = angular frequency
A = amplitude
x = displacement

± is there because velocity is a vector

Question 45

Using the equation v = ±ω√(A^2 - x^2)

Prove that at maximum displacement, the velocity is 0.

Answer 45

At maximum displacement x = A
so
v = ±ω√(A^2 - A^2)
v = ±ω√0
v = 0 ms^-1

Question 46

Using the equation v = ±ω√(A^2 - x^2)

Prove that at equilibrium position, the velocity has its maximum value.

Answer 46

At equilibrium x = 0
so
v = ±ω√(A^2 - 0^2)
v = ±ω√(A^2)
v = ±ωA
which is its maximum velocity

Question 47

A steel strip vibrates with a frequency of 20Hz and an amplitude of 5mm.
A mass of 2g is attached to the end of the strip.
Calculate:
1) the velocity of the strip as it passes through equilibrium
2) the acceleration of maximum displacement
3) the max kinetic energy of the mass
4) the max potential energy of the mass

Answer 47

1) v = ±ω√(A^2 - x^2)
ω = 2πf = 2π X 20 = 40π
x = 0 at equilibrium
v = ±40π√((5X10^-3)^2 - 0^2) = 0.63 ms^-1

2) a = -(ω)^2x
= -(40π)^2 X 5X10^-3
= -79.0 ms^-2

3) Ek = 0.5mv^2
= 0.5 X 2X10^-3 X 0.63^2
= 3.97 X 10^-4 J

4) Ek = Ep
= 3.97 X 10^-4 J

Question 48

for SHM,
describe what happens to the potential energy and kinetic energy as an object moves towards equilibrium position and away from equilibrium position

Answer 48

As the object moves towards the equilibrium position, the restoring force does work on the object and so transfers some potential energy to kinetic energy.

When the object is moving away from the equilibrium, all that kinetic energy is transferred back to potential energy again.

Question 49

for SHM

At equilibrium, what can be said about the potential energy and kinetic energy

Answer 49

At the equilibrium, the object’s potential energy is 0 and kinetic energy is maximum.

(therefore velocity is maximum)

Question 50

for SHM

what can be said about potential energy and kinetic energy at the amplitude (max displacement)

Answer 50

potential energy if maximum
kinetic energy is 0

(therefore velocity is 0)

Question 51

what is happening when a system is resonating

Answer 51

The driving frequency approaches the natural frequency.

the system gains more and more energy from the driving force and so vibrates with a rapidly increasing amplitude

Question 52

at resonance what is the phase difference between the driver and oscillator

Answer 52

90 degrees

Question 53

sketch a graph for amplitude (y) against driving frequency (x)

Answer 53

increases until the natural frequency then decreases

curved

Question 54

for the graph of driving frequency against amplitude, how does the line differ for systems more a more or less damped

Answer 54

lightly damped systems have a higher peak

their amplitude only increases dramatically when the driving frequency is very close to the natural frequency

Question 55

in SHM what equation can you use for maximum velocity

Answer 55

Vmax = ωA