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A-Level Physics - OCR A > M5, C2 Circular Motion and Oscillations > Flashcards

M5, C2 Circular Motion and Oscillations Flashcards

1
Q

define angular velocity

A

the angle an object rotates through per second

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2
Q

what are the units of angular velocity

A

rads^-1

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3
Q

what is the linear speed

A

distance / time

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

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4
Q

what does the equation v = ωr mean

A

linear speed = angular speed X radius of circle of rotation

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5
Q

derive the equation v=ωr

A

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

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6
Q

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

A

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.

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7
Q

define centripetal acceleration

A

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.

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8
Q

what does this equation mean

a = v^2 / r

A

centripetal acceleration = linear speed^2 / radius

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9
Q

what does the equation a = ω^2r mean

A

centripetal acceleration = angular speed^2 X radius

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10
Q

Relate Newton’s first law the centripetal acceleration

A

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.

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11
Q

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

A

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.

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12
Q

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

A

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

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13
Q

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

A

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

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14
Q

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

A

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
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15
Q

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

A

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.

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16
Q

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

A 
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
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17
Q

Design an experiment to investigate circular motion

A
  • 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
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18
Q

define displacement in terms of oscillations in harmonic motion

A

the distance of the object from the equilibrium

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19
Q

what are free vibrations

A

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

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20
Q

what are forced vibrations

A

when there’s an external driving force causing oscillations

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21
Q

what are damping forces

A

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

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22
Q

what’s the difference between heavy and light damping

A

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.

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23
Q

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

A

lightly damped - pendulum

critically damped - car suspension system

24
Q

what is critical damping

A

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

25
Q

what is overdamping

give an example

A

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

26
Q

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

2 answers

A

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

27
Q

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

A
  • 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.

28
Q

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

A

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

29
Q

how could you investigate simple harmonic motion using a pendulum

A
  • 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)
30
Q

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

A

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

31
Q

define simple harmonic motion

A

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

It’s acceleration is directed towards the equilibrium position

32
Q

what does the equation a = -ω^2x

A

for simple harmonic motion

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

33
Q

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

A

a α -x

so negative gradient through origin

34
Q

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
A

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.

35
Q

what is restoring force

A

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

36
Q

derive an equation for restoring force

A

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

37
Q

Draw a graph of force against displacement for simple harmonic motion

A

F α -x

so negative gradient through origin

38
Q

derive an equation for maximum acceleration in simple harmonic motion

A

max acceleration = (ω^2)A

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

39
Q

define SHM (simple harmonic motion)

A

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

40
Q

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.

A

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

41
Q

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

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

A

displacement-time:
sine graph

velocity-time:
cosine graph

acceleration-time:
-sine graph

42
Q

what does this equation mean

x = Acos(ωt)

A

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

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

43
Q

what does this equation mean

x = Asin(ωt)

A

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

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

44
Q

what does this equation mean

v = ±ω√(A^2 - x^2)

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

± is there because velocity is a vector

45
Q

Using the equation v = ±ω√(A^2 - x^2)

Prove that at maximum displacement, the velocity is 0.

A 
At maximum displacement x = A
so
v = ±ω√(A^2 - A^2)
v = ±ω√0
v = 0 ms^-1
46
Q

Using the equation v = ±ω√(A^2 - x^2)

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

A 
At equilibrium x = 0
so
v = ±ω√(A^2 - 0^2)
v = ±ω√(A^2)
v = ±ωA
which is its maximum velocity
47
Q

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

A

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

48
Q

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

A

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.

49
Q

for SHM

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

A

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

(therefore velocity is maximum)

50
Q

for SHM

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

A

potential energy if maximum
kinetic energy is 0

(therefore velocity is 0)

51
Q

what is happening when a system is resonating

A

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

52
Q

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

A

90 degrees

53
Q

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

A

increases until the natural frequency then decreases

curved

54
Q

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

A

lightly damped systems have a higher peak

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

55
Q

in SHM what equation can you use for maximum velocity

A

Vmax = ωA

A-Level Physics - OCR A (20 decks)

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