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WillinkPhysics
Thursday 26 September 2013
Tuesday 21 May 2013
Struggling with Moments?
Balancing
turning forces – the principle of moments
When an object is balanced on a pivot the
turning effect of the forces on one side of the pivot must balance the turning
effect of the forces on the other side of the pivot - if they didn’t it would
not balance.
In the picture two girls are
sitting on a see saw. They have moved until it is balanced. They are the same
weight and so to balance the see saw they must sit the same distance from the
pivot.
In the picture one of the girls gets off and a man sits on instead. They move until the see saw is balanced. The girl is much lighter than the man and so she has to sit further away from the pivot then he does so that she can balance his extra weight.
You can investigate this in the lab by using sets of weights hanging on a wooden ruler:
The principle of moments
You should remember
that the turning effect of a force is called the moment of the force and is
found by multiplying the force by its distance from the pivot. When the see saw
is balanced we say that the anticlockwise moments (those trying to turn the
object anticlockwise) equal the clockwise moments (those trying to turn the
object clockwise). In our example the man’s weight tries to turn the see saw
clockwise and the girl’s weight tries to turn it anticlockwise.
The rule for something to be balanced is called the principle of moments and is written as follows:
When an object is balanced (in equilibrium)
the sum of the clockwise moments is equal to the sum of the anticlockwise
moments.
Force
1 x distance 1 from pivot = Force 2 x
distance 2 from pivot
F1 d1
= F2 d2
Struggling with circular motion?
Circular motion
Objects move in a
straight tine at a constant speed unless a force acts on them. This is Newton’s
First Law. However, many things move in curved paths, especially circles, and
so there must be a force acting on them to pull them out of their straight line
paths and make them turn corners.
The tighter the curve that the object is made to move
in, the bigger the change of direction and so the bigger the force.
Examples of objects moving in curves are:
- The hammer swung by a hammer thrower
- Clothes being dried in a spin drier
- Chemicals being separated in a centrifuge
- Cornering in a car or on a bike
- A stone being whirled round on a string
- A plane looping the loop
- A DVD, CD or record spinning on its turntable
- Satellites moving in orbits around the Earth
- Many fairground rides
We call the force
that makes objects move in a circle the CENTRIPETAL FORCE
(the
name comes from Latin and means centre-seeking)
The centripetal force always acts towards the centre
of the circle to pull the object out of its straight-line path. Although an
object may travel round the circle at a constant speed its direction of motion
is always changing and so its velocity must be changing. Since a change of
velocity is an acceleration there must be a force acting on the object - the
centripetal force.
What produces the centripetal force?
The actual way the force is produced depends on the
particular example:
In a spin drier it is the wall of the drum pressing on
the clothes. When a car, motorbike or bicycle corners it is the friction
between the wheels and the road. (You know how difficult it is to corner on ice
where there is hardly any friction.)
When the Earth orbits the Sun it is the pull of
gravity.
When a railway train corners it is the force of the
rails on the flanged wheels.
When a stone is whirled round on a string it is the
tension in the string.
Monday 6 May 2013
Medical Applications of Physics Prezi
A link to the Prezi from last lesson. Remember sections of Ultrasound, X-rays and Total Internal Reflection were not covered in the lesson.
http://prezi.com/ivzvuu7rtck3/p3-revision-medical-applications-of-physics/
http://prezi.com/ivzvuu7rtck3/p3-revision-medical-applications-of-physics/
Test Yourself Mark Scheme - Pendulums
(a) (i) and (ii)
A suspended
object will always stop so that its centre of mass is directly below the point
of suspension.
The pendulum
bob is a symmetrical object and so the centre of mass must be on an axis of
symmetry. For a circle it will always be at the centre of the circle. The
position of the stationary pendulum bob is clearly marked and supported with a
well-explained reason.
(b) The frequency would halve.
Inversely
proportional means more than simply ‘as one quantity increases the other
quantity decreases’. If two quantities are inversely proportional, when one
doubles the other will halve.
(c) (i) Hertz
This
could also have been given in symbol form (Hz). However, if you use symbols
make sure they are correct. Hz gets a mark but hz would not.
(ii) time
period, T = 22 ÷ 10 = 2.2 s
= 0.45 Hz
The
other values for frequency have been rounded to one decimal place so it would
be okay to round this answer to 0.5.
(iii) No
it does not support the hypothesis. If the hypothesis were correct then
doubling the length from 0.5 m to 1.0 m should make the frequency go
from 0.7 Hz to 0.35 Hz and it does not.
Marks
are not awarded for simply saying yes or no. In this case 1 mark is given for a
relevant calculation supporting the ‘no’. The answer could equally have used
the example of doubling the length from 0.25 m to 0.50 m.
Test Yourself - Pendulums
(a) The diagram shows a simple pendulum at one point in an oscillation (swing). |
(i) Draw a cross (´) on the diagram so that
the centre of the cross marks the position of the centre of mass of the
pendulum bob. (1 mark)
(ii) Draw a circle on the diagram to show the
position of the pendulum bob once the pendulum stops swinging.
Give a reason
for your choice of position. (2 marks)
(b) A student has
written the following hypothesis.
‘The frequency of a simple pendulum is
inversely proportional to the length of the pendulum.’
If this
hypothesis is correct, what would happen to the frequency of a pendulum each
time the length is doubled? (1 mark)
(c) The student
investigated the hypothesis by timing 10 swings of a pendulum. The student
repeated this for several different lengths. The student’s experimental data
and calculated data are recorded in the table.
Length of pendulum in metres
|
Time for 10 swings in seconds
|
Frequency in _____
|
0.25
|
10
|
1.0
|
0.50
|
14
|
0.7
|
0.75
|
17
|
0.6
|
1.00
|
20
|
0.5
|
1.25
|
22
|
|
(i) What is the unit of frequency? (1 mark)
(ii) Calculate the frequency of the pendulum when
the length equals 1.25 m.
Write
down the equation you need to use and show how you work out your answer. (3
marks)
(iii) Do the data in the table support the student’s
hypothesis?
Support
your answer with a calculation. (1 mark)
Test Yourself Mark Scheme - Moments & Hydraulics
(a) 1
Liquids are virtually incompressible.
2 Pressure is transmitted equally in
all directions throughout a liquid.
Both
of the properties have been expressed clearly and concisely. In the first
answer if the word ‘virtually’ had been omitted the mark would still have been
given. If you forget the word ‘incompressible’ you could say ‘cannot be
squashed’.
(b) 2 ´ 105 N 3
ome
of the data have been given in standard form. You can leave them like this and
give an answer in standard form or rewrite 8 ´ 106 as 8 000 000 and give the
answer as 200 000 N. However, for some questions, mainly those
involving very big or very small numbers, you should understand and be able to
use standard form.
(c) If the maximum weight were exceeded the
moments would no longer balance. The resultant moment would cause the crane to topple.
Although
it is not clear if it is the whole crane or just the jib that would topple, the
marks have been given for the idea that a resultant moment will cause something
to rotate.
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