Transcript Teaching Aid: Circular Motion PowerPoint

Circular Motion
© David Hoult 2009
© David Hoult 2009
© David Hoult 2009
© David Hoult 2009
© David Hoult 2009
© David Hoult 2009
© David Hoult 2009
© David Hoult 2009
© David Hoult 2009
© David Hoult 2009
© David Hoult 2009
© David Hoult 2009
© David Hoult 2009
© David Hoult 2009
© David Hoult 2009
© David Hoult 2009
© David Hoult 2009
© David Hoult 2009
© David Hoult 2009
© David Hoult 2009
© David Hoult 2009
© David Hoult 2009
Acceleration of a Body Moving in a Circular Path
at Constant Speed
The magnitude of the velocity of the body is
constant but the direction is constantly changing,
therefore, the body is
© David Hoult 2009
Acceleration of a Body Moving in a Circular Path
at Constant Speed
The magnitude of the velocity of the body is
constant but the direction is constantly changing,
therefore, the body is accelerating
At any instant, the direction of the velocity is
a
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Acceleration of a Body Moving in a Circular Path
at Constant Speed
The magnitude of the velocity of the body is
constant but the direction is constantly changing,
therefore, the body is accelerating
At any instant, the direction of the velocity is
a tangent to the circular path
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The magnitude of the acceleration depends on
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The magnitude of the acceleration depends on
i) the speed of the body
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The magnitude of the acceleration depends on
i) the speed of the body
ii) the radius of the circular path
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We might suggest that
a a
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We might suggest that
a a v
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We might suggest that
a a v
and that
a a
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We might suggest that
a a v
and that
a a 1
r
and therefore
© David Hoult 2009
We might suggest that
a a v
and that
a a 1
r
and therefore
a a v
r
Consideration of the units suggests that
© David Hoult 2009
We might suggest that
a a v
and that
a a 1
r
and therefore
a a v
r
Consideration of the units suggests that
2
v
a a
r
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It can be shown that the magnitude of the
acceleration is given by
2
v
a =
r
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It can be shown that the magnitude of the
acceleration is given by
2
v
a =
r
or in terms of angular speed
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It can be shown that the magnitude of the
acceleration is given by
2
v
a =
r
or in terms of angular speed
a = r w2
or in terms time period
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It can be shown that the magnitude of the
acceleration is given by
2
v
a =
r
or in terms of angular speed
a = r w2
or in terms time period
a =
4 p2 r
T2
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The direction of this acceleration is
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The direction of this acceleration is towards the
centre of the circle
For this reason it is called a
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The direction of this acceleration is towards the
centre of the circle
For this reason it is called a centripetal
acceleration and is said to be caused by a
centripetal force
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The direction of this acceleration is towards the
centre of the circle
For this reason it is called a centripetal
acceleration and is said to be caused by a
centripetal force
v2
Fc = m
r
Fc = mrw2
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A centripetal force does not change the kinetic
energy of the body on which it acts because it
acts
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A centripetal force does not change the kinetic
energy of the body on which it acts because it
acts at 90° to the direction of the motion of the
body
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Estimate the magnitude of the force needed to
cause a car to move around a curve in a road at
50 km h-1.
What force causes the centripetal acceleration
in this situation ?
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Estimate the magnitude of the force needed to
cause a car to move around a curve in a road at
50 km h-1.
What force causes the centripetal acceleration
in this situation ?
Friction
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Estimate the magnitude of the force needed to
cause a car to move around a curve in a road at
50 km h-1.
What force causes the centripetal acceleration
in this situation ?
Friction
Estimates needed:
© David Hoult 2009
Estimate the magnitude of the force needed to
cause a car to move around a curve in a road at
50 km h-1.
What force causes the centripetal acceleration
in this situation ?
Friction
Estimates needed:
mass of car,
m
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Estimate the magnitude of the force needed to
cause a car to move around a curve in a road at
50 km h-1.
What force causes the centripetal acceleration
in this situation ?
Friction
Estimates needed:
mass of car,
m
radius of path, r
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v2
Fc = m
r
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A small mass hangs on a string inside the car. It
is observed by a passenger.
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A small mass hangs on a string inside the car. It
is observed by a passenger*.
If the car turns to the left:
* the mass is in front of the passenger.
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A small mass hangs on a string inside the car. It
is observed by a passenger.
If the car turns to the left:
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A small mass hangs on a string inside the car.
If the car turns to the left:
q
Find the angle q during the time the car is moving
round the curved path.
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© David Hoult 2009
mg
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Te cos q
mg
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Te cos q
Te sin q
mg
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Te cos q
Te sin q
mg
The vertical forces acting on the mass are in equilibrium,
therefore
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Te cos q
Te sin q
mg
The vertical forces acting on the mass are in equilibrium,
therefore
Te cos q must have the same magnitude as mg
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Te cos q
Te sin q
mg
The mass is accelerating to the left, horizontally
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Te cos q
Te sin q
mg
The mass is accelerating to the left, horizontally
The horizontal component of the tension provides the
centripetal force needed for this acceleration.
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Te cos q
Te sin q
mg
Therefore
2
v
Te sin q = m
r
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...need I say more ?
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