• The velocity-time graph (figure 15) described by equation (3.24) is linear
in t.
• The corresponding displacement-time relationship (figure 16), given by
equation (3.27), is a parabolic curve, due to the quadratic dependence of
∆s on ∆t (given by the (∆t)2 term).
3.8.3
Third Kinematic Equation
Using (3.24) in (3.27) to eliminate ∆t, then from (3.24), if
∆t =
v f − us
,
as
(3.28)
substituting in equation (3.27) gives
∆s = u
vf − us
as
1
+ as
2
It is then straightforward to show that
v f − us
as
vf2 = u2s + 2as ∆s ,
2
.
(3.29)
(3.30)
which is the third kinematic equation.
3.9
Free Fall
Why bother deriving the three kinematic equations? Motion with uniform
acceleration is important in numerous systems.
• e.g. motion under the influence of gravity – when the object is said to
be undergoing free fall.
3.9.1
Galileo’s Free-fall Experiment (c. 1590)
• Late 16th century - several European scientists became critical of the
generally accepted theories of motion proposed by the Ancient Greeks.
27
• Galileo performed a series of (real or thought) experiments (legend has
it at the Leaning Tower of Pisa) which led him to rethink the basic ideas
of motion.
G Galilei (1564–1642)
Galileo’s experiment?
Let us simultaneously drop a lead weight (mass mlead ) and a feather (mass
mfeather ) from a reasonable height, and observe their apparent speeds.:
• Aristotelean viewpoint: speed of a falling body is directly proportional to its mass – since mlead mfeather , then vlead > vfeather – i.e. lead
weight reaches ground quicker.
• Galilean/Modern viewpoint: speed of a falling body is independent
of its mass – both objects should reach the ground at the same time,
irrespective of their mass.
– It is only due to the greater air resistance experienced by the
feather why it appears to reach the ground long after the lead
weight.
– This is verifiable by repeating the experiment in a vacuum (e.g. on
the Moon 9 ),
Galileo correctly identified this phenomenon, and formulated a general conclusion for an idealised situation of motion in a vacuum:
9
see http://vesuvius.jsc.nasa.gov/er/seh/feather.html
28
• In the absence of air resistance, two objects dropped from the same
height will hit the ground at the same time and with the same speed.
• Any two objects in free-fall have the same acceleration, regardless of
their mass.
3.9.2
Acceleration due to Gravity
• The acceleration due to gravity is approximately identical everywhere
on the surface of the Earth for all materials (see figure 18
10
)- there are
only slight variations at different points.
– e.g. at the poles compared to at the Equator (this is due to the
Earth being unspherical - it bulges at the Equator due to its rotation).
• The magnitude of this acceleration is known as the acceleration due
to gravity, g:
Figure 17: Modern experiment investigating g.
• Magnitude: g = 9.80 m s−2 .
• Direction: vertically downwards, towards centre of the Earth .
10
Carusotto et al., Physical Review Letters, volume 69, pages 1722–1725, (1992)
29
Let us drop an object from some height and observe the motion under free
fall. Figure 18 shows the velocity-time graph for the resulting motion: its
velocity changes by −9.80 m s−1 every second.
Figure 18: Velocity-time graph for motion under free fall.
Note:
• g is NOT called ”gravity” - gravity is a force (see later notes), not an
acceleration.
• g is not the acceleration of free fall (i.e. a vector), but simply its magnitude.
• In the kinematic equations (3.24), (3.27) and (3.30), the y-component
of acceleration, ay = −g. In a 2- or 3-dimensional problem, this term
appears as −g ĵ.
• g is only 9.80 m s−2 on or near the Earth’s surface:
– g varies as g ∝ 1/r 2 , where r is the distance from the centre of the
Earth. We assume that g = 9.80 m s−2 in our terrestrial problems.
– g is different for other planets and moons - for example, on the
surface of the Moon, g = 0.00272 m s−2 .
30
Example:
11
a rock is released from rest at the top of a 100 m - tall building.
How long does the rock take to fall to the ground, and what is its impact
velocity?
Solution: we note that the problem is 1-dimensional. Placing the origin of
our co-ordinate system at the ground (see figure 19), we have
ay = −g = −9.80 m s−1
∆s = y1 − y0 = −100 m .
Calculating ∆t: we know ∆s, ay and us . This suggests the use of the
second kinematic equation (3.27), namely
1
∆s = u∆t + g (∆t)2 ;
2
substituting the known values, we have
−100 = 0 +
1
(−9.80) (∆t)2 ,
2
−100 = −4.9 (∆t)2 , (3.31)
Figure 19: Free fall of rock from 100 m.
i.e.
∆t =
r
100
= ±4.52 s .
4.9
• Mathematically, there are two solutions ∆t = ±4.52 s
• Physically: the solution
– ∆t = −4.52 s does not make sense - it refers to a time before the
rock was dropped.
11
Knight, Example 2.16, page 61
31
– We only retain the positive one, and state the solution
∆t = +4.52 s .
Calculating vf : in addition to ∆s, ay and us , we now know ∆t. Using the
first kinematic equation (3.24), we have
vf = us + ay ∆t = 0 + (−9.80) (4.52)
(3.32)
=−44.30 m s−1 ,
where the negative sign indicates that the velocity is downwards.
Note: alternatively, we can solve for vf using the third kinematic equation,
which doesn’t involve ∆t:
vf2 = u2s + 2ay ∆s = 0 + 2 (−9.80) (−100) = 1960 m2 s−2
i.e.
vf = ±44.30 m s−1 ,
(3.33)
and omitting the unphysical solution vf = +44.30 m s−1 while retaining the
physical one, confirms the above solution:
vf = −44.30 m s−1 .
Repeating this experiment on the Moon, where near the Moon’s surface, g =
0.00272 m s−2 . Substitution in (3.31) and (3.32) yields the results
∆t =
r
100
= 271.16 s
1.36 × 10−3
(i.e. ∼ 4.52 Earth minutes) and
vf =
p
2 (−0.00272) (−100) = −0.74 m s−1
32
(i.e. ∼ 60 times less than on Earth), respectively.
Note: as the above example shows, great care needs to be taken with the signs
of the various quantities, e.g. using a displacement of +100 m in equations
(3.31) and (3.32) would result in imaginary values for ∆t and vf .
3.10
Motion on an Inclined Plane
Consider an object moving down a straight, but frictionless, inclined plane
(e.g. a skier), as in figure 20.12
Figure 20: Motion on an inclined plane.
• From rest, object accelerates down the incline under the influence of
gravity, i.e. the situation is very similar to free fall.
• Only major difference is that in our free-fall analysis, the 1-dimensional
motion was vertically downwards – on the inclined plane, the direction
of the 1-dimensional motion is parallel to the plane itself.
The free-fall acceleration afree fall (which the object would have if the incline
suddenly disappeared) appears to result in an acceleration down the inclined
plane. Let us choose to rewrite afree fall as
12
Knight, Figure 2.36(a), page 62
33
afree fall = ak + a⊥
(3.34)
(see figure 2113 ), i.e. in terms of the respective acceleration vectors ak and a⊥
parallel and perpendicular to the incline.
Figure 21: Resolution of free-fall acceleration for motion on an inclined plane.
In particular, if afree fall = −g ĵ , from figure 21 we have
• ak is the acceleration of the object down the plane, with magnitude
ak = g sin θ .
(3.35)
• a⊥ is the acceleration of the object into the surface of the incline (the
result of which is unseen, due to the solid incline), where
a⊥ = g cos θ .
(3.36)
Thus, we see that the 1-dimensional acceleration along the incline, a s , is
as = ±g sin θ ,
(3.37)
the sign depending upon the direction of as with respect to the co-ordinate
system and which way the incline is tilted (see figure 22).
13
Knight, Figure 2.36(b), page 62
34
ay = −g sin θ
ay = +g sin θ
Figure 22: Motion on an inclined plane.
Simple Check:
• When θ = 0 → as = ±g sin 0 = 0 m s−2 .
• When θ = 90◦ → as = −g sin 90◦ = −g m s−2 .
Example:14 Starting from rest, a skier’s speed at the bottom of a 100 m
- long, frictionless, snow-covered slope is 20 m s−1 . What is the angle of the
slope?
Solution: since the skier’s motion is along the plane, this is a 1-dimensional
problem. Referring to figure 23, we place the origin at the skier’s starting
position. We know:
∆s = +100 m
us = 0 m s−1
vf = 20 m s−1
as = +g sin θ .
Figure 23: Skier moving down inclined plane.
Using the third kinematic equation (3.30), we have
vf2 = u2s + 2g sin θ ∆s ,
or, rearranging for θ,
14
Knight, Example 2.18, page 63
35
θ = sin
−1
vf2 − u2s
2g ∆s
.
Thus, substituting the known values into (3.38) gives the solution
θ = sin
−1
202 − 02
(2 × 9.80 × 100)
36
= 11.8 ◦ .
(3.38)