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Apparent Weight
Riding in a elevator– why does our weight appear to
change when we start up (increase) and slow down
(decrease)?
Our sensation of weight change is due to a force
exerted on our feet by the elevator floor (normal
force N). If force greater we feel heavier and vice
versa.
Apparent Weight
Eg. Upward accelerating elevator:
As accelerating, there must be
a net upward force.
(2nd law) Fnet = N – W = m a
But our true weight: W = m g
Apparent weight:
N = W + ma
W
N
N = m (g + a) (i.e. heavier)
If lift accelerating downwards (or decreasing upwards):
N = m (g – a) (ie. lighter)
Free-Falling
• When you jump off a wall, or throw a ball or drop a
rock in a pool, the object is free-falling ie. falling
under the influence of gravity.
• Question: What happens to our apparent weight in
free-fall?
• Nasty Exp: Cut elevator wires so its downward
acceleration a = g (i.e. free-fall)!
• Apparent weight N = m (g – a)
• But a = g, so N = 0 i.e. no normal force.
• “Weightless” is zero apparent weight.
• Everything is falling at same rate, so no normal
force is needed to support your weight.
Free-Falling
• Ex: Aircraft flying in a parabolic path can create
weightless conditions for up to 30 s!
• Spacecraft / astronauts in orbit are weightless as
they (and the spacecraft) are continuously freefalling towards the Earth!!
Circular Motion (chapter 5)
So far we have focused on linear motion or motion
under gravity (free-fall).
Question: What happens when a ball is twirled
around on a string at constant speed?
Ans: Its velocity continuously changes in direction.
This implies:
• The velocity change is caused by an acceleration.
• By Newton’s 2nd law an acceleration requires a force!
Circular Motion (chapter 5)
By Newton’s 2nd law an acceleration requires a force!
Big questions:
• What is the nature of this force / acceleration?
• What is the relationship between the acceleration and
the velocity of the ball and the radius of curvature?
• In the absence of gravity, tension provides the only
force action on the ball.
• This tension causes ball to change direction of
velocity.
v2
v3
v4
v1
v5
Instantaneous velocity vector
changing in direction but its
magnitude stays constant.
Question: What happens if you let go of the string?
Answer: Ball travels in direction of last
instantaneous vector. (Newton’s 1st law)
Let’s imagine you are on a kid’s “roundabout”…
Question: Why do we feel an outward force if it’s not
really there?
You must
pull
inwards
Your body naturally
wants to move this way
(Newton’s 1st law)
• However, to keep you in circular motion you must
apply a force inwards to change your direction.
• Your pulling inwards creates the sensation that
the roundabout is pushing you outwards!
You must
pull
inwards
Your body naturally
wants to move this way
(Newton’s 1st law)
Centripetal Acceleration
• The force (tension) causes an acceleration that is
directed inwards towards center of curvature.
• ie. The string is continuously pulling on the ball
towards the center of curvature causing its velocity to
constantly change.
• This is called centripetal acceleration (ac).
Centripetal Acceleration
 Centripetal acceleration is the rate of change
in velocity of an object due to a change in its
velocity direction only.
 It is always perpendicular to the velocity
vector and directed towards the center of
curvature.
• There is NO such thing as centrifugal (ie outward)
force.
Nature of ‘ac’
v2
Accn.
|v1| = |v2| = same speed
v1
Δv
v1
v2
r
Δv
ac =
t
v1 + Δv = v2
Acceleration is in direction of Δv.
Dependencies of ac
1. As speed of object increases the magnitude of
velocity vectors increases which makes Δv larger.
Δv
v2
v1
Δv
v2
v1
Therefore the acceleration
Δv 

ac =
 increases.
t


Dependencies of ac
2. But the greater the speed the more rapidly the
direction of velocity vector changes:
Small angle
change
Slow
r
Δv
t
increases
Large angle
change
Fast
Dependencies of ac
3. As radius decreases the rate of change of velocity
increases – as vector direction changes more
rapidly.
d
d
Small radius
Large radius
Same distance (d) moved
but
larger
angle
change.
r
Δ
v
Result:
increases as radius decreases.
t
Summary
• Points 1 and 2 indicate that the rate of change of
velocity will increase with speed.
• Both points are independent of each other and hence
ac will depend on (speed)2.
Summary
• Point 3 shows that ac is inversely proportional to
radius of curvature (i.e. ac  1 ).
r
2
v
m/s2 (towards center of curvature)
Thus: a c =
r
i.e. Centripetal acceleration increases with square of
the velocity and decreases with increasing radius.
Example: Ball on a string rotating with a velocity of
2 m/s, mass 0.1 kg, radius=0.5 m.
v
2 2
2
ac =
=
= 8 m/s
0.5
r
2
What forces can produce this acceleration?
• Tension
• Friction
• Gravitation attraction (planetary motion).
• Nuclear forces
• Electromagnetic forces
• ?
Let’s consider the ball on a string again…
If no gravity:
Center of motion
m
T
T
ac
2
m
v
• Ball rotates in a horizontal plane. T = m ac =
r
Let’s consider the ball on a string again…
With gravity:
T
Tv T
W = mg
Th
String and ball no longer in
the same horizontal
plane.
• The horizontal component of tension (Th) provides the
necessary centripetal force. (Th = mac)
• The vertical component (Tv) balances the downward
weight force (Tv = mg).
Stable Rotating Condition
T
Tv
Tv= T sin θ = mg
θ
Th
W=mg
Th= T cos θ
Th= m v2 = T cos θ
r
As ball speeds up the horizontal, tension will
increase (as v2) and the angle θ will reduce.
Stable Rotating Condition
Low speed
T
Tv
θ
Th
W=mg
High speed
Tv
T
θ
Th
W=mg
Thus, as speed changes Tv remains unaltered
(balances weight) but Th increases rapidly.
Unstable Condition
Tv
• Tv no longer balances weight.
• The ball can’t stay in this condition.
Th
T
Ex. again: Ball velocity 2 m/s, mass 0.1 kg, radius=0.5 m.
v2 2  2
ac =
=
= 8 m/s 2
0.5
r
Centripetal force Fc = mac = 0.1 x 8 = 0.8 N
Thus, horizontal tension (Th) = 0.8 N.
Now double the velocity… 2
4 4
v
=
= 32 m/s 2
Centripetal a c =
0.5
r
Fc = mac = 0.1 x 32 = 3.2 N
Thus, the horizontal tension increased 4 times!
Summary
• A centripetal force Fc is required to keep a body
in circular motion:
• This force produces centripetal acceleration that
continuously changes the body’s velocity vector.
m
v
Fc = m a c =
r
2
• Thus for a given mass the needed force:
• increases with velocity 2
• increases as radius reduces.
Example: The centripetal force needed for a car
to round a bend is provided by friction.
• If total (static) frictional force is greater than
required centripetal force, car will successfully
round the bend.
2
mv
Fs > r
Ff
• The higher the velocity and the sharper the
bend, the more friction needed!
Ff
• As Fs= μs N - the friction depends on surface type (μs).
• Eg. If you hit ice, μ becomes small and you fail to go
around the bend.
• Note: If you start to skid (locked brakes) μs changes to its
kinetic value (which is lower) and the skid gets worse!
Moral: Don’t speed around tight bends! (especially in winter)
Motion on a Banked Curve
•The normal force N depends on weight of
the car W and angle of the bank θ.
•There is a horizontal component (Nh)
acting towards center of curvature.
N
Nv= mg
Nh
θ
•This extra centripetal force can significantly
reduce amount of friction needed…
2
v
• If tan θ = rg then the horizontal Nh
provides all the centripetal force needed!
• In this case no friction is necessary and you
can safely round even an icy bend at speed… Fn
W=mg
Fc
• Ice skaters can’t tilt ice so they lean over to get a helping
component of reaction force to round sharp bends.
Vertical Circular Motion
Ball on String
Ferris Wheel
N
N>W
Feel
pulled
in and
upward
W=mg
T
T>W
W
Bottom of circle:
• Centripetal acceleration is directed upwards.
• Total (net) force is thus directed upwards:
Fnet = N - W = mac N=apparent weight (like in elevator)
Thus: N = W + mac
i.e. heavier/larger tension
N<W
N
Component of W
provides
tension
W
Feel
thrown
out and
down
W
T
N
N>W
T
W=mg
W
T>W
Top of circle:
• Weight only force for centripetal acceleration down.
N = W – m ac
i.e. lighter / less tension
If W = m ac → feel weightless (tension T=0)
i.e. a c = g =
v
2
r
or v =
rg
(larger r, higher v)
Example: The centripetal force needed for a car
to round a bend is provided by friction.
• If total (static) frictional force is greater than
required centripetal force, car will successfully
round the bend.
2
mv
Fs > r
Ff
• The higher the velocity and the sharper the
bend, the more friction needed!
Ff
• As Fs= μs N - the friction depends on surface type (μs).
• Eg. If you hit ice, μ becomes small and you fail to go
around the bend.
• Note: If you start to skid (locked brakes) μs changes to its
kinetic value (which is lower) and the skid gets worse!
Moral: Don’t speed around tight bends! (especially in winter)