apparent weight riding in a elevator– why does our weight appear to change when we start up...
Post on 19-Dec-2015
216 views
TRANSCRIPT
Apparent WeightRiding 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 WeightEg. 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 gApparent weight: N = W + ma
N = m (g + a) (i.e. heavier)
If lift accelerating downwards (or decreasing upwards):
N = m (g – a) (ie. lighter)
W
N
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 free-falling 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.
Instantaneous velocity vector changing in direction but its magnitude stays constant.
v1
v2
v3
v4 v5
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)
• 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.
Centripetal Acceleration
Nature of ‘ac’
|v1| = |v2| = same speed
Accn.
v2
v1
t
vΔac
Acceleration is in direction of Δv.
v1
v2
Δv
v1 + Δv = v2
Dependencies of ac
1. As speed of object increases the magnitude of velocity vectors increases which makes Δv larger.
t
Δvac
v1
v2
Δvv1
v2
Δv Therefore the acceleration
increases.
Dependencies of ac
2. But the greater the speed the more rapidly the direction of velocity vector changes:
Small angle change
Slow
Large angle change
Fastt
vΔ
increases
3. As radius decreases the rate of change of velocity increases – as vector direction changes more rapidly.
d d
Same distance (d) moved but larger angle change.
Large radiusSmall radius
Result: increases as radius decreases.tvΔ
Dependencies of ac
• 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. ).
Summary
Thus: m/s2 (towards center of curvature)
i.e. Centripetal acceleration increases with square of the velocity and decreases with increasing radius.
2
cr
va
r1ac
Example: Ball on a string rotating with a velocity of 2 m/s, mass 0.1 kg, radius=0.5 m.
r
va
2
c 2m/s 80.5
22
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:
TTCenter of motion
m ac
• Ball rotates in a horizontal plane. T = m ac =m v2
r
Let’s consider the ball on a string again…
With gravity:T
T
W = mg Th
Tv
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
Th= T cos θ
Tv= T sin θ = mg
Th= = T cos θm v2
r
As ball speeds up the horizontal, tension will increase (as v2) and the angle θ will reduce.
T
W=mgTh
Tv θ
Stable Rotating Condition
Thus, as speed changes Tv remains unaltered (balances weight) but Th increases rapidly.
T
W=mgTh
Tv θ
High speedT
W=mgTh
Tv θ
Low speed
Unstable Condition• Tv no longer balances weight.
Centripetal force Fc = mac = 0.1 x 8 = 0.8 N
T
ThTv
Ex. again: Ball velocity 2 m/s, mass 0.1 kg, radius=0.5 m.
r
va
2
c 2m/s 80.5
22
Thus, horizontal tension (Th) = 0.8 N.Now double the velocity…
r
va
2
c 2m/s 230.5
44
Centripetal
Fc = mac = 0.1 x 32 = 3.2 NThus, the horizontal tension increased 4 times!
• The ball can’t stay in this condition.
• 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.
• Thus for a given mass the needed force:• increases with velocity 2 • increases as radius reduces.
r
vma mF2
cc
Summary
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.
• The higher the velocity and the sharper the bend, the more friction needed!
Ff 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)
Fs > mv2
r
•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.
•This extra centripetal force can significantly reduce amount of friction needed…
Motion on a Banked Curve Nv= mg
Nh
W=mg
θ
N
• If tan θ = then the horizontal Nh
provides all the centripetal force needed!
v2 rg
Fc
Fn
• Ice skaters can’t tilt ice so they lean over to get a helping component of reaction force to round sharp bends.
• In this case no friction is necessary and you can safely round even an icy bend at speed…
Vertical Circular Motion
• Total (net) force is thus directed upwards:
N > W
Feel pulled in and upward
W=mg
N
W
T
T > W
Ferris Wheel Ball on String
Bottom of circle:
Thus: N = W + mac i.e. heavier/larger tension
Fnet = N - W = mac N=apparent weight (like in elevator)
• Centripetal acceleration is directed upwards.
T > W
Feel thrown out and down
N < W
N > W W=mg
N
N
W
W
T
T
Component of W provides
tension
Top of circle:
N = W – m ac i.e. lighter / less tension
If W = m ac → feel weightless (tension T=0)
• Weight only force for centripetal acceleration down.
(larger r, higher v)gr or v r
vga i.e.
2
c
W
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.
• The higher the velocity and the sharper the bend, the more friction needed!
Ff 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)
Fs > mv2
r