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PHYSICS 231
Lecture 16: Centripetal acceleration
Remco Zegers
Walk-in hour: Tue 4-5 pm
Helproom
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Last lecture...
 f i
 Average angular


t f  ti
t velocity (rad/s)
 Instantaneous
  lim
t 0 t
Angular velocity
 f  i
 Average angular


t f  ti
t acceleration (rad/s2)
 Instantaneous angular
  lim
t 0 t
acceleration
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2 rad  3600
10  2/360 rad
1 rad 360/2 deg
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And...
Angular and linear velocity are related:
v

r
Angular and linear acceleration are
related:
a

r
demo: tangential velocity
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Rotational motion
Angular motion
(t)= (0)+(0)t+½t2
(t)= (0)+t
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gears
If 1=3 rad/s, how fast
Is the bike going?
b) What if r1=0.1 m ?
r3=0.7 m
r1=0.3
r2=0.15
b)
V1= 1r1=3*0.3=0.9 m/s
V1=V2 (because of the chain)
2=V2/r2=0.9/0.15=6 rad/s
3=2 (connected)
V3=3r3=6*0.7=4.2 m/s
V1= 1r1=3*0.1=0.3 m/s
V1=V2 (because of the chain)
2=V2/r2=0.3/0.15=2 rad/s
3=2 (connected)
V3=3r3=2*0.7=1.4 m/s
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quiz (extra credit)
An object is moving in a circle with radius R=2 m. If the
radius is doubled and the angular velocity remains the same,
the linear velocity will:
a)
b)
c)
d)
e)
half
remain the same
double
quadruple
whatever
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for old times sake…
A man is standing on a platform resting on a frictionless
surface. At the end of the platform a wall is placed and
fixed to the platform.The man throws a ball against the
wall and it bounces back over the man as shown in the
figure. After the ball has bounced against the wall…
a) the platform moves to the left
b) the platform moves to the right
c) the platform remains at its place
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Driving a car through a bend
Is there a force that pushes
you away from the center of
the circle?
• Newton’s first law: If no net force is acting on an object,
it will continue with the same velocity (inertia of mass)
• Velocity is a vector (points to a direction)
• If no net force is acting on an object, it will not change
its direction.
• A force is acting on the car (steering+friction) but you
tend to go in the same direction as you were going!
• It is not a force that pushes you, but the lack of it!
• The side door will keep you from falling out: it exerts a
force on you and you exert a force on the door (F21=-F12)
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Centripetal acceleration
v f  vi
v
a

t f  ti
t
The change in velocity is
not the change in speed
but in direction.
Sin(/2)=(s/2)/r
Sin(/2)=(v/2)/v
(s/2)/r=(v/2)/v
v=s*(v/r)
t t
ac=v2/r
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vs/t
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Centripetal acceleration
ac=v2/r directed to the center of the circular motion
Also v=r, so ac=2r
This acceleration can be caused by various forces:
• gravity (objects attracted by earth)
• tension (object making circular motion on a rope)
• friction (car driving through a curve)
• etc
This acceleration is NOT caused by a mysterious force
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Forces that can cause centripetal acceleration.
Object swinging on a rope.
T
F=ma
T
T=mac
T=mv2/r=m2r
An object with m=1 kg is swung with a rope of length 3 m
around with angular velocity =2 rad/s. What is the tension
in the rope?
T=m2r=1*22*3=12 N
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Lifting by swinging
Swinging mass (m1) with velocity v
T
r
T
What is the relation between v and r
that will keep m2 stationary?
v out of paper
m1: T=m1a
a=-m2g/m1
m2: -T=m2g
Hanging mass (m2) Also: ac=(-)v2/r v=linear/tangential
velocity of m1
2/r= m g/m
v
2
1
Fg=m2g
demo
If m1 slows down, r must go down
so m2 sinks.
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question
Swinging mass (m1) with velocity v
T
r
v2/r= m2g/m1
v out of paper
T
Hanging mass (m2)
Fg=m2g
If the velocity of m1 is increased
to twice the original value,
the radius of orbit
a) decreases to ¼ of its original value
b) decreases to ½ of its original value
c) remains the same
d) increases to 2x its original value
e) increases to 4x its original value
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A car going through a bend
A car is passing through a bend with
radius 100 m. The kinetic coefficient
of friction of the tires on the road is
0.5. What is the maximum velocity the
car can have without flying out of the
bend?
F=ma
kn=mac
kmg=mv2/r
0.5*9.81=v2/100
v=22 m/s
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2001: A space odyssey
A space ship rotates with
a linear velocity of 50 m/s.
What should the distance
from the central axis to
the crew’s cabin’s be so that
the crew feels like they are
on earth? (the floor of the
cabins is the inside of the
outer edge of the spaceship)
The rotating spaceship has an acceleration directed towards
the center of the ship: the ‘lack’ of forces acting on the crew
pushes them against the ship.
F=ma
mg=mv2/r so r=v2/9.8
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and thus r=255 m
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Conical motion
Tcos
T


Tsin
mg
demo
What is the centripetal acceleration if
the mass is 1 kg and =20o?
Vertical direction:
F=ma
Tcos-mg=0
So T=mg/cos
Horizontal direction:
F=mac
Tsin=mac
mgsin/cos=mgtan=mac
ac=gtan=9.8*0.36=3.6 m/s2
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A general strategy
• As usual, make a drawing of the problem, if not given.
• Draw all the forces that are acting on the object(s)
under investigation.
• Decompose each of these into directions toward the
center of the circular path and perpendicular to it.
• Realize that Fto center=mac=mv2/r
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question
A point on the rim of a 0.25 m radius rotating wheel has
a centripetal acceleration of 4.0 m/s2. What is the linear
speed of the wheel? (Fto center=mac=mv2/r)
a) 1.0 m/s
b) 2.0 m/s
c) 3.2 m/s
d) 4.0 m/s
e) 8.0 m/s
ac=v2/r 4=v2/0.25 v2=16 v=4 m/s
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