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Demo spin stopper
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Is it accelerating?
Does it have Fnet?
Direction of Fnet?
What causes Fnet?
Direction of Fnet?
Is tension doing work?
Newton’s Laws of Motion:
A net external force is necessary
to make an object accelerate follow a circular path.
This force is called a
CENTRIPETAL (“center seeking”) FORCE.
It is the Fnet for an object with curved motion.
Circular or Curved Motion
Equations
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vc = d/t
= 2pr/T
ac = v2/r
Fc = mac = mv2/r.
The term “Centripetal Force” decribes
the direction of Fnet & acceleration. It is
NOT itself a force.
• Any force or forces can cause curved
motion.
Centripetal force and acceleration
may be caused by:
• gravity • friction –
• Tension in a rope or cord• EM –
• R (Fn) -
1. Write an equation for centripetal
acceleration in terms of r and T.
Substituting 2pr for v in
T
accl eq. a = v2
r
ac = 4p2r
T2.
2. The distance from moon’s center to
earth’s center = 3.8 x 108 m.
Moon’s ac = 2.8 x 10-3 m/s2. What is
moon’s period?
• 2.3 x 106 seconds
• ~ 27 days.
You feel acceleration!!
That is because of your
inertia.
The car is turning due to the inward force,
you feel as though you are being forced
leftward or outward. The car is beginning
its turning motion (to the right) while you
continue in a straight line path.
Direction
Velocity Tangent to circular path.
Accl & force toward center of circular path.
Mud sticks to tire.
Sometimes a measured in g’s.
Multiples of Earth’s a of gravity.
1g = 9.81 m/s2.
2g = 19.6 m/s2.
3g = 29.4 m/s2.
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etc.
3. An 80-kg astronaut experience
a force of 2890-N when orbiting
Earth. How many g’s does he
feel?
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ac = F/m
2890 N / 80-kg = 36 m/s2.
36 m/s2 / 9.81 m/s2
= 3.7 g.
Go to vertical circles.
Vertical Circles
Swing keys in vertical circle. Gravity Fg
acts vertically.
Speed is less at top, greater at bottom.
Fc must be less at top, greater at
bottom.
This is also true for rides which make
vertical loops.
Fc is any force pointing
into circle!!
At the top, Fc is combo of weight & Fn.
Fn
W
Taking center as +, Fc = mv2/r = mg + Fn.
Minimum v occurs just as Fn = 0, when you
lose contact with your seat, you begin to
fall.
Fc = mv2/r = mg + Fn
When Fn = 0,
mv2/r = mg. So mass is irrelevant.
Rearrange, solve for vmin,
v = (gr)1/2. v depends on radius only.
R (FN) may also be called "apparent weight" this
is the force of the seat on the rider and also
describes what the rider "feels" .
If R is less than the rider's weight, rider will
fall out. R might become negative, a safety
restraint system -- seat belts, lap bars,
shoulder restraints, are needed.
Tension in Ropes
Swinging Keys
To find the required tension at the top of the
arc: Write a free body Fnet expression:
Fc = T + mg.
Fc – mg = T
mv2/r – mg = T.
The tension is equivalent to R in the ride.
At the bottom the vel is max & Fc = T – W.
T.
Fc = T – mg.
mv2/r = T – mg.
W.
The necessary tension
to swing in a circle is: mv2/r + mg = T.
4: A 10 kg package is swinging in a
vertical circle of radius 5.0 m at the end
of a rope. If the ac is 5.0 m/s2,
a. Sketch the free body diagram at
the bottom of the loop.
b. what is the tension in the rope
when the package is at the bottom
of the loop?
• Fc = T – mg.
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mv2/r + mg = T.
m (v2/r + g) = T.
m(ac + g) = T.
T.
mg.
• 10kg (5.0 m/s2 + 10 m/s2.)
• 150 N.
5. A motorcyclist rides in a vertical
circle of 20.0 m radius. What is the
minimum speed he must have at the
top to complete the loop?
14 m/s.
• Hwk Packet.
Banked Curves
Banked Curves.
On a flat road only friction acts as the Fc to pull
car into curves.
Banked curves add a Fn component toward the
center of the circle.
If the curve is banked, there is a
critical speed where friction =0, the car
can travel the curve without slipping
out of its circular path.
Free Body Diagram Banked Curve
Normal Force Components
sinq = Fc
n
Fc
tan q = Fc
mg
This is NOT the
same vector
sketch as for Fll.
Fc = mac points horizontally toward the
center of the curve.
This component of the normal is supplying
the Fc to keep the car moving through the
banked curve.
At the critical speed, there is no need for
any friction between the car and the road's
surface.
If the speed of the car were to exceed
vcritical then the car would drift up the
incline.
If the speed of the car is less than vcritical
then the car would slip down the incline.
If friction is present both the
horizontal component of N and the
Friction force contribute to Fc.
FN tan θ + Ff = Fc.
IB Problem Packet Banked Curve
• http://www.youtube.com/watch?v=QmAd
fLlhfzw