Transcript a = a R

Sect. 5-2: Uniform Circular Motion
• The motion of a mass in a circle at a constant speed.
Constant speed
v = |v| = constant
 The Magnitude (size) of the velocity vector v is constant.
BUT the DIRECTION of v changes continually!
r
r
v  r
• A mass moving in circle at a constant speed is accelerating.
Acceleration  Rate of change of velocity
a = (Δv/Δt)
Constant speed  The Magnitude (size) of the velocity
vector v is constant. v = |v| = constant
BUT the DIRECTION of v changes continually!
 An object moving in a circle
undergoes an acceleration!!
Look at the change in velocity Δv in the limit that the time
interval Δt becomes infinitesimally small & get:
Similar triangles
 (Δv/v) ≈ (Δℓ/r)
As Δt  0, Δθ  0, A B
As Δt  0, Δv   v &
Δv is in the radial direction
 a  aR is radial!
(Δv/v) = (Δℓ/r)  Δv = (v/r)Δℓ
• Note that the (radial) acceleration is
aR = (Δv/Δt) = (v/r)(Δℓ/Δt)
As Δt  0, (Δℓ/Δt)  v
Radial Acceleration Magnitude:
Radial Acceleration Direction: Radially inward!
Centripetal  “Toward the center”
Centripetal Acceleration
 Acceleration toward the center.
• A typical figure for a particle moving in uniform circular
motion, radius r (speed v = constant):
v : Tangent to the circle always!
a = aR: Centripetal acceleration.
Radially Inward always!
 aR  v Always!!
aR = (v2/r)
Period & Frequency
• A particle moving in uniform circular motion of
radius r (speed v = constant)
• Describe in terms of the period T & the
frequency f:
• Period T  the time for one revolution (the time
to go around once), usually in seconds.
• Frequency f  the number of revolutions per
second.
 T = (1/f)
• A particle moving in uniform circular motion, radius
r (speed v = constant)
• Circumference
= distance around = 2πr
 Speed:
v = (2πr/T) = 2πrf
 Centripetal acceleration:
aR = (v2/r) = (4π2r/T2)
This 2nd form is a valid result,
but (in my opinion)
it’s not all that useful!!
Example 5-8: Acceleration of a revolving ball
A 150-g ball at the end of a string is revolving uniformly
in a horizontal circle of radius 0.600 m. The ball makes
2.00 revolutions per second. Calculate its centripetal
acceleration.
r
r
Example 5-9: Moon’s Centripetal Acceleration
The Moon’s nearly circular orbit about the
Earth has a radius of about 384,000 km and a
period T of 27.3 days. Calculate the
acceleration of the Moon toward the Earth.
Centrifuge
A centrifuge works by spinning
very fast. This means there must
be a very large centripetal force.
The object at A would go in a
straight line except for this force;
instead, it ends up at B.
Example 5-10: Ultra-Centrifuge
The rotor of an ultracentrifuge
rotates at 50,000 rpm (revolutions
per minute). A particle at the top of
a test tube is 6.00 cm from the
rotation axis. Calculate its
centripetal acceleration, in “g’s.”
Sect. 5-3: Uniform Circular Motion; Dynamics
• A particle moving in uniform circular motion, radius r
(speed v = constant):
• The Centripetal Acceleration:
aR = (v2/r) , aR  v always!!
aR is radially inward always!
• Newton’s 1st Law: Says that
there must be a force acting!
• Newton’s 2nd Law: Says that
∑F = ma = maR
= m(v2/r) (magnitude)
Direction: The total force must
be radially inward always!
For an object to be in uniform circular
motion, there must be a net
force acting on it. We already
know the acceleration, so we can
immediately write the force:
A force is required to keep
an object moving in a circle.
If the speed is constant, the force is
directed toward the center of the
circle. So, the direction of the
force is continually changing so
that it is always pointed toward the
center of the circle.
Example: A ball twirled on the
end of a string. In that case, the
force is the tension in the string.
• A particle moving in uniform circular motion, radius r
(speed v = constant)
• The acceleration: aR = (v2/r), aR  v always!!
aR is radially inward always!
• Newton’s 1st Law: There must be a force acting!
• Newton’s 2nd Law: ∑F = ma = maR= m(v2/r)
The total force ∑F must be radially inward always!
 The force which enters N’s 2nd Law  “Centripetal Force”
(Center directed force)
• NOT a new kind of force! Exactly what the centripetal
force is depends on the problem. It could be string tension,
gravity, etc. It’s a term for the right side of ∑F = ma, not the left
side! (It’s simply the form of ma for circular motion!)
Centripetal Force
You can see that the centripetal
force must be inward by thinking
about the ball on a string. Strings
only pull; they never push!!
MISCONCEPTION!!
The force on the ball is NEVER
Outward (“Centrifugal”). It is
ALWAYS inward (Centripetal) !!
Note!!
An outward force
(“centrifugal force”) is
NOT
a valid concept!
The force
ON THE BALL
is inward (centripetal).
What happens when the cord on the ball is broken or released?
There is no centrifugal force pointing
outward! What happens is that the natural
tendency of the object to move in a straight line
(Newton’s 1st Law) must be overcome.
If the centripetal force vanishes, the object flies
off at a tangent to the circle. (As it should by
Newton’s 1st Law!!)