Sect. 5-2: Uniform Circular Motion

• The motion of a mass in a circle at a constant speed.

Constant speed The Magnitude (size) of the velocity vector v is constant.

BUT the DIRECTION of v changes continually!

v r

v = |v| = constant

r

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:

As Δt 0, Δv v &

Δv is in the radial direction

a aR is radial!

Similar triangles

(Δv/v) ≈ (Δℓ/r)As Δt 0, Δθ 0, A B

(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!!

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.

Example 5-8: Acceleration of a revolving ball

r

r

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.

Example 5-9: Moon’s Centripetal Acceleration

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.

Centrifuge

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) !!

An outward force

(“centrifugal force”) is

NOT

a valid concept!

The force

ON THE BALL

is inward (centripetal).

Note!!

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!!)

What happens when the cord on the ball is broken or released?