Uniform Circular Motion. Physics of motion of a mass in a circle at Constant Speed. Constant Speed ...

Uniform Circular Motion

Uniform Circular Motion• Physics of motion of a mass in a circle at Constant Speed.

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

BUT the DIRECTION of v changes continually!

v = |v| = constant

v r

r

r

Uniform Circular Motion is Circular Motion at Constant

Speed.•The direction of thevelocity is continuallychanging.

The velocity vector is always tangent to

the circle.

• Consider details of the motion of a mass m in a circle at

Constant Speed.Question:

Is there an acceleration?• To answer this, consider both

Newton’s 1st Law&

Newton’s 2nd Law!• Recall that by Definition,Acceleration Time Rate

of Change of Velocity• That is: a = (Δv/Δt)

r

r

v = |v| = constant

•Acceleration Time Rate of Change of Velocity

a = (Δv/Δt) • Also, recall that both a & v are vectors. Constant Speed The Magnitude (size) of

the velocity vector v is Constant.BUT the DIRECTION of v changes

continually!

An object moving in a circleundergoes an acceleration!!

Centripetal (Radial) AccelerationLook at the vector velocity change 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 BAs Δt 0, Δv v & Δv is in the

radial direction a aC is radial!

Close view of the fact that, in the infinitesimal limit,

Δv 0& its direction is towards the center of the circle.

• This type of acceleration is sometimes called the

Centripetal Acceleration & sometimes called the Radial Acceleration.

Its vector direction is Radially Inward!

• The word “Centripetal” is from Greek. It means

“Towards the Center”“Centripetal Acceleration” Center

Directed or Center Seeking Acceleration

• Below is a typical figure for a particle moving in uniform circular motion, radius r (speed v = constant):

• The velocity vector vis always tangent

to the circle!• The centripetal acceleration

vector aC

is always Radially Inward! aC v always!!

Period & Frequency• Consider again a particle moving in uniform

circular motion of radius r (speed v = constant).• One common way to describe this motion is

in terms of the Period T & the frequency f. Period T The time for one revolution

(time to go around once!)

Frequency f The # of revolutionsper second.

• They are obviously related by:

T = (1/f)

• Consider again a particle moving in uniform circular motion, radius r (speed v = constant)

• Circumference = Distance Around= 2πr

Speed: v = (2πr/T) = 2πrf Centripetal Acceleration:

aC = (v2/r) = (4π2r/T2)

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 f = 2.0 revs/second (period T = 0.5 s).

Calculate its centripetal acceleration.

Example: Acceleration of a Revolving Ball

r

r




r

r

Solution

v = (2πr/T) = 7.54 m/s




r

r

Solution

v = (2πr/T) = 7.54 m/s

aC = (v2/r) = 94.7 m/s2

The Moon’s nearly circular orbit about the Earth has a radius of about 384,000 km (3.84 108 m) and a period T of 27.3 days (2.36 106 s). Calculate the acceleration of the Moon toward the Earth.

Example: Moon’s Centripetal Acceleration

The Moon’s nearly circular orbit about the Earth has a radius of about 384,000 km (3.84 108 m) & a period T of 27.3 days (2.36 106 s). Calculate the acceleration of the Moon toward the Earth.


Solutionv = (2πr/T), aC = (v2/r) = (4π2r/T2)

The Moon’s nearly circular orbit about the Earth has a radius of about 384,000 km (3.84 108 m) and a period T of 27.3 days (2.36 106 s). Calculate the acceleration of the Moon toward the Earth.


Solutionv = (2πr/T), aC = (v2/r) = (4π2r/T2)

aC = 2.72 10-3 m/s2

A bug sits on the edge of a CD, of radius r = 6 cm (0.06m), as in the figure. It undergoes uniform circular motion as the CD spins. It goes around the CD 6 times/sec.

Calculate

Examples 5.1 & 5.2: A Bug on a CD

a. The period of the motion.

b. The speed of the bug.

c. The centripetal acceleration of the bug.


Calculate


a. The period of the motion. Because f = 6 rev/s, T = 1/f = 0.17s

b. The speed of the bug.



Calculate



b. The speed of the bug. v = (2πr/T) = 2.3 m/s



Calculate



b. The speed of the bug. v = (2πr/T) = 2.3 m/s

c. The centripetal acceleration of the bug. aC = (v2/r) = 88.2 m/s2

Newton’s Laws + Circular Motion

• It’s straightforward to see how Newton’s 2nd Law can be applied to circular motion:

• Since the acceleration is directed toward the center of the circle, the net force must be in that direction also!

• This “Centripetal Force” can be supplied by a variety of physical objects or forces

• Also, the “circle” does not need to be a complete circle.

Uniform Circular Motion; Dynamics• Consider a particle moving in uniform circular motion at radius r & speed v = constant.

•Centripetal Acceleration is: aC = (v2/r) , aC v always!!aC is radially inward always!

Newton’s 1st Law There must be a force acting!

Newton’s 2nd Law ∑F = ma = maC = m(v2/r)

Direction:The total force must be radially inward always!

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. The direction of the force is continually changing so that it is always pointed toward the center of the circle.

∑F = ma = maC = m(v2/r)Example: A ball twirled on the end of a string. In that case, the force is the tension in the string.

“Centripetal Force”Newton’s 2nd Law:

∑F = ma = maC= m(v2/r) • The total force ∑F must be radially inward always!

• The force which enters Newton’s 2nd Law in this case is often called a “Centripetal Force”.

(It is a center directed force)• The “Centripetal Force” is NOT a new kind of

force! It could be string tension, gravity, etc. • It’s the right side of ∑F = ma, not the left side!• It’s the form of “ma”, for circular motion!

Centripetal ForceYou can understand 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 (“centrifugal”)

force ON THE BALL is NOT a valid concept! The string tension force

ON THE BALLis Inward (centripetal).

The string tension force F on the ball is INWARD toward the center of the circle!

FF

There is no centrifugal force pointing outward on the ball!

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

For the ball to move in a circle, there must be an inward (Centripetal) force pointed towards the circle center so that the natural tendency of the object to move in a straight line (Newton’s 1st Law!) will be overcome. If the centripetal force goes to zero, the ball will fly off in a direction tangent to the circle

(Newton’s 1st Law again!)

Example (Estimate)

Estimate the force a person must exert on a string attached to a 0.15 kg ball to make the ball revolve in a horizontal circle of radius 0.6 m. The ball makes 2 rev/s.

m = 0.15 kg, r = 0.6 m, f = 2 rev/s T = 0.5 sAssumption: Circular path is in horizontal plane, so

φ 0 cos(φ) 1 Newton’s 2nd Law:

∑F = ma FTx = max= maR = m(v2/r) v = (2πr/T) = 7.54 m/s

So, the tension is (approximately) FTx 14 N

Example: Revolving Ball (Vertical Circle)A ball, mass m = 0.15 kg on the end of a (massless) cord of lengthr = 1.1 m cord is swung in a vertical circle. Calculate:a. The minimum speed the ball must have at the top of its arc so that the ball continues moving in a circle. b. The tension in the cord at the bottom of the arc, assuming that there the ball is moving at twice the speed found in part a.

Hint:The minimum speed at the top will happen for the minimum tension FT1

Note: Here, string tension & gravity are acting together (both enter Newton’s 2nd Law!) to produce centripetal acceleration.

Problemr = 0.72 m, v = 4 m/s, m = 0.3 kg

Use: ∑F = maR

• At the top of the circle, Newton’s 2nd Law is:

(down is positive!)

FT1 + mg = m(v2/r)

FT1 = 3.73 N• At the bottom of the circle:

Newton’s 2nd Law is:(up is positive!)

FT2 - mg = m(v2/r)

FT2 = 9.61 N

A bug sits on the edge of a CD, of radius r = 0.06m, as in the figure. It undergoes uniform circular motion as the CD spins. It goes around the CD 6 times/sec. Results from before:

Example: Back to a Bug on a CDFrom Examples 5.1 & 5.2

T = 0.17s, v = 2.3 m/s, aC = 88.2 m/s2

Given: Mass m = 5 10-3 kg Calculate: The force which keeps the bug on the CD

F = maC = 0.1 N

Conceptual ExampleA Ferris wheel rider moves in a vertical circle of radius r at constant speed v. Is the normal force FN1 that the seat exerts on the rider at the top of the wheel a. less than, b. more than, or c. equal to the normal force FN2 that the seat exerts on the rider at the bottom of the wheel?

UseNewton’s 2nd Law:

∑F = maC at top & bottom.Solve for normal force & compare.

1

2

Conceptual Example• A tether ball is hit so that it

revolves around a pole in a circle of radius r at constant speed v.

• In what direction is the acceleration?

• What force causes it? Newton’s 2nd Law: ∑F = ma

x: ∑Fx = max

FTx = maC = m(v2/r)

y: ∑Fy = may = 0

FTy - mg = 0, FTy = mg

Centripetal Force ExampleOutside the International

Space Station• Whirls a ball on a string in a

perfect circle. • The centripetal acceleration is

produced by the tension in the string.

• If the string breaks, the object would move in a direction tangent to the circle at a constant speed.

Uniform Circular Motion. Physics of motion of a mass in a circle at Constant Speed. Constant Speed ...

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