Uniform Circular Motion Uniform Circular Motion Physics of

Uniform Circular Motion

Uniform Circular Motion • Physics of motion of a mass in a circle at Constant Speed The Magnitude (size) of the velocity vector v is Constant. BUT the DIRECTION of v changes continually! v r r r v = |v| = constant

• Uniform Circular Motion is Circular Motion at Constant Speed. • The direction of the velocity is continually changing. The velocity vector is always tangent to the circle.

• Consider the details of the motion of mass m in a circle at v = |v| = constant Constant Speed. Question: Is there an acceleration? • To answer this, consider both Newton’s 1 st Law r & Newton’s 2 nd Law! • Recall that by Definition, Acceleration Time Rate of Change of Velocity • That is: a = (Δv/Δt) r

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 circle undergoes an acceleration!!

Centripetal (Radial) Acceleration • Look 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 B As Δt 0, Δv v & Δv is in the radial direction a a. C 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 v is always tangent to the circle! • The centripetal acceleration vector a. C is always Radially Inward! a. C 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 revolutions per 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: a. C = (v 2/r) = (4π2 r/T 2)

Example: 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 f = 2. 0 revs/second (period T = 0. 5 s). Calculate its centripetal acceleration. r r

Example: 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 f = 2. 0 revs/second (period T = 0. 5 s). Calculate its centripetal acceleration. r r Solution v = (2πr/T) = 7. 54 m/s

Example: 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 f = 2. 0 revs/second (period T = 0. 5 s). Calculate its centripetal acceleration. r r Solution v = (2πr/T) = 7. 54 m/s a. C = (v 2/r) = 94. 7 m/s 2

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 towards 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 towards the Earth. Solution v = (2πr/T), a. C = (v 2/r) = (4π2 r/T 2) a. C = 2. 72 10 -3 m/s 2

Examples 5. 1 & 5. 2: A Bug on a CD • A bug sits on the edge of a CD, of radius r = 6 cm (0. 06 m), as in the figure. It undergoes uniform circular motion as the CD spins. It goes around the CD 6 times/sec. Calculate a. The period of the motion. b. The speed of the bug. c. The centripetal acceleration of the bug.

Examples 5. 1 & 5. 2: A Bug on a CD • A bug sits on the edge of a CD, of radius r = 6 cm (0. 06 m), as in the figure. It undergoes uniform circular motion as the CD spins. It goes around the CD 6 times/sec. Calculate a. The period of the motion. Because f = 6 rev/s, T = 1/f = 0. 17 s b. The speed of the bug. c. The centripetal acceleration of the bug.

Examples 5. 1 & 5. 2: A Bug on a CD • A bug sits on the edge of a CD, of radius r = 6 cm (0. 06 m), as in the figure. It undergoes uniform circular motion as the CD spins. It goes around the CD 6 times/sec. Calculate a. The period of the motion. Because f = 6 rev/s, T = 1/f = 0. 17 s b. The speed of the bug. v = (2πr/T) = 2. 3 m/s c. The centripetal acceleration of the bug.

Examples 5. 1 & 5. 2: A Bug on a CD • A bug sits on the edge of a CD, of radius r = 6 cm (0. 06 m), as in the figure. It undergoes uniform circular motion as the CD spins. It goes around the CD 6 times/sec. Calculate a. The period of the motion. Because f = 6 rev/s, T = 1/f = 0. 17 s b. The speed of the bug. v = (2πr/T) = 2. 3 m/s c. The centripetal acceleration of the bug. a. C = (v 2/r) = 88. 2 m/s 2

Newton’s Laws + Circular Motion

• It’s straightforward to see how Newton’s 2 nd 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: a. C = (v 2/r) , a. C v always!! a. C is radially inward always! Newton’s 1 st Law There must be a force acting! Newton’s 2 nd Law ∑F = ma. C = m(v 2/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 towards the center of the circle. ∑F = ma. C = 2 m(v /r) • Example: A ball twirled on the end of a string In that case, the force is the tension in the string.

“Centripetal Force” nd 2 Newton’s Law: 2 ∑F = ma. C= m(v /r) • The total force ∑F must be radially inward always! • The force which enters Newton’s 2 nd 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 Force • You 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 BALL F F is Inward (centripetal). • The string tension force F on the ball is INWARD toward the center of the circle!

• 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 1 st 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 1 st Law again!) There is no centrifugal force pointing outward on the ball!

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 s Assumption: Circular path is in horizontal plane, so φ 0 cos(φ) 1 Newton’s 2 nd Law: ∑F = ma FTx = max= ma. R = m(v 2/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 length r = 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 FT 1 Note: Here, string tension & gravity are acting together (both enter Newton’s 2 nd Law!) to produce centripetal acceleration.

Problem r = 0. 72 m, v = 4 m/s, m = 0. 3 kg Use: ∑F = ma. R • At the top of the circle, Newton’s 2 nd Law is: (down is positive!) FT 1 + mg = m(v 2/r) FT 1 = 3. 73 N • At the bottom of the circle: Newton’s 2 nd Law is: (up is positive!) FT 2 - mg = m(v 2/r) FT 2 = 9. 61 N

Example: Back to a Bug on a CD From Examples 5. 1 & 5. 2 A bug sits on the edge of a CD, of radius r = 0. 06 m, as in the figure. It undergoes uniform circular motion as the CD spins. It goes around the CD 6 times/sec. Results from before: T = 0. 17 s, v = 2. 3 m/s, a. C = 88. 2 m/s 2 Given: Mass m = 5 10 -3 kg Calculate: The force which keeps the bug on the CD F = ma. C = 0. 1 N

Conceptual Example A Ferris wheel rider moves in a vertical circle of radius r at constant speed v. Is the normal force FN 1 that the seat exerts on the rider at the top of the wheel 1 a. less than, b. more than, or c. equal to the normal force FN 2 that the seat exerts on the rider at the bottom of the wheel? Use Newton’s 2 nd Law: ∑F = ma. C at top & bottom. Solve for normal force & compare. 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 2 nd Law: ∑F = ma x: ∑Fx = max FTx = ma. C = m(v 2/r) y: ∑Fy = may = 0 FTy - mg = 0, FTy = mg

Centripetal Force Example Outside 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.