Chapter 6: Circular Motion & Other Applications of Newton’s Laws.

Chapter 6: Circular Motion & Other Applications of Newton’s Laws

Recall a Ch. 4 ResultThe Acceleration of a Mass Moving in Circle

(At Constant Speed)

Consider a particle moving in a circle

of radius r, at a constant speed v.

The velocity vector is tangent to the

circle. There is a Centripetal

Acceleration, a = ac. The acceleration

vector is directed radially inward.

ac v always

ac = (v2/r)

Newton’s Laws + Circular MotionCentripetal Acceleration

ac = (v2/r) vBy Newton’s 1st Law

There must be a net force acting.By Newton’s 2nd Law:

∑F = ma = mac = m(v2/r)(magnitudes)

DirectionThe total force ∑F

must be radially inward.

• For a particle moving in uniform circular motion around a circle of radius r (speed v = constant): The acceleration is:

ac = (v2/r), ac v always!!ac is radially inward always!

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

• By Newton’s 2nd Law: ∑F = ma Fr = mac= m(v2/r)

The total force ∑F must be radially inward always! The total force on the right side of

Newton’s 2nd Law The Centripetal Force ∑F Fr (A center directed force)

• Fr is NOT a new kind of force. Exactly what it is depends on the problem. It could be string tension, gravity, etc. It is the right side of ∑F = ma, not the left side! (It is the form of ma for circular motion)

• 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:

Centripetal ForceYou 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) !!

Example: A ball twirled on a string in a circle at constant speed. The centripetal force Fr is the tension in the string.

MISCONCEPTION!!The force on the ball is NEVER outward (“centrifugal force”). The force on the ball is ALWAYSinward (centripetal force). An outward force (“centrifugal”) isNOT a valid concept!

The force ON THE BALL is inward (centripetal). What happens when the ball is released? (Fr = 0). Newton’s 1st Law says it

should move off in a straight line at constant v.

An outward force

(“centrifugal force”) is

NOT a valid concept!

The force

ON THE BALL is inward (centripetal).

Note!!

Example 6.1: Conical Pendulum A ball, mass m, is suspended from a string of length L. It revolves with constant speed v in a horizontal circle of radius r. The angle L makes with the horizontal is θ. Find an expression for v.

T ≡ tension in the string. Fig. (b) shows horizontal & vertical components of T: Tx = Tsinθ, Ty = Tcosθ. Newton’s 2nd Law: ∑Fx = Tsinθ = mac= m(v2/r) (1) ∑Fy = Tcosθ – mg = 0; Tcosθ = mg (2) Dividing (1) by (2) gives: tanθ = [v2/(rg)] , or v = (rg tanθ)½ From trig, r = L sinθ so, v = (Lg sinθ tanθ)½ (Reminder: ½ power means the square root)

Curve radius: r = 35 m. Static friction coefficient between tires & road: μs = 0.523. The centripetal force that keeps the car on the road is the static friction force fs between the tires & the road.

Calculate the maximum speed vmax for the car to

stay on the curve. Free body diagram is (b).

Newton’s 2nd Law (let + x be to left) is:

∑Fx = fs = mac = m(v2/r) (1)∑Fy = 0 = n – mg; n = mg (2)The maximum static friction force is (using (2))

fs(max) = μsn = μsmg (3)

If m(v2/r) > fs(max), so vmax is the

solution to μsmg = m[(vmax)2/r]

Or, vmax = (μsgr)½ Putting in numbers gives: vmax = 13.4 m/s

Example 6.3: Car Around a Curve

Example 6.4: Banked CurvesEngineers design curves which are banked(tilted towards the inside of the curve) to keep cars on the road. If r = 35 m & we need v = 13.4 m/s, calculate the angle θ of banking needed (without friction). From free body diagram, the horizontal (radial) & vertical components of the force n normalto the surface are:nx = n sinθ, ny = n cosθ,

Newton’s 2nd Law∑Fx = n sinθ = m(v2/r) (1)

∑Fy = 0 = n cosθ – mg; n cosθ = mg (2)

Dividing (1) by (2) gives: tanθ = [(v2)/(gr)] Putting in numbers gives: tanθ = 0.523 or

θ = 27.6°

Example: “Loop-the-Loop”!A pilot, mass m, in a jet does a “loop-the-loop. The plane, Fig. (a), moves in a vertical circle, radius r = 2.7 km = 2,700 m at a constantspeed v = 225 m/s. a) Calculate the force, nbot (normal force), exerted by the seat on the pilot at the bottom of the circle, Fig. (b). b) Calculate this force, ntop, at the top

of the circle, Fig. (c). TOP: Fig. (b). Newton’s 2nd Law in the radial (y) direction (up is “+”).

∑Fy = nbot – mg = m(v2/r) so nbot = m(v2/r) + mg or nbot = mg[1 + (v2/rg)] = 2.91 mg (putting in numbers) he feels “heavier”.

BOTTOM: Fig. (c). Newton’s 2nd Law in the radial (y) direction (down is “+”).

∑Fy = ntop + mg = m(v2/r) so ntop = m(v2/r) - mg orntop = mg[(v2/rg) - 1] = 0.913 mg (putting in numbers) he feels “lighter”.

Example (Estimate)

m = 0.15 kg, r = 0.6 m, f = 2 rev/s T = 0.5 sAssumption: Circular path is in horizontal plane, so θ 0 cos(θ) 1

Newton’s 2nd Law: ∑F = ma FTx = max= mac = m(v2/r)

v =(2πr/T) = 7.54 m/s FTx = 14 N (tension)

Example

Problemr = 0.72 m, v = 4 m/s

m = 0.3 kg

• Use: ∑F = mac

• Top of circle:

Vertical forces:(down is positive!)

FT1 + mg = m(v2/r)

FT1 = 3.73 N

• Bottom of circle:

Vertical forces:(up is positive)

FT2 - mg = m(v2/r)

• FT2 = 9.61 N

Example

n

n