Presentation on theme: "Chapter 6: Circular Motion & Other Applications of Newton’s Laws"— Presentation transcript:
1 Chapter 6: Circular Motion & Other Applications of Newton’s Laws
2 Recall a Ch. 4 Result The Acceleration of a Mass Moving in Circle (At Constant Speed) Consider a particle moving in a circleof radius r, at a constant speed v.The velocity vector is tangent to thecircle. There is a CentripetalAcceleration, a = ac. The accelerationvector is directed radially inward. ac v alwaysac = (v2/r)
3 Newton’s Laws + Circular Motion Centripetal Accelerationac = (v2/r) vBy Newton’s 1st LawThere must be a net force acting.By Newton’s 2nd Law:∑F = ma = mac = m(v2/r)(magnitudes)DirectionThe total force ∑Fmust be radially inward.
4 ac = (v2/r), ac v always!! ac is radially inward always! 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 ofNewton’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)
5 Newton’s 1st Law: Newton’s 2nd Law: ∑F = ma = maR = m(v2/r) 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 theacceleration, so we canimmediately write the force:
6 ALWAYS inward (Centripetal) !! 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 NEVEROutward (“Centrifugal”). It isALWAYS inward (Centripetal) !!
7 MISCONCEPTION!! Example: A ball twirled on a string in a circle at constant speed. The centripetalforce Fr is the tension in the string.MISCONCEPTION!!The force on the ball is NEVERoutward (“centrifugal force”).The force on the ball is ALWAYSinward (centripetal force).An outward force (“centrifugal”) isNOT a valid concept!The force ON THE BALL isinward (centripetal).What happens when the ball is released?(Fr = 0). Newton’s 1st Law says itshould move off in a straightline at constant v.Figure 6.1: An overhead view of a ball moving in a circular path in a horizontal plane. A force Fr directed toward the center of the circle keeps the ball moving in its circular path.
8 Note!! An outward force (“centrifugal force”) is NOT a valid concept! The forceON THE BALLis inward (centripetal).Note!!
9 Example 6.1: Conical Pendulum A ball, mass m, is suspended froma string of length L. It revolveswith constant speed v in ahorizontal circle of radius r. Theangle L makes with the horizontalis θ. Find an expression for v.T ≡ tension in the string. Fig. (b) shows horizontal & verticalcomponents 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)½From trig, r = L sinθ so, v = (Lg sinθ tanθ)½(Reminder: ½ power means the square root)
10 Example 6.3: Car Around a Curve Curve radius: r = 35 m. Static friction coefficientbetween tires & road: μs = The centripetalforce that keeps the car on the road is the staticfriction force fs between the tires & the road.Calculate the maximum speed vmax for the car tostay 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 thesolution to μsmg = m[(vmax)2/r]Or, vmax = (μsgr)½Putting in numbers gives: vmax = 13.4 m/sFigure 6.4: (Example 6.3) (a) The force of static friction directed toward the center of the curve keeps the car moving in a circular path. (b) The free-body diagram for the car.
11 Example 6.4: Banked Curves Engineers design curves which are banked(tilted towards the inside of the curve) to keep carson the road. If r = 35 m & we needv = 13.4 m/s, calculate the angle θ ofbanking needed (without friction). From freebody 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 = orθ = 27.6°Figure 6.5: (Example 6.4) A car rounding a curve on a road banked at an angle θ to the horizontal. When friction is neglected, the force that causes the centripetal acceleration and keeps the car moving in its circular path is the horizontal component of the normal force.
12 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, radiusr = 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 thebottom of the circle, Fig. (b).b) Calculate this force, ntop, at the topof the circle, Fig. (c).Figure 6.5: (Example 6.4) A car rounding a curve on a road banked at an angle θ to the horizontal. When friction is neglected, the force that causes the centripetal acceleration and keeps the car moving in its circular path is the horizontal component of the normal force.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 ornbot = 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] = mg (putting in numbers) he feels “lighter”.
13 Example (Estimate) Assumption: Circular path is in horizontal plane, 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) 1Newton’s 2nd Law: ∑F = ma FTx = max= mac = m(v2/r)v =(2πr/T) = 7.54 m/sFTx = 14 N (tension)