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

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Recall a Ch. 4 Result The 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 = a c. The acceleration vector is directed radially inward. a c v always a c = (v 2 /r)

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Newton’s Laws + Circular Motion Centripetal Acceleration a c = (v 2 /r) v By Newton’s 1 st Law There must be a net force acting. By Newton’s 2 nd Law: ∑F = ma = ma c = m(v 2 /r) (magnitudes) Direction The total force ∑F must be radially inward.

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For a particle moving in uniform circular motion around a circle of radius r (speed v = constant): The acceleration is: a c = (v 2 /r), a c v always!! a c is radially inward always! By Newton’s 1 st Law: There must be a force acting! By Newton’s 2 nd Law: ∑F = ma F r = ma c = m(v 2 /r) The total force ∑F must be radially inward always! The total force on the right side of Newton’s 2 nd Law The Centripetal Force ∑F F r (A center directed force) F r 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)

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A particle moving in uniform circular motion, radius r (speed v = constant): The Centripetal Acceleration: a R = (v 2 /r), a R v always!! a R is radially inward always! Newton’s 1 st Law: Says that there must be a force acting! Newton’s 2 nd Law: Says that ∑F = ma = ma R = m(v 2 /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:

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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) !!

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Example: A ball twirled on a string in a circle at constant speed. The centripetal force F r is the tension in the string. MISCONCEPTION!! The force on the ball is NEVER outward (“centrifugal force”). The force on the ball is ALWAYS inward (centripetal force). An outward force (“centrifugal”) is NOT a valid concept! The force ON THE BALL is inward (centripetal). What happens when the ball is released? (F r = 0). Newton’s 1 st Law says it should move off in a straight line at constant v.

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An outward force (“centrifugal force”) is NOT a valid concept! The force ON THE BALL is inward (centripetal). Note!!

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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: T x = Tsinθ, T y = Tcosθ. Newton’s 2 nd Law: ∑F x = Tsinθ = ma c = m(v 2 /r) (1) ∑F y = Tcosθ – mg = 0; Tcosθ = mg (2) Dividing (1) by (2) gives: tanθ = [v 2 /(rg)], or v = (rg tanθ) ½ From trig, r = L sinθ so, v = (Lg sinθ tanθ) ½ (Reminder: ½ power means the square root)

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Curve radius: r = 35 m. Static friction coefficient between tires & road: μ s = The centripetal force that keeps the car on the road is the static friction force f s between the tires & the road. Calculate the maximum speed v max for the car to stay on the curve. Free body diagram is (b). Newton’s 2 nd Law ( let + x be to left) is: ∑F x = f s = ma c = m(v 2 /r) (1) ∑F y = 0 = n – mg; n = mg (2) The maximum static friction force is (using (2)) f s (max) = μ s n = μ s mg (3) If m(v 2 /r) > f s (max), so v max is the solution to μ s mg = m[(v max ) 2 /r] Or, v max = (μ s gr) ½ Putting in numbers gives : v max = 13.4 m/s Example 6.3: Car Around a Curve

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Example 6.4: Banked Curves Engineers 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 normal to the surface are: n x = n sinθ, n y = n cosθ, Newton’s 2 nd Law ∑F x = n sinθ = m(v 2 /r) (1) ∑F y = 0 = n cosθ – mg; n cosθ = mg (2) Dividing (1) by (2) gives: tanθ = [(v 2 )/(gr)] Putting in numbers gives: tanθ = or θ = 27.6 °

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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 constant speed v = 225 m/s. a) Calculate the force, n bot (normal force), exerted by the seat on the pilot at the bottom of the circle, Fig. (b). b) Calculate this force, n top, at the top of the circle, Fig. (c). TOP: Fig. (b). Newton’s 2 nd Law in the radial (y) direction (up is “+”). ∑F y = n bot – mg = m(v 2 /r) so n bot = m(v 2 /r) + mg or n bot = mg[1 + (v 2 /rg)] = 2.91 mg (putting in numbers) he feels “heavier”. BOTTOM: Fig. (c). Newton’s 2 nd Law in the radial (y) direction (down is “+”). ∑F y = n top + mg = m(v 2 /r) so n top = m(v 2 /r) - mg or n top = mg[(v 2 /rg) - 1] = mg (putting in numbers) he feels “lighter”.

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Example (Estimate) m = 0.15 kg, r = 0.6 m, f = 2 rev/s T = 0.5 s Assumption: Circular path is in horizontal plane, so θ 0 cos(θ) 1 Newton’s 2 nd Law: ∑F = ma F Tx = ma x = ma c = m(v 2 /r) v =(2πr/T) = 7.54 m/s F Tx = 14 N (tension)

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Example

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Problem r = 0.72 m, v = 4 m/s m = 0.3 kg Use: ∑F = ma c Top of circle: Vertical forces: (down is positive!) F T1 + mg = m(v 2 /r) F T1 = 3.73 N Bottom of circle: Vertical forces: (up is positive) F T2 - mg = m(v 2 /r) F T2 = 9.61 N

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Example n n

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