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

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 = ac. The acceleration vector is directed radially inward.  ac  v always ac = (v2/r)

Newton’s Laws + Circular Motion Centripetal Acceleration ac = (v2/r)  v By Newton’s 1st Law There must be a net force acting. By Newton’s 2nd Law: ∑F = ma = mac = m(v2/r) (magnitudes) Direction The total force ∑F must be radially inward.

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 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)

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 the acceleration, so we can immediately write the force:

ALWAYS inward (Centripetal) !! 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) !!

MISCONCEPTION!! 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 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? (Fr = 0). Newton’s 1st Law says it should move off in a straight line 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.

Note!! 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)

Example 6.3: Car Around a Curve 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 Figure 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.

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: 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° 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.

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, 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). 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 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 or ntop = mg[(v2/rg) - 1] = 0.913 mg (putting in numbers) he feels “lighter”.

Example (Estimate) Assumption: Circular path is  in horizontal plane, 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 2nd Law: ∑F = ma  FTx = max= mac = m(v2/r) v =(2πr/T) = 7.54 m/s FTx = 14 N (tension)

Example

Problem Use: ∑F = mac FT1 + mg = m(v2/r) FT2 - mg = m(v2/r) r = 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: (up is positive) FT2 - mg = m(v2/r) FT2 = 9.61 N

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