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Circular Motion

Explore the concept of uniform circular motion and the centripetal acceleration acting on objects moving in circles. Learn about the relationship between velocity, acceleration, frequency, and period in circular motion. Dive into centripetal force and distinguish it from centrifugal force. Solve problems involving centripetal acceleration and force, and understand the physics behind circular motion.

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Circular Motion

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  1. Circular Motion

  2. Uniform Circular Motion An object moving uniformly (at the same speed) in a circle is constantly changing direction which means it is constantly changing its velocity which means it is constantly accelerating.

  3. Instantaneous velocity Δv change in velocity Acceleration With Uniform Motion In the diagram, v2 + (– v1)= ∆v. The triangle formed by v1 , v2 and ∆v is similar to the triangle formed by r1, r2 and d. As the angle θ approaches zero, ∆v is at a right angle to the then coinciding sides v1 and v2 . Thus the change in velocity is directed towards the centre of the circle.

  4. Centripetal Acceleration Since the change in velocity has a direction towards the circle’s centre, the acceleration also is directed towards the circle’s centre because a = ∆v/t. The name given to the acceleration of an object moving in a circle is centripetal acceleration, “centre-seeking acceleration”.

  5. Centripetal Acceleration Formula In the diagram (a), triangle ABC is similar to the triangle in (b). This means that Δv/Δl = v/r (equation 1). Rearranging this expression gives Δv = vΔl/r (equation 2). Acceleration is defined as a = Δv/Δt (equation 3). Substituting the Δv in eq.3 with its equivalent in eq. 2 produces a = vΔl/rΔt. But Δl/Δt is linear speed, v of the object: v = Δl/Δt. So, a = vv/r = v2/r. The acceleration of an object (centripetal acceleration) in uniform circular motion is directed towards the centre at:

  6. Circular Velocity and Acceleration Circular velocity is always a tangent to the circle of motion but circular acceleration is always directed towards the centre of the circle of motion. When its accelerating force is removed, an object flies off at a tangent to its former motion.

  7. The Perpendicularity of Circular Velocity and Acceleration An object moving in circular motion constantly has its velocity in a direction perpendicular to its acceleration.

  8. Frequency, Period and Velocity in Circular Motion The frequency (f) is the number of revolutions per unit time and is the reciprocal of period (T) which is the time for one revolution ( f = 1/T ). Circular speed (v) is v = d/t. Since the distance travelled is the circumference of the circle, v = 2πr/T.

  9. Acceleration, Frequency, Period and Velocity in Circular Motion From a = v2/r, substitute v = 2πr/T into a = v2/r. So now a = v2/r becomes a = (2πr/T)2/r . This is a = 4π2r2/T2r . A r in the numerator cancels with r in denominator to give a = 4π2r/T2.

  10. F = 2.00 Hz R = .600 m Problem 1: A Revolving Ball’s Accel;eration A 150 g ball at the end of a string is revolving uniformly, horizontally with a radius of .600 m. The ball has a frequency of 2.00 Hz (makes 2.00 revs/s). What is its centripetal acceleration? Thus T = 1/2.00 Hz = .500s . v = 2πr/T = 2(3.14)(0.600m)/.500s = 7.54 m/s The a = (7.54 m/s)2/0.600 m = 94.8 m/s2 .

  11. Problem 2: The Moon’s Centripetal Acceleration The moon has an average orbital radius of 384,000 km and a period of 27.3 days (T). Determine the acceleration of the moon towards the earth. a = v2/r = (2πr)2/T2r = [2(3.14)(3.84E8 m)]2/(2.36E6 s)2(3.84E8) a = .00272 m/s2 = 2.72E-3 m/s2 . Comparing this to g at earth: 2.72E-3 m/s2(1 g/9.80 m/s2) = 2.78E-4 (Fraction of g at earth)

  12. Centripetal Force From Newton’s Second Law, the sum of the forces on an object (unbalanced force) equals the product of the mass and the acceleration of the object (ΣF = ma) . Thus the centripetal force on a object with circular motion is Fc = mv2/R . This force’s direction is towards the centre of the circular motion.

  13. Circular Motion Requires A Constant Centripetal Force Without a constant force towards the centre of circular motion, the motion becomes linear motion, motion in a straight line.

  14. There is no Centrifugal Force A common misconception is that there is an outward, so-called, centrifugal (“centre-fleeing”) force. Consider a ball circling on a string. For the action force of the string on the ball, there is an outward reaction force of the ball on the string but not on the ball itself. If there were a centrifugal force, a ball would fly outward from a tangent when released from its circular motion. If centrifugal was Actual tangential real motion after release

  15. The Cause of the Apparent Centrifugal Force The supposed outward force felt is a result of an object’s inertia, its tendency to “want” to move in a straight line.

  16. Problem 3 Calculate the centripetal force acting on a 1.5 kg object whirling at a speed of of 2.3 m/s in a horizontal circle of radius 0.60 m. Fc = mv2/r Fc = (1.5 kg)(2.3 m/s)2/0.60m Fc = 13 N

  17. Problem 4 A car travelling at 14 m/s goes around an unbanked curve in the road that has a radius of 96 m. What is the centripetal acceleration? ac = v2/r = (14 m/s)2/96 m = 2.0 m/s2

  18. Problem 5 A plane makes a complete circle with a radius of 3,622 m in 2.10 min. What is the speed of the plane? V = 2πr/T = (2π)(3,622 m)/(2.10 min)(60 s/min) = 181 m/s

  19. Tension force in rope Tension component balancing weight Weight = mg Tension component causing centripetal force The Forces on a Tetherball The tension force in the rope has components that balance the weight and supply a centripetal force.

  20. Vertical Circular Motion At the top position, the centripetal force equals the tension force (T) plus the weight (Fg = ma). Fc = T + Fg At the bottom position, the centripetal force equals the tension force minus the weight. Fc = T – Fg

  21. Minimum Speed to Maintain Circular Motion at Top At the top, FT + mg = mv2/r (centripetal force). When FT = 0 (tension in rope is zero), circular motion stops. Thus the minimum speed at the top needed to continue circular motion is mg = mv2/r which simplifies to g = v2/r and v2 = gr and v = √(gr). A 0.150 kg ball on a 1.10 m cord, swung in a vertical circle would take a minimum speed of 3.28 m/s to keep circling. (v = √(9.80 m/s2)(1.10 m) = 3.28 m/s). Note mass has no effect

  22. The Cord Tension at the Bottom of a Vertical Motion A 0.150 kg ball is swung in a vertical circle in a 1.10 m radius. What would be the tension on the cord at the bottom of its motion if it is swung at a speed of 6.56 m/s? At the bottom FT – mg = mv2/r ; so, FT = mg + mv2/r FT = (0.150 kg)(9.80 m/s2) + (0.150 kg)(6.56 m/s)2/1.10 m = 7.34 N

  23. The Cord Tension at the Top of a Vertical Motion A 0.150 kg ball is swung in a vertical circle in a 1.10 m radius. What would be the tension on the cord at the top of its motion if it is swung at a speed of 6.56 m/s? At the top FT + mg = mv2/r ; so, FT = mv2/r - mg FT = (0.150 kg)(6.56 m/s)2/1.10 m - (0.150 kg)(9.80 m/s2) = 5.87 N

  24. Centripetal force Weight or gravity Centripetal force Weight or gravity Seat force Seat Force Seat Force in a Ferris Wheel At the top of a Ferris wheel, the seat force up and weight force down add to the centripetal force which is down. At the bottom of a Ferris wheel, the seat force up and the weight force down add up to the centripetal force which is up. Thus a Ferris wheel seat exerts less force on a person at the top of its motion and more force on a person at the bottom of its motion.

  25. Tension = 135 N Centripetal force Weight = mg Speed Breaking a String With Vertical Circular Motion A string requires a 135 N force to break it. A 2.00 kg mass is tied to this string and whirled in a vertical circle with a radius of 1.10 m. What is the maximum speed that this mass can be whirled without breaking the string? At the bottom of the motion, the tension force is the greatest. At the bottom FT – mg = mv2/r ; so, FT = mg + mv2/r 135 N = (2.00 kg)(9.80 m/s2) + (2.00 kg)(x m/s)2/1.10 m 135 N = 19.6 N + 1.8181(x m/s)2 x2 = (135 - 19.6) kgm/s2 /1.8181 kg = 63.4729 m2/s2 x = √(63.4729) = 7.97 m/s

  26. Tension Force From Period What is the tension force at the top of a vertical circular motion for a 1.7 kg object moving at the end of a 0.60 m string that takes 1.1 s for a revolution? At the top FT + mg = mv2/r ; so, FT = mv2/r – mg Recall that v = 2πr/T so, FT = (1.7 kg)(2π(0.60m)/1.1 s)2/0.60 m - (1.7 kg)(9.80 m/s2) = FT = 33.279 N – 16.66 = 16.619 N = 17 N Centripetal force Weight = mg Tension force

  27. Force Exerted on Curved Road Surface A 826 kg vehicle travelling at 14.0 m/s goes over a hill with a 61.0 m radius of curvature. What is the force exerted on the road by the car at the crest of the hill? At the top of the hill, Fcar = mv2/r – mg , since Fc = FN + mg. Thus Fcar = (826 kg)(14.0 m/s)2 / 61.0 m – (826 kg)(9.8 m/s2) Fcar = 2654.03 N – 8094.8 N = -5440.77 N or 5 440 N down Centripetal force Car force is down but equal to the normal force Weight = mg

  28. Supplying Centripetal Force for Vehicles on Curves For a vehicle to make a curve, it must have an inward centripetal force. On a flat road, this force is supplied by the road reaction force to the friction force applied by the vehicle’s tires on the road.

  29. When Is Friction too Low to Keep a Vehicle Turning? A 1000 kg car rounds a curve on a flat road of radius 50 m at a speed of 50 km/h (14 m/s). Will the car make the turn if the pavement is dry and the coefficient of static friction is µ = 0.60? F = mv2/r = (1000 kg)(14 m/s)2/(50 m) = 3900 N The normal force equals the weight of the vehicle = (9.80 m/s2)(1000 kg) = 9800 N . Ffr = µFN = (0.60)(9800 N) = 5900 N Since the friction force is much greater than the required centripetal force, the centripetal force is easily supplied by friction and the car turns the curve with no issues.

  30. When Is Friction too Low to Keep a Vehicle Turning? A 1000 kg car rounds a curve on a flat road of radius 50 m at a speed of 50 km/h (14 m/s). Will the car make the turn if the pavement is icy and the coefficient of static friction is µ = 0.20? F = mv2/r = (1000 kg)(14 m/s)2/(50 m) = 3900 N The normal force equals the weight of the vehicle = (9.80 m/s2)(1000 kg) = 9800 N . Ffr = µFN = (0.20)(9800 N) = 2000 N Since the friction force is less than the required centripetal force, the centripetal force can not be supplied by friction and the car will skid.

  31. Banked Curves Banking or angling curves creates a force component inward that can act as the centripetal force to keep a vehicle moving around a curve, even without friction. This inward-acting force is a component of the normal force (reaction of road to vehicle weight) and can be determined by Nsinθ. The angle θ between the normal and vertical is the same as the angle which the road is banked at.

  32. Friction Force on a Banked Curve for a Still Vehicle For a non-moving vehicle on a banked curve, the friction force of the tires on the road acts in a direction up the incline to oppose the motion tendency to slide down the incline.

  33. Friction Force on a Banked Curve for a Moving Vehicle For a moving vehicle on a banked curve, the friction force of the tires on the road acts in a direction down the incline to oppose the motion tendency of the moving car to slide up the incline due to its inertia.

  34. Forces Providing The Centripetal Force For Turning Together, the inward force from the normal force and the inward force from the friction force provide the centripetal force for turning a vehicle. Friction, F Component of friction adding to centripetal force Fy = Fcos θ , F = μFN θ Angle, θ

  35. Formula For Banked Curves The normal force plus the weight equals the centripetal force. From the force diagram, tanθ = Fc/ mg tan θ = mv2/r / mg = v2/rg tanθ = v2/rg Note that tan θ is rise/run or slope Normal Fc Weight = mg θ θ

  36. Banked Curve Problem At what angle should a frictionless curve be banked if a car is to safely round the curve of radius 475 m at a speed of 79 km/h (22 m/s)? tan θ = v2/rg = (22m/s)2/(475 m)(9.80 m/s) tan θ = 0.10 ; Θ = tan-1(0.10) = 5.9o

  37. Banked Curve Problem A car is rounding a 515 m frictionless curve. If the curve is banked at an angle of 12.0 degrees, what is the correct speed for the car? tan θ = v2/rg v = √(rgtan θ) = √[(515 m)(9.80 m/s2)(tan 12.0)] = 32.8 m/s v = 32.8 m/s (118 km/h)

  38. Summary of Banked Curve Relationships

  39. A AQuestiona highway curve that has a radius of curvature of 100 meters is banked at an angle of 15 degrees. What is the vehicle speed which this curve is appropriate if there is no friction between the road and the tires? Also, on a dry day, when friction is present, the automobile negotiates the curve at a speed of 25 meters per second. What is the minimum value of the coefficient of friction that will keep the car from siding?

  40. Factors Affecting Tension and Centripetal Forces A

  41. A AAnswerBanked CurveIn the first case let’s assume that the curve is frictionless [much like the conditions that might exist during a freezing rain]. As always, the first step will be to make the freebody diagram describing all of the relevant forces acting on the car as it passes through the curve. In this case the gravitational force Fg is straight down, while the normal force FN is directed up and toward the left, perpendicular to the roadway. The acceleration is centripetal and is directed left, toward the center of the curve. Here I want to show you a slightly different way of solving a Newton’s 2nd Law problem. Since this is a Newton’s 2nd Law problem the sum of the forces in the problem must be equal to the product of the mass and the acceleration. If you make a vector diagram, attached, adding the two forces acting on the car [Fg and FN], the sum of these two vectors should be equal to the mass times the acceleration. The acceleration here is centripetal and so is directed toward the left, while the gravitational force is directed down. Therefore, the resulting triangle is a right triangle and so these forces can be related using sin, cosine or tangent.Using this triangle, we can determine the critical velocity of the car as it passes through this banked, frictionless curve usingtan(Q) = m*ac/Fg = [m*v2/R]/[m*g] = v2/[R*g]v = sqrt[R*g*tan(Q)]What if there IS friction? What would be the maximum safe speed for a car to pass through this same curve if the coefficient of friction between the car and the road is  and the curve is still banked at an angle ? Now the freebody diagram also includes the frictional force Ff directed down and parallel to the incline.Now, the resulting vector diagram includes three forces adding up, tip to tail, to the resulting mass times the acceleration. As before, the acceleration is centripetal and is directed toward the left. The added frictional force Ff is perpendicular to the normal force FN. Since the frictional force is parallel to the surface that provides it, the frictional force meets the horizontal at the same angle Q as the incline.The normal force, therefore, meets the vertical at the same angle Q, since the normal force is perpendicular to the frictional force.The problem can now be completed by resolving the normal force FN and the frictional force Ff into components that are either parallel to or perpendicular to the direction of the centripetal acceleration.Having broken the frictional force Ff and the normal force FN into their horizontal and vertical components, the vectors in the horizontal direction can be added up to equal the mass times the acceleration:FN*sin(Q) + Ff*cos(Q) = m*acWhile the forces in the vertical direction can be made equal to one another since there is no acceleration in the that direction.FN*cos(Q) = Fg + Ff*sin(Q) = m*g + FN*mu*sin(Q)Typically, in a problem of this type there will be two unknown quantities, the normal force FN and the maximum safe velocity v.If you solve each equation above for the normal force FN FN = [m*v2/R]/[sin(Q)+mu*cos(Q)FN = [m*g]/[cos(Q)- mu*sin(Q)]and then make the two equations equal to one another, it is relatively easy to solve for the maximum safe speed v.[m*v2/R]/[sin(Q) + mu*cos(Q)] = [m*g]/[cos(Q) - mu*sin(Q)]Solve for v = sqrt{[(R*g*(sin(Q)+mu*sin(Q)]/[cos(Q)-mu*sin(Q)]}  

  42. A The relationship between maximum speed, radius and the bank angle can be found from considering the forces FNsinθ>mv2r FNcosθ=mg mgsinθcosθ=mv2r tanθ=v2rg or with friction FNsinθ+μFNcosθ=mv2r FNcosθ−μFNsinθ=mg sinθ+μcosθcosθ−μsinθ=v2rg A

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