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Circular Motion • • • • • • • General Requirements Special Case of Newton’s 2nd Law Kinematics of circular motion Dynamics of circular motion Examples Frequency and Period More examples Circular Motion – General Requirements • What is criterion for circular motion? • Must accelerate to side at certain rate! • Rate determined by velocity, radius Circular Motion and 2nd Law • Circular motion is special case of 2nd Law 𝐹 = 𝑚𝑎 Forces that produce that acceleration go here Required acceleration goes here • Much like elevator (chapter 4) 𝐹 = 𝑚𝑎 FT ma Tension – weight goes here Elevator acceleration goes here 696 N What is required acceleration? • Two Isosceles Triangles Radius and velocity triangles Velocity triangle Δr v r r • From similar triangles: Δv/v = Δr/r • a = Δv/ Δt = v/r (Δr/ Δt) = v2/r v Δv Δv = v Δr/r Circular Motion Basics • For circular motion a = v2/r • Then right side 2nd Law must be mv2/r 𝐹 = 𝑚𝑎 Real forces that provide this motion go here • If acceleration or force: – Equal, stays in circle – Less, spirals out of circle – More, spirals into circle Required for circular motion here Circular Motion Caution • “Reverse” type problem • mv2/r is not a force, it is a requirement! • a =v2/r goes on right side of F=ma, work backward to get necessary forces • Something else must supply actual force – Tension – Normal force – Gravity • Guaranteed way of losing 7-8 points on exam (installment #2) Again, remember 𝐹 = 𝑚𝑎 mv2/r does not go here - Goes here Ftension + Fnormal + Fgravity + Ffriction = mv2/r Example - Chapter 5 Problem 1 A child sitting 1.1 m from the center of a merry-go-round moves with a speed of 1.25 m/s. Calculate a) the centripetal acceleration of the child and b) the net horizontal force exerted on the child (mass = 25 kg) • Acceleration 𝑣2 1.25 𝑚/𝑠 𝑎𝑟 = = 𝑟 1.1𝑚 2 = 1.42 𝑚/𝑠 2 • Required Force 𝑚𝑣 2 25 𝑘𝑔 ∗ 1.25 𝑚/𝑠 = 𝑟 1.1𝑚 2 = 35.5 𝑁 1.25 m/s • What real force provides that force? – Friction? – Tension in string? 1.1 m Example • A horizontal force of 210 N is exerted on a 2 kg discus as it rotates uniformly in a horizontal circle (at arm’s length) of radius 0.90 m. Calculate the speed of the discus • Available Force 𝑚𝑣 2 2 𝑘𝑔 ∗ 𝑣 = 𝑟 0.9𝑚 2 = 210 𝑁 • What velocity v will require that force? • 𝑣= 0.9 𝑚∗210 𝑁 2 𝑘𝑔 = 9.7 𝑚 𝑠 = 189 𝑘𝑔 𝑚2 𝑠 2 2 𝑘𝑔 v m/s 0.9 m Frequency and period • Alternate way of describing velocity • Frequency – – – – Number of revolutions per unit time Rpm, etc always use seconds v = 2πr f • Period – – – – Time to make one revolution 24 hour earth rotation Always use seconds v = 2πr /T • Frequency Period reciprocals f = 1/T 2𝜋𝑟 Circular Motion Examples 1 • Example 5-1, 5-3 – A 150-g ball at the end of a string is revolving uniformly in a horizontal circle of radius 0.6 m. The ball makes 2.0 rev/sec. What is the velocity and centripetal acceleration? What force must be exerted on the ball to keep it revolving this way? 2𝜋𝑟 – v = 7.54 m/s, ar = 94.7 m/s2 – F = 14 N • Example 5-2 – What is the velocity of the moon going around the earth? What is the centripetal acceleration? (r = 384,000 km = 384 x 106 m, T = 27.3 days = 2.36 x 106 s) – 𝑣= 2𝜋𝑟 𝑇 – 𝑎= 𝑣2 𝑟 = 1022.35 𝑚/𝑠 𝑚 = .0027 𝑠2 = 2.78 ∙ 10−4 𝑔 Circular Motion Examples 2 • Problem 4 – What is the acceleration of a spec of clay on the edge of a potter’s wheels turning at 45 rev/min if the wheel’s diameter is 35 cm? 2𝜋𝑟 – v=? • Space station – What is the velocity and radial acceleration of the International Space Station orbiting at an altitude of 400 km? – r = 6.78 x 106 m (6.38 x 106 m + 400 km) – T = 1.5 hours = 5400 s – v = 7889 m/s = 28,400 km/h – a = 9.19 m/s2 • Dachel on the slippery floor