Simple Harmonic Motion Springs So the equation force

Simple Harmonic Motion

Springs • So the equation force of a spring is as follows: F = kx (Hooke’s Law) F – the force supplied by the spring k – the spring constant (depends on how the spring is made) x – displacement of the spring from its equilibrium position

Spring Example 1 I have a slinky with a spring constant of 100 N/m. If I stretch the slinky 5 meters from its equilibrium position, with what force will the spring pull on my hand? F = kx F = (100 N/m)(5 m) = 500 N F = 500 N

Period • The period of a spring system is given by the equation below: T= T – the period of motion m – Mass of the body attached k – spring contant

Spring Example 2 • What is the mass of my car if the shocks have a spring constant of 6000 N/m and it oscillates with a period of 2 seconds when I hit a bump in the road? T= (T/2π)2 = m/k (6000 N/m)(2 s/2π)2 = m 607. 93 kg = m

Restoring Force The restoring force for a pendulum is the component of gravity that is tangent to its circular motion. In most cases: F = mgsin(Θ)

Pendulum Example 1 • What restoring force does gravity supply to a 0. 5 kg pendulum that is at 30 degrees? F = mgsin(Θ) F = (. 5 kg)(9. 8 m/s 2)sin(30) F = 2. 45 N • What is the restoring force at 90 degrees? F = mgsin(Θ) F = (. 5 kg)(9. 8 m/s 2)sin(90) F = 4. 9 N

Period • The period of a pendulum is given by the following formula: T= L – Length of the Pendulum g – Acceleration due to Gravity T – Period

Pendulum Example 2 • I find an old grandfather clock with a period of 1. 0 s per swing. If the grandfather clock is located in Montgomery (g = 9. 8 m/s 2), how long is the pendulum? T= (T/(2π))2 = L/g g(T/(2π))2 = L (9. 8 m/s 2)(1 s/(2π))2 = L 0. 25 m

Example 1: A 2 kg body is tied to the end of a cord and whirled in a horizontal circle of radius 2 m. If the body makes three complete revolutions every second, determine its period and linear speed m = 2 kg r=2 m f = 3 rev/s = 0. 33 s = 37. 70 m/s

Example 2: A ball is whirled at the end of a string in a horizontal circle 60 cm in radius at the rate of 1 revolution every 2 s. Find the ball's centripetal acceleration. = 5. 92 m/s 2

Example 3: A 1000 -kg car rounds a turn of radius 30 m at a velocity of 9 m/s a. How much centripetal force is required? m = 1000 kg r = 30 m v = 9 m/s = 2700 N b. Where does this force come from? Force of friction between tires and road.

Example 4: A geosynchronous satellite is one that stays above the same point on the equator of the Earth. Determine: a. The height above the Earth’s surface such a satellite must orbit and FUG = FC m. E = 5. 98 x 1024 kg RE = 6. 38 x 106 m T = 1 day = 86400 s GMET 2 = 4 π2 r 3 r = 4. 23 x 107 m from the Earth’s center

height = r - r. E = 4. 23 x 107 - 6. 38 x 106 m = 3. 59 x 107 m b. The satellite’s speed = 3072. 54 m/s

Example 5: A 2 -kg mass swings in a horizontal circle at the end of a cord of length 10 m. What is the constant speed of the mass if the rope makes an angle of 300 with the vertical? 1. Draw & label sketch. q = 300 2. Recall formula for pendulum. L q h T Find: v = ? R 3. To use this formula, we need to find R = ? R = L sin 300 = (10 m)(0. 5) R=5 m

Example 5(Cont. ): Find v for q = 300 4. Use given info to find the velocity at 300. R=5 m Solve for v=? g = 9. 8 m/s 2 q = 300 T L q R=5 m h R v = 5. 32 m/s

Example 5 B: Now find the tension T in the cord if m = 2 kg, q = 300, and L = 10 m. T cos q L q 2 kg T T= q h mg R SFy = 0: mg cos q T cos q - mg = 0; = T (2 kg)(9. 8 m/s 2) cos 300 T sin q T cos q = mg T = 22. 6 N

Example 5 extra: Find the centripetal force Fc for the previous example. q = 300 2 kg T cos q L q h T Fc q mg R T T sin q m = 2 kg; v = 5. 32 m/s; R = 5 m; T = 22. 6 N Fc = mv 2 R or Fc = T sin 300 Fc = 11. 32 N

HOMEWORK: 1 -4

Circular Motion-2

Example 6: A car negotiates a turn of radius 70 m when the coefficient of static friction is 0. 7. What is the maximum speed to avoid slipping? m v Fc R ms = 0. 7 Fc = mv 2 R fs = msmg From which: v = msg. R g = 9. 8 m/s 2; R = 70 m v = 21. 91 m/s

Example 8: A car negotiates a turn of radius 80 m. What is the optimum banking angle for this curve if the speed is to be equal to 12 m/s? m/s n mg tan q = v 2 g. R = (12 m/s)2 (9. 8 m/s 2)(80 m) q tan q = 0. 184 How might you find the centripetal force on the car, knowing its mass? q = 10. 410

Example 9: A 2 -kg rock swings in a vertical circle of radius 8 m. The speed of the rock as it passes its highest point is 10 m/s. What is tension T in rope? mg + T = At Top: v mg T R T= v T = 25 N - 19. 6 N mv 2 R - mg - T = 5. 40 N The force are least here

Example 9 B: A 2 -kg rock swings in a vertical circle of radius 8 m. The speed of the rock as it passes its lowest point is 10 m/s. What is tension T in rope? v T - mg = At Bottom: R T T= v mv 2 R + mg mg T = 25 N + 19. 6 N T = 44. 6 N The force are greatest here

Example 10: What is the critical speed vc at the top, if the 2 kg mass is to continue in a circle of radius 8 m? 0 v At Top: mg T vc occurs when T = 0 R v v= g. R = mg + T = mg = mv 2 R (9. 8 m/s 2)(8 m) vc = mv 2 R g. R vc = 8. 85 m/s

Example 11: What is the apparent weight of a 60 -kg person as she moves through the highest point when R = 45 m and the speed at that point is 6 m/s? n + v mg R Apparent weight will be the normal force at the top: mg - n= mv 2 R v n = mg - mv 2 R n = 540 N

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