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Rotational Motion

Rotational Motion. Arc length: s = r θ. s. If θ is in radians, then the arc length, s , is θ times r . This follows directly from the definition of a radian. One radian is the angle made when the radius of a circle is wrapped along the circle. θ. r. r.

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Rotational Motion

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  1. Rotational Motion

  2. Arc length: s = rθ s If θ is in radians, then the arc length, s, is θtimesr. This follows directly from the definition of a radian. One radian is the angle made when the radius of a circle is wrapped along the circle. θ r r When the arc length is as long as the radius, the angle subtended is one radian. (A radian is really dimensionless, since it’s found by dividing a length by a length.) 1 radian r

  3. Angular Speed, ω B Linear speed is how fast you move, measured as distance per unit time. Angular speed is how fast you turn, measured as an angle per unit time. The symbol for angular speed is the small Greek letter omega, ω, which looks like a curvy “w”. Units for angular speed include: degrees per second; radians per second; and rpm (revolutions per minute). 110° A 3 m Suppose an object moves steadily from A to B along the circle in 5 s. Then ω = 110°/5 s = 22°/s. The distance it covers is (110°/360°)(2π)(3) = 5.7596 m. So its linear speed is v = (5.7596 m)/(5 s) = 1.1519 m/s.

  4. v = rω M´ The Three Stooges go to the park. Moe and Larry are on a merry-go-round ride of radius r that Larry is pushing counterclockwise, running at a speed v. In a time t, Moe goes from M to M´ and Curly goes from C to C´. Moe is twice as far from the center as Curly. The distance Moe travels is rθ,where θ is in radians. Curly’s distance is ½ rθ. Both stooges sweep out the same angle in the same time, so each has the same angular speed. However, since Moe travels twice as far, his linear speed in twice as great. C´ θ v/2 C v M s = rθ rθ t θ t st r = = v = rω

  5. Ferris Wheel Problem Schmedrick is working as a miniature ferris wheel operator (radius 2.1 m). He gets a little overzealous and cranks it up to 75 rpm. His little brother Poindexter flies out at point P, when he is 35° from the low point. At the low point the wheel is 1 m off the ground. A 1.5 m high wall is 27 m from the low point of the wheel. Does Poindexter clear the wall? P Strategy outlined on next slide

  6. Ferris Wheel Problem-Solving Strategy 1. Based on his angular speed and the radius, calculate Poindexter’s linear speed. 2. Break his launch velocity down into vertical and horizontal components. 3. Use trig to find the height of his launch. 4. Use trig to find the horizontal distance from the launch site to the wall. 5. Calculate the time it takes him to go that far horizontally. 6. Calculate height at that time. 7. Draw your conclusion. 16.4934 m/s vx = 13.5106 m/svy0 = 9.4602 m/s 1.3798 m 25.7955 m 1.9093 s 1.5798 m He just makes it by about 8 cm!

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