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Ch 7 - Circular Motion • Circular motion: Objects moving in a circular path. Measuring Rotational Motion • Rotational Motion – when an object turns about an internal axis. – Ex. Earth’s is every 24 hrs. • Axis of rotation – the line about which the rotation occurs • Arc length – the distance (s) measured along the circumference of the circle • Radian- an angle whose arc length is equal to its radius, which is approximately equal to 57.3° • When the arc length “s” is equal to the length of the radius, “r”, the angle θ swept by r is equal to one rad. • Any angle θ is radians if defined by: • θ = s/r • When a point moves 360°, θ=s/r = 2πr/r = 2π rad Therefore, to convert from rads to degrees θ(rad) = π θ(deg) 180° For angular displacement, Δθ=Δs/r Angular displacement (in radians)= change in arc length/distance from axis Example Problem • While riding on a carousel that is rotating clockwise, a child travels through an arc length of 11.5 m. If the child’s angular displacement is 165°, what is the radius of the carousel? • Ans. 3.98 m Angular Substitutes for Linear Quantities • Linear (Straight Line) – Displacement = x – Velocity = v – Acceleration = a • Rotational – Displacement = θ – Velocity = ω – Acceleration = α • Angular speed – the rate at which a body rotates about an axis, expressed in radians per second • Symbol = ω (omega) Unit = rad/s • ω(avg) = Δθ/Δt • ω can also be in rev/s • To convert: 1 rev = 2π rad Example Problem • A child at an ice cream parlor spins on a stool. The child turns counter-clockwise with an average angular speed of 4.0 rad/s. In what time interval will the child’s feet have an angular displacement of 8.0 rad • Ans. 6.3 s • Angular Acceleration – the time rate of change of angular speed, expressed in radians per second per second • avg = ω2-ω1/t2-t1 =Δω/Δt • Average angular acceleration = change in angular speed / time interval Example Problem • A car’s tire rotates at an initial angular speed of 21.5 rad/s. The driver accelerates, and after 3.5 s the tire’s angular speed is 28.0 rad/s. What is the tire’s average angular acceleration during the 3.5 s time interval? • Ans. 1.9 rad/s2 Frequency vs. Period • Frequency – # of revolutions per unit of time. Unit: revolutions/second (rev/s). • Period – time for one revolution. Unit = second (s). • Inversely related: t = 1/f and f = 1/t Tangential Velocity • Speed that moves along a circular path. • Right angles to the radii. • Direction of motion is always tangent to the circle. Rotational Speed • The number of rotations per unit of time. • All parts of the object rotate about their axis in the same amount of time. • Units: RPM (revolutions per minute). Tangential vs. Rotational • If an object is rotating: – All points on the object have the same rotational (angular) velocity. – All points on the object do not have the same linear (tangential) velocity. • Tangential speed is greater on the outer edge than closer to the axis. A point on the outer edge moves a greater distance than a point at the center. • Tangential speed = radial distance x rotational speed v r Centripetal Acceleration • The acceleration of an object moving in a circle points toward the center of the circle. • Means “center seeking” or “toward the center”. ac r 2 v ac r 2 7.3 Forces that maintain circular motion • Consider a ball swinging on a string. Inertia tends to make the ball stay in a straight-line path, but the string counteracts this by exerting a force on the ball that makes the ball follow a circular path. • This force is directed along the length of the string toward the center of the circle. The force that maintains circular motion (formerly known as centripetal force) • Fc = (mvt2)/r • Force that maintains circular motion = mass x (tangential speed)2 ÷ distance to axis of motion • Fc = mrω2 • Force that maintains circular motion = mass x distance to axis x (angular speed)2 • Because this is a Force, the SI unit is the Newton (N) Practice Problem • A pilot is flying a small plane at 30.0 m/s in a circular path with a radius of 100.0 m. If a force of 635N is needed to maintain the pilot’s circular motion, what is the pilot’s mass? • Answer: m = 70.6 kg Common Misconceptions • Inertia is often misinterpreted as a force • Think of this example: How does a washing machine remove excess water from clothes during the spin cycle? Newton’s Law of Universal Gravitation • Gravitational force: a field force that always exists between two masses, regardless of the medium that separates them; the mutual force of attraction between particles of matter • Gravitational force depends on the distance between two masses Newton’s Law of Universal Gravitation Gm1m2 F 2 r Where F = Force M1 and m2 are the masses of the two objects R is the distance between the objects And G = 6.673 x 10-11 Nm2/kg2 (constant of universal gravitation) Practice Problem • Find the distance between a 0.300 kg billiard ball and a 0.400 kg billiard ball if the magnitude of the gravitational force is 8.92 x 10-11 N. • Answer: r = 3.00 x 10-1 m