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Rotational Motion & Angular Momentum

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Rotational Motion & Angular Momentum Rotational Motion & Angular Momentum Rotational Motion Every quantity that we have studied with translational motion has a rotational counterpart TRANSLATIONAL ROTATIONAL Displacement x Angular Displacement Velocity v Angular velocity acceleration a Angular acceleration Mass m Inertia I Momentum p Angular Momentum L Force F Torque Angular Position s Arclength Radius (from axis of r rotation) Measured in radians – all angular quantities will be measured in radians (NOT degrees) Angular Position • In translational motion, x = 3 position is represented by a x point, such as x. 0 5 linear p/2 • In rotational motion, r position is represented p 0 by an angle, such as , and a radius, r. angular 3p/2 Displacement Dx = 4 • Linear displacement x is represented by the 0 5 vector Dx. linear p/2 • Angular displacement is D 60 represented by D, which is not a vector, but p 0 behaves like one for small values. angular 3p/2 Angular Displacement Compare to D풙 = 풙 − 풙풐 D 0 Which direction is positive (by convention)? Positive – Counterclockwise Negative - Clockwise EXAMPLE: (a) What is the angular position, , if we go around a circle two times? s 2(2p푟) Ans. = = = 4π r r (b) Say you go a quarter turn more, what is D? 9휋 휋 Ans. D = = − 4π = o 2 2 (c) What is the arclength covered in total between Probls. (a) and (b) if the radius of the circle is 3 m? 휋 9휋 27휋 Ans. s = r = (3 m)(4p + ) =3m( ) = m 2 2 2 P.O.D. 1: A Boy on a horse is hunting a goose around a strange world of radius 25 m. The angular separation between the hunter/hunted is a constant 5. What is the angular distance (in m) between the boy/horse and the goose? v Speed and velocity s T • The same particle rotates with an D r avg. angular velocity given by vT D훉 = D퐭 • Linear (Tangential) and angular speeds are related by the equation v = r Angular Velocity Compare to ∆풙 풗 = D 풂풗품 ∆풕 avg Dt Units – rad/s, rev/s Direction – Same as displacement (positive is counterclockwise, negative is clockwise) EXAMPLE: A space station ring of radius 500 m spins twice in 12 minutes. (a) Find its angular velocity (in rads/s). (b) Find the linear (tangential) velocity (in m/s) of a point on the outer edge of the ring. Ans. (a) Twice means two revolutions = 2(2p) = 4 p radians 휃 4휋 = = 푠푒푐 = 0.017 rad/sec 푡 12 min × 60 Τ푚푛 (b) tangential velocity, vT = r = 0.017 (500 m) = 8.73 m/s P.O.D. 2: A figure skater spins through five revolutions in a time of 2.4 s. (a) Find her angular velocity (in rads/s). (b) Find the linear (tangential) velocity (in m/s) of her foot if it is 0.3 meters away from her body. v Acceleration s T • The same particle rotates with an D r avg. angular acceleration given by vT D = D퐭 • Linear (Tangential) and angular accelerations are related by the equation a = r Angular Acceleration D Compare to ∆풗 풂 = 풂풗품 ∆풕 avg Dt • Angular Acceleration is how the angular velocity changes with time Units – rad/s2 Centripetal Acceleration A pendulum is swinging back and forth. At the bottom of the swing the force of gravity is pulling it downwards but it doesn’t fall down. This means there must be a force pulling upwards to balance it out. Fc This is the centripetal force. Since F = ma, the center-seeking acceleration is called centripetal Fg acceleration and is given by: v2 a 2r c r Sample Problem A compact disk rotates about an axis according to the formula: (t) = t2 – 6 (a) What is the linear speed of a point 20 cm from the center at t = 5 s? (b) What is the linear acceleration at 0.5 m at t = 5 s? Ans. (a) To find the linear speed, use the relationship, where (5) we obtained from the function (t) = t2 – 6 (5)= 52 –6=19 v = r = (19 rad/s)(0.20 m) = 3.8 m/s. (b) To find the linear acceleration, use the relationship 2 2 2 ac = r = (19 rad/s) (0.20 m) = 72.2 m/s P.O.D. 3: Diana Prince starts rotating with an angular velocity of 5 rad/s, where the negative sign indicates a clockwise rotation, to transform into Wonder Woman. After 5 seconds she has completed the transformation and her angular velocity is 2 rad/s. (a) Find Wonder Woman’s angular acceleration (in rad/s2). (b) Find the centripetal acceleration (in m/s2) of her indestructible bracelets (at = 2 rad/s) if they are 0.8 m away from her body when she is spinning. Constant Angular Acceleration • Our kinematics equations have angular equivalents • Just as with their linear counterparts, these only work for constant acceleration First Kinematic Equation 2 • x = vot + ½ at (linear form) – Substitute angle for position. – Substitute angular velocity for linear velocity. – Substitute angular acceleration for linear acceleration. 2 = ot + ½ t (angular form) Second Kinematic Equation • v = vo + at (linear form) – Substitute angular velocity for linear velocity. – Substitute angular acceleration for linear acceleration. = o + t (angular form) Sample Problem A bicycle starts from rest and for 10.0 s has a constant linear acceleration of 0.8 m/s2 to the right. During this period, the tires do not slip. The radius of the tires is 0.50 m. At the end of the 10.0 s interval what is the angle through which each wheel has rotated? Ans. The angular acceleration can be found from the formula 풎 풂 ퟎ.ퟖ ퟐ a = r 휶 = 휶 = 풔 = 1.6 rad/s2 풓 ퟎ.ퟓퟎ 풎 The angular acceleration should be negative because the tire spins clockwise. To find the angular displacement: 2 2 2 (t) = ot + ½t = 0(10 s) + ½(1.6 rad/s )(10 s) = 80 rad P. O. D. 4: an extreme diver rotates at an angular velocity of +3.75 rad/s while doing his first set of flips. When he does his second set of flips, he accelerates and reaches a greater angular velocity of +4.4 rads/s. If his angular acceleration is 1.7 rad/s2... (a) How long does this flipping part of the dive take (in s)? (b) Find his angular displacement after the time found in (a). ANGULAR MOMENTUM! Angular Momentum Angular momentum depends on linear momentum and the distance from a particular point. It is a vector quantity with symbol L. If r and v are then the magnitude of angular momentum w/ resp. to point Q is given by L = rp = mvr. In this case L points out of the page. If the mass were moving in the opposite direction, L would point into the page. The SI unit for angular momentum is the kgm2 / s. (It has no special v name.) Angular momentum is a conserved quantity. A torque is needed to change L, just a force is r m needed to change p. Anything spinning has angular has angular Q momentum. The more it has, the harder it is to stop it from spinning. Angular Momentum: General Definition If r and v are not then the angle between these two vectors must be taken into account. The general definition of angular momentum is given by a vector cross product: L = r p This formula works regardless of the angle. From cross products, the magnitude of the angular momentum of m relative to point Q is: L = r p sin = m v r. In this case, by the right-hand rule, L points out of the page. If the mass were moving in the opposite direction, L would point into the page. v r m Q Moment of Inertia vs. Angular Momentum Any moving body has inertia. (It wants to keep moving at constant v.) The more inertia a body has, the harder it is to change its linear motion. Rotating bodies possess a rotational inertia called the moment of inertia, I. The more rotational inertia a body has, the harder it is change its rotation. For a single point-like mass w/ respect to a given point Q, I = mr 2. For a system, I = the sum of each mass m times its respective distance from the point of interest. r m2 Q m1 2 r1 I = mr r2 Q 2 2 2 I = mi ri = m1 r1 + m2r2 Moment of Inertia of various shapes Moment of Inertia Example Two merry-go-rounds have the same mass and are spinning with the same angular velocity. One is solid wood (a disc), and the other is a metal ring. Which has a bigger moment of inertia relative to its center of mass? r r m m Ans. I is independent of the angular speed. Since their masses and radii are the same, the ring has a greater moment of inertia. This is because more of its mass is farther from the axis of rotation. Since I is bigger for the ring, it would more difficult to increase or decrease its angular speed. Torque & Angular Acceleration nd Newton’s 2 Law, as you know, is Fnet = ma nd The 2 Law has a rotational analog: net = I A force is required for a body to undergo acceleration. A “turning force” (a torque) is required for a body to undergo angular acceleration. The bigger a body’s mass, the more force is required to accelerate it. Similarly, the bigger a body’s rotational inertia, the more torque is required to accelerate it angularly.
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