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Comparing rotational and linear motion

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Angular Velocity Angular velocity, is the rate of change in angular displacement. (radians per second.) fAngular frequency f (rev/s). f Angular frequency f (rev/s). Angular velocity can also be given as the frequency of revolution, f (rev/s or rpm): Angular velocity in rad/s. Angular velocity in rad/s. tttt

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Angular Acceleration Angular acceleration is the rate of change in angular velocity. (Radians per sec per sec.) The angular acceleration can also be found from the change in frequency, as follows:

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Torque and Angular Acceleration When an object is subject to a net force, it undergoes an acceleration. (Newton’s 2 nd ) When a rigid object is subject to a net torque, it undergoes an angular acceleration When a rigid object is subject to a net torque, it undergoes an angular acceleration. Force and Linear Acceleration

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Inertia of Rotation Consider Newton’s second law for the inertia of rotation F = 20 N a = 4 m/s 2 Linear Inertia, m = F/a m = = 5 kg 20 N 4 m/s 2 F = 20 N R = 0.5 m = 2 rad/s 2 Force Force does for translation what torque torque does for rotation : Rotational Inertia, I I = = = 2.5 kg m 2 (20 N)(0.5 m) 2 rad/s 2

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Moment of Inertia This mass analog is called the moment of inertia, I, of the object is defined relative to rotation axis SI units are kg m 2

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More About Moment of Inertia I depends on both the mass and its distribution. If an object’s mass is distributed further from the axis of rotation, the moment of inertia will be larger.

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Common Moments of Inertia Common moments of inertia are on page 251.

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Example 1 A circular hoop and a disk each have a mass of 3 kg and a radius of 20 cm. Compare their rotational inertias. R I = mR 2 Hoop R I = ½mR 2 Disk I = 0.120 kg m 2 I = 0.0600 kg m 2

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Important Analogies For many problems involving rotation, there is an analogy to be drawn from linear motion. x f R 4 kg 50 rad/s = 40 N m A resultant force F produces negative acceleration a for a mass m. I m A resultant torque produces angular acceleration of disk with rotational inertia I.

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Example 2 Treat the spindle as a solid cylinder. a) What is the moment of Inertia of the spindle? b) If the tension in the rope is 10 N, what is the angular acceleration of the wheel? c) What is the acceleration of the bucket? d) What is the mass of the bucket? M

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Solution a) What is the moment of Inertia of the spindle? Given: M = 5 kg, R = 0.6 m M = 0.9 kgm 2

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Solution b) If the tension in the rope is 10 N, what is ? Given: I = 0.9 kg m 2, T = 10 N, r = 0.6 m M (0.6m)(10 N)/(0.9 kg∙m 2 ) = 6.67 rad/s 2 c) What is the acceleration of the bucket? Given: r=0.6 m, = 6.67 rad/s a=4 m/s 2 a = (6.67 rad/s 2 )(0.6 m)

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Solution d) What is the mass of the bucket? Given: T = 10 N, a = 4 m/s 2 M M = 1.72 kg

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Comparing rotational and linear motion

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Combined Rotation and Translation v cm First consider a disk sliding without friction. The velocity of any part is equal to velocity v cm of the center of mass. v R P Now consider a ball rolling without slipping. The angular velocity about the point P is same as for disk, so that we write: Or

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Two Kinds of Kinetic Energy v R P Kinetic Energy of Translation: K = ½mv 2 Kinetic Energy of Rotation: K = ½I 2 Total Kinetic Energy of a Rolling Object: KE of center-of-mass motion KE due to rotation

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Example 3 What is the kinetic energy of the Earth due to the daily rotation? Given: M earth =5.98 x10 24 kg, R earth = 6.63 x10 6 m. First, find = 7.27 x10 -5 rad/s = 2.78 x10 29 J

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Summary – Rotational Analogies QuantityLinearRotational DisplacementDisplacement x Radians InertiaMass (kg)I (kg m 2 ) ForceNewtons NTorque N·m Velocity v “ m/s ” Rad/s Acceleration a “ m/s 2 ” Rad/s 2 Momentum mv (kg m/s) I (kg m 2 rad/s)

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Analogous Formulas Linear MotionRotational Motion F = ma = I K = ½mv 2 K = ½I 2 Work = Fx Work = Power = Fv Power = I Fx = ½mv f 2 - ½mv o 2 = ½I f 2 - ½I o 2

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Example 4 A solid sphere rolls down a hill of height 40 m. What is the velocity of the ball when it reaches the bottom? (Note: We don’t know r or m!) v = 23.7 m/s

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Angular Momentum Analogy between L and p Angular MomentumLinear momentum L = I p = mv = L/ tF = p/ t Conserved if no net outside torques Conserved if no net outside forces Rigid body Point particle

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Example 5 A 65-kg student sprints at 8.0 m/s and leaps onto a 110-kg merry-go-round of radius 1.6 m. Treating the merry-go-round as a uniform cylinder, find the resulting angular velocity. Assume the student lands on the merry-go-round while moving tangentially.

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Solution Known: M, R, m, v 0 Find: F First, find L 0 Next, find I tot Now, given I tot and L 0, find = 2.71 rad/s

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Summary of Formulas: I = mR 2 mgh o ½ ½mv o 2 = mgh f ½ f ½mv f 2 Height?Rotation?velocity?Height?Rotation?velocity? Conservation:

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