Chapter 8 Rotational Equilibrium and Rotational Dynamics Wrench Demo Torque Torque is the tendency of a force to rotate an object about some axis Fd F is the force d is the lever arm or moment arm Units are Newton m Direction of Torque Torque is a vector quantity Direction determined by axis of twist Perpendicular to both r and F In 2 d torque points into or out of paper Clockwise torques point into paper Defined as negative Counter clockwise torques point out of paper Defined as positive Non perpendicular forces Only the ycomponent Fsin produces a torque Non perpendicular forces Fr sin F is the force r is distance to point where F is applied is the angle between F and r Torque and Equilibrium Forces sum to zero no linear motion Fx 0 and Fy 0 Torques sum to zero no rotation 0 Meter Stick Demo Axis of Rotation Torques require point of reference Point of reference can be anywhere Use same point for all torques Pick the point to make problem least difficult Example Given M 120 kg Neglect the mass of the beam a Find the tension in the cable b What is the force between the beam and the wall Solution a Given M 120 kg L 10 m x 7 m Find T Basic formula 0 Axis TL Mgx 0 Solve for T 824 N Solution b Given M 120 kg L 10 m x 7 m T 824 N Find f Basic formula F y 0 T Mg f 0 Solve for f 353 N f Alternative Solution b Given M 120 kg L 10 m x 7 m Find f Basic formula 0 Mg L x fL 0 Solve for f 353 N f Axis Another Example Given W 50 N L 0 35 m x 0 03 m Find the tension in the muscle W x Basic formula 0 L Fx WL 0 F 583 N Center of Gravity Gravitational force acts on all points of an extended object However it can be considered as one net force acting on one point the center of gravity X m g x i i i X mi g i m x X m i i i i i Weighted Average Example Consider the 400 kg beam shown below Find TR Solution Find TR Basic formula 0 to solve for TR choose axis here MgX TR L 0 TR 1 121 N X 2m L 7m One Last Example Given x 1 5 m L 5 0 m wbeam 300 N wman 600 N Find T Fig 8 12 p 228 Slide 17 x L Solution Given x L wbeam wman Find T First find Ty Basic formula Fig 8 12 p 228 0 Slide 17 wman x wbeam L 2 Ty L 0 Ty 330 N Now find T Ty T sin 53 T 413 N x L Baton Demo Moment of Inertia Demo Torque and Angular Acceleration When 0 rigid body experiences angular acceleration Relation between and is analagous to relation between F and a F ma I Moment of Inertia Moment of Inertia mass analog is called the moment of inertia I of the object This I mi ri 2 i r is defined relative to rotation axis SI units are kg m2 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 Demo Moment of Inertia Olympics Moment of Inertia of a Uniform Ring Divide ring into segments The radius of each segment is R I mi ri MR 2 2 Example What is the moment of inertia of the following point masses arranged in a square a about the x axis b about the y axis c about the z axis Solution a Find I about the x axis Given M2 2 M3 3 kg L 0 6 m Basic formula I mi ri 2 r 0 6 sin 45 First find distance to 2 kg masses I M 2r M 2r 2 2 0 72 kg m2 Solution b Find I about the y axis r 0 6 sin 45 Same as before except you use the 3 kg masses I M 3r 2 M 3r 2 1 08 kg m2 Solution c Find I about the z axis r 0 6 sin 45 Use all the masses I M 2r M 2r M 3r 2 M 3r 2 2 2 1 8 kg m2 Other Moments of Inertia Example 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 Solution a What is the moment of Inertia of the spindle Given M 5 kg R 0 6 m Moments of Inertia 2 MR cylindrical shell 1 MR 2 solid cylinder 2 2 MR 2 solid sphere 5 2 MR 2 spherical shell 3 1 ML2 rod about end 3 1 ML2 rod about middle 12 M 1 I MR 2 0 9 kgm2 2 Solution b If the tension in the rope is 10 N what is Given I 0 9 kg m2 T 10 N r 0 6 m Basic formula rF I rT I Solve for 6 67 rad s2 c What is the acceleration of the bucket Given r 0 6 m 6 67 rad s Basic formula a r a 4 m s2 M Solution d What is the mass of the bucket Given T 10 N a 4 m s2 Basic formula F ma Ma Mg T T M g a M 1 72 kg M Example A cylindrical space station of radius R 12 m and mass M 3400 kg is designed to rotate around the axis of the cylinder The space station s moment of inertia is 0 75 MR2 Retrorockets are fired tangentially to the surface of space station and provide an impulse of 2 9x104 N s a What is the angular velocity of the space station after the rockets have finished firing b What is the centripetal acceleration at the edge of the space station Solution Given M 3400 R 12 I 0 75 MR2 3 67x105 F t 2 9x104 a Find Basic formula I I t Basic formula rF t I RF t I Solve for 0 948 rad s b Find centripetal acceleration Basic formula 2 a r a 2R 10 8 m s2 Rotational Kinetic Energy Each point of a rigid body rotates with angular velocity 1 1 1 2 2 2 2 KE mi vi mi ri I 2 2 2 Including the linear motion 1 2 1 2 KE mv I 2 2 KE due to rotation KE of center of mass motion Example What is the kinetic energy of the Earth due to the daily rotation Given Mearth 5 98 x1024 kg Rearth 6 63 x106 m Basic formula 2 T Solid sphere …
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