PROLEM 7. The rotor of an electric motor has an angular velocity of 600 rpm when the load and power are cut off. The 0-lb rotor, which has a centroidal radius of gyration of 9 in., then coasts to rest. Knowing that the kinetic friction of the rotor produces a couple of magnitude.5 lb ft, determine the number of revolutions that the rotor executes before coming to rest. Kinetic energy. Position. 600 rpm 0 rad/s 0 9 W W m I k.958 lbs ft g g. T I.958 0 6.55 0 ft lb Position. 0 T 0 Work. M.5 lbft U M.5 Principle of work and energy. T U T : 6.55 0.5 0 5460 radians 8690 rev
PROLEM 7. Each of the gears and has a mass of.4 kg and a radius of gyration of 60 mm, while gear C has a mass of kg and a radius of gyration of 50 mm. couple M of constant magnitude 0 N m is applied to gear C. Determine (a) the number of revolutions of gear C required for its angular velocity to increase from 00 to 450 rpm, (b) the corresponding tangential force acting on gear. Moments of inertia. Gears and : I I mk (.4)(0.06) 8.640 kg m I C ()(0.5) 700 kg m Kinematics. Kinetic energy. Position. Gear : Gear : r r r C C 00 C.5C 80.5 C T I : 0 C 00 rpm rad/s 5 50 rpm rad/s 5 ( T ) (8.64 0 ).9609 J 5 ( T ) (8.64 0 ).9609 J 0 ( T ) C (700 ) 4.8044 J System: T ( T) ( T) ( T) C 0.76 J Position. 450 rpm 5 rad/s C 7.5 rad/s
PROLEM 7. (Continued) Gear : Gear : ( T ) (8.64 0 )(7.5 ) 59.957 J ( T ) (8.64 0 )(7.5 ) 59.957 J ( T ) C (700 )(5 ) 99.789 J System: T ( T) ( T) ( T) C 49.7 J Work of couple. U MC 0 C Principle of work and energy for system. T U T C : 0.76 0 49.7 C 9.898 radians (a) Rotation of gear C. C 6.5 rev Rotation of gear. (.5)(9.898) 99.744 radians Principle of work and energy for gear. ( T ) M ( T ) :.9609 M (99.744) 59.957 (b) Tangential force on gear. M 0.574 N m M 0.574 Ft Ft 7.4 N r 0.08
PROLEM 7. Solve Problem 7., assuming that the 0-N m couple is applied to gear. PROLEM 7. Each of the gears and has a mass of.4 kg and a radius of gyration of 60 mm, while gear C has a mass of kg and a radius of gyration of 50 mm. couple M of constant magnitude 0 N m is applied to gear C. Determine (a) the number of revolutions of gear C required for its angular velocity to increase from 00 to 450 rpm, (b) the corresponding tangential force acting on gear. Moments of inertia. Gears and : I I mk (.4)(0.06) 8.640 kg m I C ()(0.5) 700 kg m Kinematics. Kinetic energy. Position. Gear : Gear : r r r C C 00 C.5C 80.5 C T I : 0 C 00 rpm rad/s 5 50 rpm rad/s 5 ( T ) (8.64 0 ).9609 J 5 ( T ) (8.64 0 ).9609 J 0 ( T ) C (700 ) 4.8044 J System: T ( T) ( T) ( T) C 0.76 J
PROLEM 7. (Continued) Position. C 450 rpm 5 rad/s 7.5 rad/s Gear : Gear : ( T ) (8.64 0 )(7.5 ) 59.957 J ( T ) (8.64 0 )(7.5 ) 59.957 J ( T ) C (700 )(5 ) 99.789 J System: T ( T) ( T) ( T) C 49.7 J Work of couple. U M 0 Principle of work and energy for system. (a) Rotation of gear C. Rotation of gear. T U T : 0.76 0 49.7 Principle of work and energy for gear. 9.898 radians 9.898 C 5.959 radians C.54 rev.5 9.898 radians ( T ) M ( T ) :.9609 M (9.898) 59.957 (b) Tangential force on gear. F t M.485 N m M.485 Ft 7.86 N r 0.08
PROLEM 7.4 The 0-kg turbine disk has a centroidal radius of gyration of 75 mm and is rotating clockwise at a constant rate of 60 rpm when a small blade of weight 0.5 N at Point becomes loose and is thrown off. Neglecting friction, determine the change in the angular velocity of the turbine disk after it has rotated through (a) 90, (b) 70. Mass of blade. m 5 grams 0.05 kg Weight of blade. mg (0.05)(9.8) 0.5 N Moment of inertia about O. O 0(0.75) 5 0 (0.) 0.946 kg m I mk m r Location of mass center for the position shown. mr ( mm) x mr x m m Position. 0, 60 rpm rad/s Kinetic energy: I O T Center of gravity lies at the level of Point O. h 0 Potential energy: V ( mg mg) h 0 (a) Position. 90 Kinetic energy: Center of gravity lies a distance mr m m T I O above Point O. h mr m m Potential energy: V ( mg mg) h mgr (0.5)(0.) 0.50 N m
PROLEM 7.4 (Continued) Conservation of energy. T V T V : IO 0.946 IO I V 0 O V ()(0.5) ( ) 6.5706 rad/s 6.5706 0.067 rad/s 0.50 rpm (b) Position. 70 Kinetic energy: Center of gravity lies a distance mr m m T I O below Point O. h mr m m Potential energy: V ( mg mg) h mgr (0.5)(0.) 0.5 N m Conservation of energy. T V T V : IO I V 0 O V ()( 0.5) ( ) IO 0.946 6.0946 rad/s 6.0946 0.0606 rad/s 0.49 rpm
PROLEM 7.7 Greek engineers had the unenviable task of moving large columns from the quarries to the city. One engineer, Chersiphron, tried several different techniques to do this. One method was to cut pivot holes into the ends of the stone and then use oxen to pull the column. The 4-ft diameter column weighs,000 lbs, and the team of oxen generates a constant pull force of 500 lbs on the center of the cylinder G. Knowing that the column starts from rest and rolls without slipping, determine (a) the velocity of its center G after it has moved 5 ft, (b) the minimum static coefficient of friction that will keep it from slipping. Given: Diagram of two positions: 000 lb m =7.67 slugs. ft/s I mr, r ft, F 500 lbs, x 5 ft Work Energy Equation: T V g V e, ' U T V g V e, U ' 5 0 Fdx 7500 ftlb T mv I Kinematics: v r (a) Solving for Velocity at positon : 7500 000 v 000 g 4g r v r v 5.8 ft/s
PROLEM 7.7 (Continued) Free ody Diagram: Kinetics: Fx max Fy may MG 500 f ma f 500 ma x x N mg 0 fr I N 000 lb I Kinematics: Substitute into moment equation: Substitute into x-direction equation: (b) Minimum Necessary Coefficient of Friction: ax r ax fr mr r f ax m f f 500 m m f 500 lbs f s s 0.04 N
PROLEM 7.7 5-m long ladder has a mass of 5 kg and is placed against a house at an angle 0. Knowing that the ladder is released from rest, determine the angular velocity of the ladder and the velocity of when 45. ssume the ladder can slide freely on the horizontal ground and on the vertical wall. Kinematics: Let v v, v v, and ω. Locate the instantaneous center C by drawing C perpendicular to v and C perpendicular to v. Triangle GC is isosceles. G G GC L/. The velocity of the mass center G is Kinetic energy: v v L / G T mv I I ml 4 Since the ladder can slide freely, the friction forces at and are zero. Use the principle of conservation of energy. T V T V : Potential energy: Use the ground as the datum. V mgh where L h cos
PROLEM 7.7 (Continued) Position. 0 ; rest (T 0) Position. 45 ;? L L 0 mg cos 0 I ml mg cos 45 4 Data: ssume m5 kg. L5 m g 9.8 m/s (5 kg)(5 m).5 kg m I ml I ml.5kgm (5kg)(5m) 5kg m 4 4 (5 kg)(9.8 m/s )(.5 m)(cos 0cos45 ) (5 kg m ).690 rad /s.7004 rad/s ngular velocity. ω.70 rad/s Velocity of end. v Lcos (.7004 rad/s)(5 m)cos0 v 5.07 m/s