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Download Dentist Dentist Dentist Dentist Dentist DentistDentist Dentist Dentist Dentist and more Lecture notes Dentistry in PDF only on Docsity! Dynamics LINEAR MOTION (1) s (2) V (3) a (4) v, = X ·X 1 0 - x1 . Xo - t V1 • V0 -- t - Vo + - !J.x -- 6t !J.v -- !J,t at (5) s - V 0 t + .l.. a t2 - 2 (6) v 2 = 1 V 2 + 0 2as (7) s - 1 ( V 0 + V 1 ) t - - 2 Variable Symbol Units Position X m Displacement s m Velocity (initial) Vo m/s Velocity (final) v, m/s Accelerati on a m/s2 Time t s F=ma F=ks For ce (Total) F N Mass M kg Spring stiffness k N/mm a C = c.;2 r F C = mCJ2 r Centrifugal Accel. a, m!s' Centrifugal Force Fe N W=FS PE=mgh KE=0.5 m v2 SE= 0.5 k x2 P=W/t P=Fv Work w J Potential Energy PE J Kinetic Energy KE J Spring Energy SE J Power p 'N Conservation of Energy (11) (12) (13) (14) (15) (16) ROTARY MOTION 0 = 0 -0 1 0 (l) = 0 1 -0 0 60 -- t 6t a.= (01 • (00 6ro -- t 6t (l) = 1 (l) + 0 a.t 0 - coo t + .l.. a. t2 - 2 (l) 2 = 1 (l) 2 +2a.0 0 (17) 0 - 1 ( ro0 + ro 1 ) t - - 2 Rot Variable Symbol Units Angle 0 rad Angular Displacement 0 rad Angula r Velocity (Initia l) COo rad/s Angular Velocity (F inal) (01 rad/s Angular Acceleration a. rad/s2 Time t s T=la, Torque T Nm Mass Moment of Inertia I kgm2 Torsional Spring stiffness /Not used/ (Not used / a C = v2 / r F C = mv2 / r Centrifugal Acceleration a, m!s' Centrifugal Force Fe N W=T8 - KE= 0.5 I CJ2 - P=W/t P=TCJ Work w J (No circular gravity) . . Kinetic Energy (rotary) KE J (Nor used) . . Power p vv PEI + KEl + SEl + ,vin - ,vout = PE2 + KE2 + SE2 Mo1llentn1ll = Ill v l1llpnlse =Ft= Change of Illomentn1ll = Ill(v1-v0) Coefficient of Restitution c = v1/v0 1 ft= 0.3048 m Prefix Sy1nbol 1 in = 25 .4 Illlll ' G g1ga 1 Rev = 21T rad Illega M 1 rad= 180/1T = 57.3 ° kilo k 1 1ll/s = 3.6 k1n/h 1llilli Ill 1 RPM= 21T/60 rad/s 1ll1cro µ Value 109 106 103 10 -3 1 o - 6 To solve a linear or rotary problem: (1) Pick 3 known + 1 unknown variable (2) Choose formula with this set of 4 variables (3) Re-arrange to solve for unknown variable Trigonometric Rules a2 = b2 + c2 - 2bc cos A a sin A = b - - C sin C sin B Newton's Laws of Motion (1) If no net force then constant velocity (2) F = ma (3) Every action has equal & opposite reaction Friction on Inclined Plane How to do i t (typical) (1) Resolve into normal and parallel components (2) If in equil, then I normal = 0 and I parallel = 0 (3) Apply Fr= µF nonce you have 2 unknowns Friction Force: Fr= µF 0 Weight Force: F w = m g Normal Force: F = F cos 8 n w Parallel Force: F = F sin 8 p w µ= Coefficient of friction 4>= Angle of friction 8= Angle of inclination MOMENTS OF INERTIA I thin hoop or r ing of thick ring of inner radius solid cylinder or disc radius R & mass M: R1 , outer radius R2, and of radius R and mass M: mass M: M*R•2 M*(R1•2 • R2•2)/2 (M*R•2)/2 solid sphere of radius thin-walled hollow sphere slender rod of length L R and mass M: of radius R & mass M: and mass M, spinning around center: (2/5)*M*R•2 (2/3)*M*R•2 (M*L • 2)/12 N = RPM Shaft Power P=Tm= 21rNT 60 T = Torque (Nm) V Belts Pitch diameter � = e(µetsin /J) Velocity Ratio VR =S E / S L Sheave outside diameter S E = Distance moved by effort S L = Distance moved by load Mechanical Advantage MA= F L / F E F E = Force of effort F L = Force of load Wedge angle Belt ride-out Groove depth C flat plate with sides of length A and B and mass M: GVG/Pl)/l,1 slender rod of length L and mass M, spinning around end: (M*L • 2)/3 Ft Ff Ltt.r--t.11r25N VR of block and tackle fc =100 Ni Efficiency Number of ropes 7/ =MA/ VR attached to the load h=10 cm