Moment of inertia ROTARY ACTUATOR ● The load will create inertial forces (kinetic energy) when Moment of inertia moving the load with Rotary Actuator. In order to stop the moving load,It is necessary to use stopper or Shock ● Moment of inertia shows, it is not easy to rotate the stationary Absorbers to absorb the kinetic energy of load. object; the same which means it is difficult to stop the rotating object. ● The moving load with actuator can be distinguished as ● Rotary Actuators in the allowable kinetic energy has its limitations, following it can be calculated moment of inertia to calculate minimum rotation 1. Linear motion (air cylinder), Fig.(1) of moment of inertia described as following. 2. Rotation motion (rotary actuator), Fig.(2) r ● Calculate the kinetic energy by using the formula in FIG. L I = m • r2 Linear motion m : Mass m r : Radius of rotation E : Kinetic energy m m : Load mass V : Speed V Above figure represents the moment of inertia for the distance r from 1 2 E = ― • m • V ....(1) rotary shaft to mass m of the object. 2 The formula for moment of Inertia is not the same if the shapes of the object are different. Fig. (1) Linear motion The following examples are the calculated on the basis of specific moment of inertia. Rotation motion 1. Thin shaft r Position of rotational axis: Perpendicular to the shaft through the E : Kinetic energy shaft through the center of gravity. 2 m I : Moment of inertia(=m.r ) ω : Speed m : Mass r : Radius of rotation a 2 I = m • —a 12 1 2 1 2 2 E = ― • I • ω = ― • m • r • ω ....(2) 2 2 Fig. (2) Rotation motion ● For linear motion, if the velocity V of formula (1) is constant, 2. Thin rectangular plate the kinetic energy E and mass m is proportional ; The rotation Position of rotational axis: Parallel to side b and through the motion, formula (2) shows even the angular velocity ω and center of gravity. mass m is constant, kinetic energy E will also proportional with r². Therefore, even the mass is small but the rotation radius r is large, when the moment of Inertia I=m·r² is large, kinetic energy E will becomes larger, it will cause bearing damage or other a accidents. 2 I = m • —a ● Therefore when there is a rotation motion, the product selection 12 should be base on moment of inertia instead of mass. Moment of inertia ROTARY ACTUATOR 3. Thin rectangular plate (Including Use the following formula to calculate the rotation time rectangular parallelepiped) t : Rotation time (s) Position of rotational axis: Perpendicular to the plate 2 through the center of gravity. t ≥ 2 • I • θ E : Kinetic energy (J) I : Moment of inertia (kg.m2) E θ : Rotation angle (rad) b a The meaning of this formula is the critical rotation time for not cause damage of the cylinder. Therefore the rotation time must be set on 2 2 or over the t seconds calculated in above formula. I = m • ———a + b 12 After calculated the moment of inertia by load shape, use the following formula to calculate the kinetic energy of the load. 4. Round plate (Including column) Position of rotational axis: Through the center axis. E : Kinetic energy (J) E = 1/2 • I • ω2 I : Moment of inertia (kg.m2) ω : Angle speed (rad/s) r Angle speed 2 I = m • —r 2 t : Rotation time (s) ω = 2 θ / t (1) 2 .... I : Moment of inertia (kg.m ) ω = θ / t.....(2) θ : R otation angle (rad) 5. Solid sphere However, when the rotation time for 90° becomes longer than 2 Position of rotational axis: Through the center of diameter. seconds, use formula (2). Calculation example r Load form: Cuboid Rotation angle θ: 180° Length of a part : 0.12 m 2 I = m • —2r Rotation time t: 1 s/180° Length of b part : 0.06 m 2 Mass (m) : 0.1 kg 2 2 a + b b I = m • ——— a 6. Thin round plate 12 Position of rotational axis: Through the center of diameter. (Step 1) Find the angle speed ω. 2θ 2 r ω = — = — × π = 6.28 rad/s t 1 2 (Step 2) Find the moment of inertia I. I = m • —r 4 2 2 I = m • ———a + b 12 -4 -4 = 0.1 × —144 × 10 + 36 × 10 7. Gear transmission 12 (B) (A) No. of teeth=a = 1.5 × 10-4 kg.m2 1.Find the moment of inertia (Step 3) Find the kinetic energy E. IB for the rotation of shaft (B). 1 2 1 -4 2 2.IB is converted to the E = — • I • ω = — × 1.5 × 10 × 6.28 2 2 moment of inertia IA for the rotation of the shaft (A). = 0.002958 J 2 IA= ( —a ) • IB No. of teeth=b b.