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Newton’s second law for rotation (转动定律) 应用物理系—张春华 Key words: Particle (质点) Rigid body(刚体) Rotational inertia(转动惯量) Torque(力矩) 1. The rotational variables: REVIEW! s r (1) Angular position: (角位置) (2) Angular displacement: (角位移) (3) Angular velocity (角速度) f i Rotation axis d dt (4) Angular acceleration (角加速度) d dt C m R Reference line Zero angular position 2. Relationship of the linear and angular variables Distance: Speed: REVIEW! s r v r Tangential acceleration: at r Radial acceleration: an r 2 REVIEW! 3. Rotational inertia (转动惯量): I mi ri 2 axis I r dm 2 For rigid body : dm For particle: I=mr2 M Unit: kg.m2 The rotational inertia involves both mass and its distribution. r Newton’s second law for rotation (转动定律) (1)Torque (力矩) Torque is a turning or twisting action on a body about a rotation axis due to a force. rF z F Ft O r d rF sin rFt Fd Where and is the angle between r and F , Ft is the component of perpendicular F to . r Unit: N.m P fji Notes: F2 R O r1 fij F1 mi mj d o (1) Torque is a vector. Its direction always along the axis (for rotating about fixed axis), either positive or negative. (right hand rule) (2) If several torque act on a rigid body that rotate about a fixed axis, the net torque is the sum of individual torque. net 1 2 3 i (3) The net torque of internal forces of rigid body is zero. f d 0 i i i (2) Newton’s second law for rotation (转动定律) Newton’s second law: F net ma For rigid body: Treat the rigid body as a collection of particles, for one mass element mi : F i and f i are the external and internal forces acting on the mass element . Then we have: z ω,α Fit ri O Fi F i f i mi a i mi fit fi Only the tangential component of the force can accelerate the particle along the path. So: Fit f it mi ait Fit ri f it ri mi ait ri mi ri 2 For whole rigid body: Fit ri fit ri mi ri2 That is: i 0 I net I net I net I Compare with: F net ma Review & summary: net I rF I mi ri2 net i I r 2 dm d dt Sample problem: Fig. Shows a uniform disk, with mass M=2.5kg and radius R=20cm, mounted on a fixed horizontal axle. A block with mass m=1.2kg hangs from a mass-less cord that is wrapped around the rim of the disk. Find the acceleration of the falling block, the angular acceleration of the disk, and the tension in the cord. The cord does not slip and there is no friction at the axle. (I=1/2MR2) R Axis O M m α M ·R T ′= - T a T m mg Solution: For M: = TR = I mg – T = ma For m: a = R α (1) (2) M· R (3) T From above equations, we find the answer: a=4.8 m/s2 T ′= - T =24 rad/s2 T=6.0 N Home work: a m mg