Phys05 - Unit 5 Circular Motions 5.1 5.2 5.3 5.4 5.5...

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1 Unit 5 Circular Motions 5.1 Circular Motion 5.2 Rotation with constant angular acceleration 5.3 Centripetal acceleration 5.4 The banked curve 5.5 The rotor 5.1 Circular Motion For a rigid body to rotate around a fixed axis, each particle in the rigid body undergoes a circular motion. Take one of the particle as an example, θ is the angular displacement, s is the distance it moves and r is the distance between the axis and the particle. = s r (1) The average angular velocity is defined as ω av tt t = = 21 (2) Instantaneous angular velocity == lim t t d dt 0 (3) Instantaneous tangential velocity v T = r . For a circular motion, the arc length is given by sr = , where r is constant. Differentiate both sides, we obtain ds d r dt dt = . Since the instantaneous speed is ds v dt = and the angular velocity is d dt = , we obtain vr = . (4) Angular acceleration α 2 2 0 lim dt d dt d t t = = = s=r r O
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2 (5) Tangential acceleration a T = r α Remark: Although every point in a rigid body travels different distance during the rotation, the angular displacements are the same. At any instant, every point in a rigid body has the same angular velocity. In advanced study, ω is quite often defined as a vector with its direction pointed by the right-hand rule. This is common in the area of science and engineering. 5.2 Rotation with constant angular acceleration Circular Motions Linear kinematics t + = 0 vua t = + 2 0 1 2 tt θ ωα =+ 2 1 2 s ut at 22 0 2 2 vu a s Remarks: (1) t tan cons dt d = = t + = 0 , where 00 ()| t t = = . (2) dt d = = t dt d 0 0 2 1 2 θω + (*) When 0 = 0, we have 2 0 1 2 . (3) Squaring on both sides of 0 t = + gives 2 2 2 ωω αα + . Rewrite the equation as 1 2( ) 2 αω + and using (*). Hence we obtain ) . When 0 = 0, we have 0 2 .
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3 5.3 Centripetal acceleration Consider the case of uniform circular motion. Uniform circular motion: Speed is a constant, e.g. v v v = = 2 1 G G . Obviously, triangles OP 1 P 2 and Op 1 p 2 are similar. 1 v v r s G = or s r v s r v v = = 1 G Acceleration t s r v t v a av = = G r v t s r v a a t av t 2 0 0 lim lim = = = In fact, the centripetal acceleration is given by r r v a ˆ 2 = G , where r ˆ is the unit vector pointed from the origin. The acceleration can be rewritten as r r a ˆ 2 ω = G , since r v = . When we consider only the magnitude of the centripetal force, we have 2 2 v ar r == . Remarks: For a non-uniform circular motion, the acceleration consists of two parts. (1) Tangential component of acceleration a dv dt r d dt r tan ≡= = α (Non-uniform circular motion) Its direction is always tangential to the circular path of the particle.
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Phys05 - Unit 5 Circular Motions 5.1 5.2 5.3 5.4 5.5...

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