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## Kinematics of Rotation of Rigid Bodies

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**Kinematics of Rotation of Rigid Bodies**Angular displacement Δθ = θ – θ0 Δθ > 0 if rotation is counterclockwise Δθ < 0 if rotation is clockwise z Angle of rotation s s’ Angle of rotation θ is a dimensionless quantity, but it is a vector r’ r If θ is in radians, then s=rθ and s’=r’θ with the same θ.**Example:**A total eclipse of the sun**Angular Velocity and Acceleration**z z Average angular velocity Instantaneous angular velocity ωz>0 ωz<0 Instantaneous angular speed is a scalar Average angular acceleration Instantaneous angular acceleration**Kinematic Equations of Linear and Angular Motion**with Constant Acceleration (1) (3) (4) (2)**Relations between Angular and Tangential Kinematic**Quantities Ice-skating stunt “crack-the-whip”**Centripetal and Tangential Accelerations in Rotational**Kinematics**Rotational Kinetic Energy and Moment of Inertia**Kinetic energy of one particle Rotational kinetic energy is the kinetic energy of the entire rigid body rotating with the angular speed ω ω Definition of the moment of inertia Total mechanical energy h Translational kinetic energy Rotational kinetic energy Potential energy 0cm Parallel-Axis Theorem Proof: =0**Rotation about an Axis Shifted from a Center of Mass**Position**Exam Example 23:Blocks descending over a massive**pulley(problem 9.83) ω Data: m1, m2, μk, I, R, Δy, v0y=0 m1 R Find: (a) vy; (b) t, ay; (c) ω,α; (d) T1, T2 Solution: (a)Work-energy theorem Wnc= ΔK + ΔU, ΔU = - m2gΔy, Wnc = - μk m1g Δy , since FN1 = m1g, ΔK=K=(m1+m2)vy2/2 + Iω2/2 = (m1+m2+I/R2)vy2/2 since vy = Rω x 0 m2 ay Δy y (b) Kinematics with constant acceleration: t = 2Δy/vy , ay = vy2/(2Δy) (c) ω = vy/R , α = ay/R = vy2/(2ΔyR) (d) Newton’s second law for each block: T1x + fkx = m1ay → T1x= m1 (ay + μk g) , T2y + m2g = m2ay → T2y = - m2 (g – ay)**Moments of Inertia of Various Bodies**Scaling law and order of magnitude I ~ ML2**Moment of Inertia Calculations**Cylinder rotating about axis of symmetry The total mass of the cylinder is Result: