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Sect. 11.3: Angular Momentum Rotating Rigid Object. Translation-Rotation Analogues & Connections. Translation Rotation Displacement x θ Velocity v ω Acceleration a α Force (Torque) F τ Mass (moment of inertia) m I
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Translation-Rotation Analogues & Connections Translation Rotation Displacement x θ Velocity v ω Acceleration a α Force (Torque) F τ Mass (moment of inertia)m I Newton’s 2nd Law ∑F = ma ∑τ = Iα Kinetic Energy (KE) (½)mv2 (½)Iω2 Work (constant F,τ) Fd τθ Momentum mv ? CONNECTIONS: s = rθ,v = rω, at= rα ac = (v2/r) = ω2r , τ = Fdsin, I = ∑(mr2)
Recall:Angular MomentumLof point massmat position rmoving with momentump = mv(see figure)the cross product: • Units:(kg.m2)/s • Magnitude:L = mvr sin =angle betweenp& r • Direction:Perpendicular to the plane formed byr&p. If r& pare in thexy-plane, Lis in thez-direction.
Special Case:Particle of mass m moving with velocity v in a circular path of radius r (see figure).Momentum p = mv. Note also that for this case, v = rω, where ω = angular velocity of m around the center of the circle. • Angular Momentum Magnitude L = mvr sin(90o) = mvr = mr2ω vis perpendicular tor anglebetweenp& r is = 90º sin = sin(90o) = 1 • Angular Momentum Direction Vectoris pointed out of the diagram (towards the reader). • So, a particle in uniform circular motion has a constant angular momentum L = mvr = mr2ω about an axis through the center of its path.
Generalize to a Rigid Object rotating about an axis passing through it’s center of mass (see figure).Each particle in the object rotates in the xy plane about the z axis with angular speed w.Angular momentum of an individual particle is Li = mi ri2w. The vectors & are directed along the z axis.
To find the total angular momentum Lof the object, sum up the angular momentaLiof the individual particles: (1) From Ch. 10, I ∑miri2is the object’s moment of inertia. Combining (1) with Newton’s 2nd Law for Rotations: Gives: Newton’s 2nd Law for Rotations: (for rigid objects) α =angular acceleration of the object
Example 11.5: Bowling Ball • A bowling ball (assume a perfect sphere, mass M = 7 kg, radius R = 0.12 m), spins at frequencyf = 10 rev/s ω = 2πf = 20π rad/s From table (Ch. 10) moment of inertia is I = (2/5)MR2 • Calculate the magnitude of the angular momentum, Lz = Iω • Direction of L is clearly the +z direction. Lz = Iω = (2/5)MR2ω = (2/5)(7)(.12)2(20π ) = 2.53 kg m2/s
Example 11.6: Seesaw • A father, mass mf& his daughter, mass md sit on opposite sides of a seesaw at equal distances from the pivot point at the center. Assume seesaw is a perfect rod, mass M & length l. At some time, their angular speed is ω. • Table (Ch. 10) gives moment of inertia of seesaw:Is = (1/12)Ml2. Moments of inertia of the father & daughter about pivot point. Treat them like point masses:If = mf[(½)l]2, Id = md[(½)l]2 (A) Find an expression for the magnitude of the angular momentum (about pivot point) at time when angular speed is ω. Note: Total moment of inertia is:I = Is + If + Id = (1/12)Ml2 + (¼)mfl2 + (¼)mdl2 The total angular momentum is: L = Iω = (¼)l2[(⅓)M + mf + md]ω
Example 11.6: Seesaw continued mdgcosθ (B) Find an expression for the magnitude of the angular acceleration when the seesaw makes an angle θ with the horizontal.Newton’s 2nd Law for Rotations:∑τext= Iα ∑τext= [(1/12)Ml2 + (¼)mfl2 + (¼)mdl2]α (1) Left side: ∑τext= τs+τf +τd mfgcosθ Torque definition: τ = rFsin = Fd; = angle between force F & vector r from point of application to pivot point. Moment arm d = rsin Seesaw: τs= Mgds; Mgpasses through pivot point. Moment armds = 0 τs= 0 Father: τf= wfdf; wf = component of father’s weight: wf = mfgcosθdf = (½)lsin; = 90º, sin = 1, df = (½)l τf= (½)lmfgcosθ(counterclockwise, positive torque!) Daughter: τd= wddd; wd = component of daughter’s weight: wg = mfgcosθ; dd = (½)lsin; = 90º, sin = 1, dd = (½)l τd= - (½)lmfgcosθ(clockwise, negative torque!) So, from (1),α = [∑τext]/I = [2(mf – md)cosθ](l)[(⅓)M +mf+ md]