Transcript Part II
Sect. 11.3: Angular Momentum Rotating Rigid Object Translation-Rotation Analogues & Connections Translation Rotation Displacement x Velocity v Acceleration a Force (Torque) F Mass (moment of inertia) m Newton’s 2nd Law ∑F = ma Kinetic Energy (KE) (½)mv2 Work (constant F,τ) Fd Momentum mv CONNECTIONS: θ ω α τ I ∑τ = Iα (½)Iω2 τθ ? s = rθ, v = rω, at= rα ac = (v2/r) = ω2r , τ = Fdsin, I = ∑(mr2) • Recall: Angular Momentum L of point mass m at position r moving with momentum p = mv (see figure) the cross product: L =r p • Units: (kg.m2)/s • Magnitude: L = mvr sin = angle between p & r • Direction: Perpendicular to the plane formed by r & p. If r & p are in the xy-plane, L is in the z-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 v is perpendicular to r angle between p & r is = 90º sin = sin(90o) = 1 • Angular Momentum Direction Vector L = r p is 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. L =r p • 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 L & w are directed along the z axis. • To find the total angular momentum L of the object, sum up the angular momenta Li of the individual particles: (1) Lz = Li = mi ri 2 w = Iw i i From Ch. 10, I ∑miri2 is the object’s moment of inertia. Combining (1) with Newton’s 2nd Law for Rotations: dL tot = ext dt Gives: Newton’s 2nd Law for Rotations: (for rigid objects) α = angular acceleration of the object ext dLz dw = =I = I dt dt Example 11.5: Bowling Ball • A bowling ball (assume a perfect sphere, mass M = 7 kg, radius R = 0.12 m), spins at frequency f = 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 (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 Left side: (1) mdgcosθ mfgcosθ ∑τext = τs + τf + τd 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; Mg passes through pivot point. Moment arm ds = 0 τs = 0 Father: τf = wfdf; wf = component of father’s weight: wf = mfgcosθ df = (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]