Sect. 11.3: Angular Momentum Rotating Rigid Object

Translation-Rotation Analogues & Connections Translation Rotation Displacementx θ Velocityvω Acceleration aα Force (Torque) Fτ Mass (moment of inertia) mI Newton’s 2 nd Law ∑F = ma ∑τ = Iα Kinetic Energy (KE) (½)mv 2 (½)Iω 2 Work (constant F,τ) Fd τθ Momentummv? CONNECTIONS: s = r θ, v = rω, a t = rα a c = (v 2 /r) = ω 2 r, τ = Fdsin , I = ∑(mr 2 )

Recall: Angular Momentum L of point mass m at position r moving with momentum p = mv (see figure) the cross product: Units: (kg. m 2 )/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(90 o ) = mvr = mr 2 ω v is perpendicular to r  angle  between p & r is  = 90º  sin  = sin(90 o ) = 1 Angular Momentum Direction V ector is pointed out of the diagram (towards the reader). So, a particle in uniform circular motion has a constant angular momentum L = mvr = mr 2 ω 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  Angular momentum of an individual particle is L i = m i r i 2  The vectors  are directed along the z axis.

To find the total angular momentum L of the object, sum up the angular momenta L i of the individual particles: (1) From Ch. 10, I  ∑m i r i 2 is the object’s moment of inertia. Combining (1) with Newton’s 2 nd Law for Rotations: Gives: Newton’s 2 nd Law for Rotations: (for rigid objects) α = angular acceleration of the object

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 )MR 2 Calculate the magnitude of the angular momentum, L z = Iω Direction of L is clearly the +z direction. Example 11.5: Bowling Ball L z = Iω = ( 2 / 5 )MR 2 ω = ( 2 / 5 )(7)(.12) 2 (20π ) = 2.53 kg m 2 /s

A father, mass m f & his daughter, mass m d 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: I s = ( 1 / 12 )Ml 2. Moments of inertia of the father & daughter about pivot point. Treat them like point masses: I f = m f [(½)l] 2, I d = m d [(½)l] 2 Example 11.6: Seesaw (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 = I s + I f + I d = ( 1 / 12 )Ml 2 + (¼)m f l 2 + (¼)m d l 2  The total angular momentum is: L = Iω = (¼)l 2 [(⅓)M + m f + m d ]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 2 nd Law for Rotations: ∑τ ext = Iα ∑τ ext = [( 1 /12)Ml 2 + (¼)m f l 2 + (¼)m d l 2 ]α (1) Left side: ∑τ ext = τ s + τ f + τ d Torque definition: τ = rFsin  = Fd;  = angle between force F & vector r from point of application to pivot point. Moment arm d = rsin  Seesaw: τ s = Mgd s ; Mg passes through pivot point. Moment arm d s = 0  τ s = 0 Father: τ f = w f  d f ; w f  =  component of father’s weight: w f  = m f gcosθ d f = (½)lsin  ;  = 90º, sin  = 1,  d f = (½)l  τ f = (½)lm f gcosθ (counterclockwise, positive torque!) Daughter: τ d = w d  d d ; w d  =  component of daughter’s weight: w g  = m f gcosθ; d d = (½)lsin  ;  = 90º, sin  = 1,  d d = (½)l  τ d = - (½)lm f gcosθ (clockwise, negative torque!) So, from (1), α = [∑τ ext ]/I = [2(m f – m d )cosθ]  (l)[(⅓)M +m f + m d ]  m f gcosθ m d gcosθ 