Relating Translational and Rotational Variables Rotational position and distance moved s = θ r (only radian units) Rotational and translational speed d s v = dt v = ω r = ds dt = d θ dt r Relating period and rotational speed T = πr v ωt = π = π ω units of time (s) [ distance = rate time]
Relating Translational and Rotational Variables acceleration is a little tricky Rotational and translational acceleration a) from v = ω r dv dt = d ω r dt a t = d ω r = α r dt from change in angular speed tangential acceleration b) from before we know there s also a radial component a r = v r = ω r radial acceleration c) must combine two distinct a tot = a r + a rotational accelerations t = ω r + αr
Relating Translational and Rotational Variables Rotational position and distance moved s = θ r (only radian units) Rotational and translational speed v = d r dt = ds dt = d θ dt r v = ω r Rotational and translational acceleration a t = d ω r = α r tangential acceleration dt a r = v r = ω r a tot = a r + a t = ω r + αr radial acceleration v = ω r a t = α r a r = ω ω r ( ) a tot = a t + a r
θ (t) ( θ 1 ) Δ θ = θ s = θ r ω (t) = d θ (t) dt v = d θ dt r = ω r v = ω r α (t) = d ω (t) dt = d θ (t) dt a tot = a r + a t = ω r + αr a tot = a t + a r = a r ω ω r a t = α r ( ) Rotational Kinematics: ( ONLY IF α = constant) ω = ω 0 + αt θ = θ + ω 0 t + 1 αt
Example #1 A beetle rides the rim of a rotating merry-go-round. If the angular speed of the system is constant, does the beetle have a) radial acceleration and b) tangential acceleration? a tot = a r + a t If ω = constant, α = 0 = ω r + 0 a tot = ω r( ˆ r ) only radial If the angular speed is decreasing at a constant rate, does the beetle have a) radial acceleration and b) tangential acceleration? If α = neg constant, v r ˆ ω = ω 0 + αt a r a tot = a r + a t = ω r + αr a tot = a t + a r a tot = ((ω 0 + αt) r) + ( αr) both radial and tangential a tot a t
Checkpoint #1 A ladybug sits at the outer edge of a merry-goround, and a gentleman bug sits halfway between her and the axis of rotation. The merry-go-round makes a complete revolution once each second. The gentleman bug s angular speed is 1. half the ladybug s.. the same as the ladybug s. 3. twice the ladybug s. 4. impossible to determine
Checkpoint # A ladybug sits at the outer edge of a merrygo-round, that is turning and speeding up. At the instant shown in the figure, the tangential component of the ladybug s (Cartesian) acceleration is: 1. In the +x direction. In the - x direction 3. In the +y direction 4. In the - y direction 5. In the +z direction 6. In the - z direction 7. zero
Checkpoint#3 A ladybug sits at the outer edge of a merry-goround that is turning and is slowing down. The vector expressing her angular velocity is 1. In the +x direction. In the - x direction 3. In the +y direction 4. In the - y direction 5. In the +z direction 6. In the - z direction 7. zero
Problem 10 7: The wheel in the picture has a radius of 30cm and is rota6ng at.5rev/sec. I want to shoot a 0 cm long arrow parallel to the axle without hi?ng an spokes. (a) What is the minimum speed? (b) Does it mager where between the axle and rim of the wheel you aim? If so what is the best posi6on. The arrow must pass through the wheel in less time than it takes for the next spoke to rotate Δt= 1 8 rev.5rev / s = 0.05s (a) The minimum speed is v mn = 0cm 0.05s = 400cm / s = 4m / s (b) No there is no dependence upon the radial position.
Problem 10 30: Wheel A of radius r A =10 cm is coupled by belt B to wheel C of radius r C =5 cm. The angular speed of wheel A is increased from rest at a constant rate of 1.6 rad/s. Find the 6me needed for wheel C to reach an angular speed of 100 rev/min. If the belt does not slip the tangential acceleration of each wheel is the same. r A α A = r C α C α C = r A r C α A = 0.64rad / s The angular velocity of C is. ω C = α C t so t= ω C α C = ω Cr C r A α A with ω= 100rev/sec t=16s c
TRANSLATION Review: Newton s nd Law Kinetic energy of Rotation F net a proportionality is inertia, m constant of an object energy associated with state of translational motion KE trans = 1 mv motion of particles with same v mass is translational inertia What about rotation? What is energy associated with state of rotational motion? KE system = rotational inertia (moment of inertia) about some axis of rotation 1 m v i i ( ) 1 = m ωr i i = 1 m [ ( i r i )]ω I where v i = ωr i trans rot All particles have same ω Energy of rotational motion KE rot = 1 Iω [ KE trans = 1 ] mv
Moment of Inertia I For a discrete number of particles distributed about an axis of rotation I m i r i all mass units of kg m Simple example: 4 I = m i r i = m 1 r 1 + m r + m r + m 1 r 1 i=1 = m 1 r 1 + m r what about other axis? - Rotational inertia (moment of inertia) only valid about some axis of rotation. - For arbitrary shape, each different axis has a different moment of inertia. - I relates how the mass of a rotating body is distributed about a given axis. - r is perpendicular distance from mass to axis of rotation
Moment of inertia: comparison I 1 = m i r i = (5 kg)( m) + (7 kg)( m) = 5 kgm I = (5 kg)(0.5 m) + (7 kg)(4.5 m) ( ) kgm = 144 kgm = 1.3 + 14 Note: 5 kg mass contributes <1% of total - mass close to axis of rotation contributes little to total moment of inertia. If rigid body = few particles I = m i r i If rigid body = too-many-to-count particles Sum Integral
Moment of inertia: continuum mass I = m i r i r dm with ρ = m V I = ρ r dv Example: moment of inertia of thin rod with perpendicular rotation through center I = r dm = ρ r dv where dv = area dx = A dx I = ρa r dx r = x in this case L x + L + L I = ρa x dx - L = ρa 1 + ( 3 ) L x3 - L = ρa( 1 ( 1 3 8 L3 ) 1 ( 1 3 8 L3 )) I = m V A L3 1 = 1 1 m A A L L 3 ( ) = 1 1 ml INDEPENDENT OF AREA
Some Rotational Inertias EACH OF THESE Rotational Inertias GO THROUGH THE Center of Mass!
10-7: Parallel-Axis Theorem If h is a perpendicular distance between a given axis and the axis through the center of mass (these two axes being parallel). Then the rota6onal iner6a I about the given axis is I = I COM + Mh Proof: ( ) I= r dm = { x a ( ) }dm + y b ( )dm a x dm I = x + y b ydm + a + b I = I COM + 0 + 0 + h M COM ( ) dm
Example #1 Moment of Iner4a: Look at the drawing of the simple rod of length L. (a) What is I through an axis parallel to the rod? (b) What is I through an axis perpendicular to the rod through the CM? (c) What is I for an axis at the end but perpendicular to the rod? (a) I through the axis parallel to the rod. (b) I through the axis perpendicular to the rod. I= mr for a solid cylinder R 0, I 0 I= ml 1 from table (C) I through the axis perpendicular to the rod at the end. I= ml 1 + mh = 4mL 3
Example #: Moment of inertia of a Pencil It depends on where the rotation axis is considered I = 1 3 ML I = 1 1 ML I = 1 MR I = 0.3 10 6 kg m I = 150 10 6 kg m I = 38 10 6 kg m Consider a 0g pencil 15cm long and 1cm wide somewhat like MASS, you can feel the difference in the rotational inertia
Example #3 A bicycle wheel has a radius of 0.33m and a rim of mass 1. kg. The wheel has 50 spokes, each with a mass 10g. What is the moment of inertial about axis of rotation? What is moment of inertia about COM? I tot,com = I rim,center + 50I spoke What is I spoke (parallel-axis)? Ispoke = Irod.com + Mh = 1 1 ML + M ( 1 L ) = 1 3 ML = 1 0.01kg 3 ( )( 0.33m) 3.6 10 4 kg m Putting together I tot,com = I rim,center + 50I spoke = M wheel R + 50I spoke = ( 1.kg) ( 0.33m) + 50(3.6 10 4 kg m ) = 0.149kg m
Example #4 What is the kinetic energy of the earth s rotation about its axis? Energy of rotational motion is found from: KE rot = 1 Iω What is earth s moment of inertia, I? Iearth = Isphere = 5 Mr = 6 10 4 kg 5 ( )( 6.4 10 6 m) 1 10 38 kg m What is earth s angular velocity, ω? From T = π ω Now plug- n-chug: KE rot = 1 1 10 38 kg m 7.3 10 5 rad /s ω = π radians day = π ( 3600 4) = 7.3 10 5 rad /s ( ) =.6 10 9 J