Physics 2101 S t ec i tti on 3

Size: px

Start display at page:

Download "Physics 2101 S t ec i tti on 3"

Madeleine Thornton
7 years ago
Views:

1 Physics 101 Section 3 March 5 rd : Ch. 10 Announcements: Grades for Exam # posted next week. Today Ch , 10, Next Quiz is March 1 on SHW#6 Class Website:

2 Summary: Translational and Rotational Variables Rotational position and distance moved s = θ r (only radian units) Rotational and translational speed v = dr dt = ds dt = d θ dt r v = ω r Rotational and translational acceleration a t = d ω tangential acceleration dt r = α r v a r = = ω r radial/centripetal acceleration r a = a tot r t + a t v = ω r a t = α r a r = ω ω r ( ) = ω r + αr a tot = a t + a r

3 Example: Beetle on a merry-go-round A beetle rides the rim of a rotating merry-go-round. If the angular speed of the system stem is constant, does the beetle have a) radial acceleration and b) tangential acceleration? If ω = constant, α = 0 a tot = a r + a t = 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 = negative but constant, v r ˆ ω = ω0 + αt a r a = a + a a tot = a r + a tot t r t = ω r + αr a tot = ((ω 0 + αt) r) + ( αr) both radial and tangential a tot a t

4 A ladybug sits at the outer edge of a merry-go-round, and a gentleman bug sits halfway between her and the axis of rotation. ti The merry-go-round makes a complete revolution once each second. The gentleman bug s angular speed is gentleman bug ladybug 1. half the ladybug s.. the same as the ladybug s. 3. twice the ladybug s. 4. impossible to determine

5 A ladybug sits at the outer edge of a merry-go-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

6 A ladybug sits at the outer edge of a merry-go-round 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

7 Sec. 3.7 Multiplying vectors: 1) Dot Product or Scalar Product (Vector Vector = scalar) A A θ B = A B cosθ If A B then A B=0 = A x B x + A y B y + A z B z If A B then A B=max B ) Cross Product (Vector Vector = Vector) = i ˆ ˆ j k ˆ C A B = A x A y A z B x B y B z C = A B sinθ If A B then A B=max If A B then A B=0 Two vectors A and B define a plane. Vector C is perpendicular to plane according to right-hand hand rule.

8 Kinetic Energy of Rotation TRANSLATION Review: Newton s nd Law F net a energy associated with state of translational l motion proportionality is inertia, m constant of an object KE trans v mass is translational inertia motion of particles with ihsame v What about rotation? What is energy associated with state of rotational motion? KE = 1 m system v i where v i = ωr i = m i 1 m i ( ωr i ) ( ) trans rot = 1 [ m i r i ]ω All particles have rotational inertia (moment of inertia) about some axis of rotation I same ω Energy of rotational motion KE rot = 1 Iω [ KE trans = 1 ] mv

9 Moment of Inertia 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 1 1 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

10 Moment of inertia: comparison I 1 = m i r i 1 i i = (5 kg)( m) + (7 kg)( m) I = (5 kg)(0.5 m) + (7 kg)(4.5 m) = 5 kgm = ( ) kgm =144 kgm Note: 5 kg mass contributes <1% of total - Mass close to axis of rotation contributes little to total moment of inertia. - How does a larger moment of finertia i feel l? [ F net = ma ] If rigid body = few particles I = m i r i If rigid body = too-many-to-count particles Sum Integral

11 Moment of inertia: continuum mass I = m m i r i r dm with ρ = I = ρ r V dv Example: moment of finertia i of thin rod with ihperpendicular 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 1 3 dx ρ A( x ) - L L + = = ρa 1 3 ( 1 8 L3 ) 1 3 ( 1 8 L3 ) 3 -L ( ) I = m V 1 = 1 1 A L3 m A A L L 3 () = 1 1 ml INDEPENDENT of cross section area

12 Some Rotational Inertias Each of these rotational inertias GO THROUGH the center of mass!

13 Parallel-Axis Theorem I = I COM + Mh Proof: ( ) ( ) }dm I= r dm = { x a + y b ( )dm a xdm ( ) I = x + y b ydm+ a + b I = I COM h M dm

14 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 = kg m I = kg m I = kg m Consider a 0g pencil 15cm long and 1cm wide somewhat like MASS, you can feel the difference in the rotational ti inertia

15 Example #1 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 + M 1 ( ML 1 L ) = 1 3 ML = kg 3 ( )0.33m ( ) kg m Putting together I tt tot,com = Irim,center + 50I spoke = M wheel R + 50I spoke = ( 1.kg)0.33m ( ) + 50( kg m ) = 0.149kg m

16 Example # How much work did Superman exert on earth in order to stop it? What is the kinetic energy of the earth s rotation about its axis? Energy of rotational motion is found from: KErot = Iω 1 What is earth s moment of inertia, I? Iearth = Isphere = 5 Mr = 6 5( kg)6.4 ( m) kg m What is earth s angular velocity, ω? From T = π ω Now plug- n-chug: π ω = π radians = π day KE rot = 1 kg m rad /s ( ) = rad /s ( ) = J

11. Rotation Translational Motion: Rotational Motion:

11. Rotation Translational Motion: Motion of the center of mass of an object from one position to another. All the motion discussed so far belongs to this category, except uniform circular motion. Rotational