HOMEWORK 4 SOLUTIONS CHAPTER 9 9.C1 9.· HOMEWORK 4 SOLUTIONS CHAPTER 9 9.C1 Tangential acceleration

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Transcript of HOMEWORK 4 SOLUTIONS CHAPTER 9 9.C1 9.· HOMEWORK 4 SOLUTIONS CHAPTER 9 9.C1 Tangential acceleration

  • HOMEWORK 4 SOLUTIONS CHAPTER 9 9.C1 Tangential acceleration of a point on a rotating object is the component of points acceleration vector that is perpendicular to the radial component. Unlike the radial acceleration, the tangential acceleration is zero for a uniformly rotating object (i.e one whose angular velocity is constant). 9. MC1

    B. One turn is 2 radians = 2 x 3.14 = 6.28 6. 9. MC2 Angular motion is mathematically similar to linear motion. In particular have for the angular distance

    that an object with angular acceleration rotates through at time t is

    2

    21 t = (1)

    (assuming that objects starts with zero angular velocity at angle of zero) [This is analogous to 2

    2

    1 atx = for linear motion.]

    We know that t = 2T from the problem. All we need now is . Have

    T

    initialfinal

    = The change in angular acceleration over the first revolution (2)

    2

    initialfinal

    average

    += The definition of average angular velocity (3)

    We are told that 0=initial

    .

    We are told that T

    revolutionaverage

    1= .

    Inserting this into (3) gives T

    srevolutionfinal

    2= .

    Inserting this into (2) gives 2

    2

    T

    srevolution=

    Inserting this into (1) gives 222

    1 )2(2

    TT

    srevolution

    = = 4 revolutions. [C]

  • 9. MC10

    From the definition of the moment of inertial I

    222 6)2()2( mddmdmI =+=

    After the masses switch positions, the new moment of inertial I is

    222 9)2)(2(' mdmddmI =+=

    Solve first equation for d: m

    Id

    6= .

    Insert this into the second equation:

    Im

    ImI

    2

    3

    69'

    2

    =

    = [C]

    9. MC11

    The general formula for moment of inertial I is

    2

    i

    i

    irmI = .

    If the ri s double, I increases by a factor of 4. If the mi s double, I increase by a factor of 2. 4 x 2 = 8.

    [C] 9. MC15

    Let 2MRI = where 5

    2= for the solid sphere, and

    3

    2= for the hollow sphere.

    Initial potential energy completely converted to kinetic energy. Have energy conservation initial potential energy = (final kinetic energy of linear motion) + (final kinetic energy of rotational motion)

    22

    12

    2

    1 IMvMgh +=

  • For rotational motion have Rv = . Also, 2MRI =

    So

    ( )2

    2

    212

    21

    +=

    R

    vMRMvMgh

    22

    1 )1( Mv+=

    Solve for v

    +

    =1

    2ghv

    The larger is , the smaller is v. So the hollow sphere (3

    2= ) is moving more slowly than the solid

    sphere (5

    2= ) at the bottom of the ramp.

    Answer is A. The solid sphere is faster. CHAPTER 10

  • Chapter 11 Multiple Choice

  • Chapter 11 Problems