SOLUTIONS PROBLEM SET 5 NEWTON’S LAWS WITH CIRCULAR MOTION; TORQUE...

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1 SOLUTIONS PROBLEM SET 5 NEWTON’S LAWS WITH CIRCULAR MOTION; TORQUE # 1 the radius of rotation is 2 2 1.25 1 0.75 R m = = 1 cos 0.8 1.25 θ = = and 0.75 sin 0.6 1.25 θ = = T 1 T 2 mg R y x T 1 T 2 T 2 cosq T 2 sinq q T 1 cosq T 1 sinq mg y x Here 1.00 cos 0.8 1.25 θ = = a) total vertical force should be zero 1 1 2 2 cos 60(0.8) 4 9.81 cos cos 0 11 cos 0.8 y T mg F T T mg T N θ θ θ θ Σ = = = = = b) the magnitude of the centripetal force: 2 1 2 sin sin x c v F T T ma m R θ θ Σ = + = = 1 2 ( sin sin ) 0.75(60(0.6) 11(0.6)) 2.83 / 4 RT T v ms m θ θ + + = = =

Transcript of SOLUTIONS PROBLEM SET 5 NEWTON’S LAWS WITH CIRCULAR MOTION; TORQUE...

  • 1

    SOLUTIONS PROBLEM SET 5 NEWTON’S LAWS WITH CIRCULAR MOTION; TORQUE

    #1

    the radius of rotation is 2 21.25 1 0.75R m= − =

    1cos 0.81.25

    θ = = and 0.75sin 0.61.25

    θ = =

    T1

    T2mg

    R

    y

    xT1

    T2

    T2cosq

    T2sinq

    q

    T1cosqT1sinq

    mg

    y

    x

    Here 1.00cos 0.81.25

    θ = =

    a) total vertical force should be zero

    1

    1 2 2cos 60(0.8) 4 9.81cos cos 0 11cos 0.8y

    T mgF T T mg T Nθθ θθ− − ⋅

    Σ = − − = → = = =

    b) the magnitude of the centripetal force:

    2

    1 2sin sinx cvF T T ma mR

    θ θΣ = + = =

    1 2( sin sin ) 0.75(60(0.6) 11(0.6)) 2.83 /4

    R T Tv m smθ θ+ +

    = = =

  • 2

    #2

    m1 m2

    R1R2

    m1 m2

    R1

    R2

    Top view

    a) for m1 :

    21

    1 1 2 1 1 11

    x cvF T T m a mR

    Σ = − = = (1)

    1 1 1 0y NF F m gΣ = − = (2) for m2 :

    22

    2 2 2 2 22

    x cvF T m a mR

    Σ = = = (3)

    2 2 2y NF F m gΣ = = (4) if v2 = 3m/s then from (3):

    2 22

    2 22

    33 181.5

    vT m NR

    = = =

    b) 1 1v Rω= and 2 2v Rω= therefore: 1 1 2 112 2 2

    3 1 2 /1.5

    v R v Rv m sv R R

    ⋅= → = = =

    from (1) 2 2

    1 11 2

    1

    2 2 18 261

    m vT T NR

    ⋅= + = + =

    y

    x

    m1

    FN1

    m1g

    T1

    T2

    a1

    y

    x

    m2

    FN2

    m2g

    T2

    a2

  • 3

    #3

    a) If the speed is low, the car tends to slip down. The vertical force should be zero.

    FNFNcosq

    FNsinq

    mgq

    ffs

    ffscosqffssinqar q

    q

    FN

    ffs

    mg

    y

    x

    cos sin 0y N fsF F f mgθ θΣ = + − = (1) 2

    sin cosx N fs cvF F f ma mR

    θ θΣ = − = = (2)

    the larger the static friction force (ffs), the smaller the speed v. For maximum static friction, v is minimum.

    From (2) 2minsin cosN s NvF F mR

    θ µ θ− = (3)

    From (1) cos sinN s NF F mgθ µ θ+ = (4)

    (3)/(4): 2 2min min

    minsin cos sin10 0.1cos10 8.58 /cos sin cos10 0.1sin10 9.81 100

    s

    s

    v v v m sgR

    θ µ θθ µ θ− −

    = = = → =+ + ⋅

    b) If the speed is high, the car tends to slip up. The vertical force should be zero.

    FNFNcosq

    FNsinq

    mgq ffs

    ffscosqffssinq

    arq

    q

    FN

    ffs

    mg

    y

    x

  • 4

    cos sin 0y N fsF F f mgθ θΣ = − − = (1) 2

    sin cosx N fs cvF F f ma mR

    θ θΣ = + = = (2)

    the larger the static friction force (ffs), the larger the speed v. For maximum static friction, the speed is maximum.

    From (2) 2maxsin cosN s NvF F mR

    θ µ θ+ = (3)

    From (1) cos sinN s NF F mgθ µ θ− = (4)

    (3)/(4): 2 2min min

    minsin cos sin10 0.1cos10 16.5 /cos sin cos10 0.1sin10 9.81 100

    s

    s

    v v v m sgR

    θ µ θθ µ θ+ +

    = = = → =− − ⋅

    the range of speeds the car can have without slipping up or down: 8.58 / 16.5 /m s v m s≤ ≤

    #4

    2

    2

    (1)

    net r c

    N

    N

    F ma ma

    vmg F mr

    vF m gr

    = =

    + =

    ⎛ ⎞= −⎜ ⎟

    ⎝ ⎠

    Top of loop:

    (force of seat on pilot is the normal force

    2

    2

    (2)

    net r c

    N

    N

    F ma ma

    vF mg mr

    vF m gr

    = =

    − =

    ⎛ ⎞= +⎜ ⎟

    ⎝ ⎠

    Bottom of loop:

    mg FNar

    mg

    FNar

  • 5

    2 (3)

    net r c

    N

    F ma mavF mr

    = =

    = Side of loop:

    Comparing (1), (2) and (3) the normal force is greatest at the bottom of the loop. The least force is at the top of the loop. #5

    9545 0.45N m

    L cm mτ = ⋅

    = =

    with r F r Lτ = × =rr r

    magnitude of torque is sin 95rFτ θ= = θ is the angle between rr and F

    r

    95 229.3sin sin 0.45 sin 67

    N mF Nr L mτ τθ θ

    ⋅= = = =

    ⋅ o

    #6

    balance torques at the right end of the board and balance forces. Clockwise torque is negative.

    mg

    FNar

    F

    67o

    L

  • 6

    support

    support

    support

    00

    500 3 280 1.5 1 0 0

    500 3 280 1.5 19201

    net

    girl board end

    N m N m F m

    F N

    τ

    τ τ τ τ

    =

    + + + =

    − ⋅ − ⋅ + ⋅ + =

    ⋅ + ⋅= =

    rr r r r

    support

    0

    0

    ˆ1920 280 500 1140

    y

    end board girl

    end

    F

    F F F F

    F Nj

    Σ =

    + + + =

    = − − = −

    r r r r

    r

    #7

    22.7700 700(2 ) / 60 73.3 /I kg mrpm rad sπ

    = ⋅

    = =

    use Iτ α= but what is α ?

    2 2

    2 2 22

    2 2

    1( ) ( )2

    (73.3 / ) 0 17.11 /2( ) 2(2 (25) 0)

    2.7 17.11 / 46.2

    o o

    o

    o

    rad s rad s

    kg m rad s N m

    α θ θ ω ω

    ω ωα

    θ θ π

    τ

    − = −

    − −= = =

    − −

    = ⋅ ⋅ = ⋅

    #8

    m2

    T2

    ff2

    FN2

    m2gcosq

    m2gsinq

    m1T1

    FN1

    m1g

    ff1

    m2g

    T1T2

    R

    y

    x x +

    consider m1:

    1 1 1

    1 1 1 1

    1 1 1 1 1 1

    0x f

    y N N

    F T f m aF F m g F m gT m g m a T m a m gµ µ

    Σ = − =

    Σ = − = → =

    − = → = +

  • 7

    consider m2: 2 2 2 2

    2 2 2 2

    2 2 2 2 2 2 2 2

    sin 0

    cos30 0 cos30

    cos30 sin 0 sin30 cos30

    x f

    y N N

    F T f m g m aF F m g F m gT m g m g m a T m a m g m gµ µ

    Σ = − − − 3 =

    Σ = − = → =

    − − + 3 = → = − + −

    consider pulley:

    2 tan1 2

    1 2

    1 2

    ( is clockwise, so negative)12

    12

    12

    IaI MRR

    T R T R MRa

    T T Ma

    τ α α

    τ τ α

    Σ =

    ⎛ ⎞− = = −⎜ ⎟⎝ ⎠

    − = −

    − = −

    r

    substitute expression for T1 and T2

    1 1 2 2 2

    1 2 2 1 2

    2 22 1 2

    1 2

    2

    1( ) ( sin 30 cos30 )2

    130 cos302

    sin30 cos30 6 9.81 / sin 30 0.360 9.81 / (2 6 cos30)0.5 (2 6 0.5 10)

    0.309 /

    m a m g m g m g m a Ma

    m g m gsin m g M m m a

    m g m g m g kg m s m s kg kgaM m m kg

    a m s

    µ µ

    µ µ

    µ µ

    + − − − = −

    ⎛ ⎞− + = − − −⎜ ⎟⎝ ⎠

    − − ⋅ ⋅ − ⋅ ⋅ += = =

    + + + + ⋅

    = then find T1 and T2 1 1

    2 21 1

    2 2

    2 2 22 2

    ( )

    2 (0.309 / 0.360 9.81 / ) 7.67

    ( sin30 cos30 )

    6 (9.81 / sin30 0.360 9.81 / cos30 0.309 / ) 9.21

    T m a g

    T kg m s m s N T

    T m g g a

    T kg m s m s m s N T

    µ

    µ

    = +

    = ⋅ + ⋅ = =

    = − −

    = ⋅ = ⋅ − = =