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Finding Velocity

  1. May 11, 2009 #1
    1. The problem statement, all variables and given/known data

    The soap-box car has a weight of 110 lb, including the passenger but excluding its four wheels. Each wheel has a weight of 5 lb, radius of .5 ft, and a radius of gyration k = .3 ft, computed about an axis passing through the wheels axle. Determine the cars speed after it has traveled 100 ft starting from rest. the wheels roll without slipping. Neglect Air.


    2. Relevant equations

    T1 = [tex]\sum[/tex]U1-2 = T2


    3. The attempt at a solution

    T1 = 0 because it starts from rest
    for T2 the mass would be 110lb car pluss the 4 5lb wheels, converted to slugs of course, and the v is what i am looking for

    im not sure about the middle term

    I = mk3 = .1554*.32 = .013986

    What else do i need
     
  2. jcsd
  3. May 11, 2009 #2

    LowlyPion

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    What force is acting on the soapbox?

    Incline? Angle?
     
  4. May 11, 2009 #3
    Fy = 130 lb
     
  5. May 11, 2009 #4

    LowlyPion

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    Isn't that just the weight?
     
  6. May 12, 2009 #5
    T2 = .5mv2 + .5Iw2

    =.5(130/32.2)v2 + .5(130/32.2).32w2

    T2 = 325 / 161 v2 + 117 / 644 w2

    v = .5w2

    T2 = 221/322 w2

    [tex]\sum[/tex]U1-2 = V1 - V2, V = Wy

    V1 = Wy = W(100sin30)

    V2= 0

    Therefore

    100Wcos30 = 221 / 322 w2

    Something aint right
     
  7. May 12, 2009 #6

    LowlyPion

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    First they have given you the weights. You need the mass.

    Second what is sin30? Is that the angle I previously inquired about being given?
     
  8. May 12, 2009 #7
    oops sorry, i didnt understand when u asked the incline angle

    it is 30 degrees

    and they hypotenus is 100ft, so 100sin30 = y = 50

    &

    I did convert to mass when i divided the weight 130 by 32.2

    Also...

    When do use the info on the wheels, r = .5, k = .3, w = 5 lb
     
  9. May 12, 2009 #8

    LowlyPion

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    Oh, I see. You got it. I see the 32.2 now.

    OK the radius of gyration basically gives you the I = m*Rg2

    But for your linear translational kinetic energy you have a r = .5 ft
     
  10. May 15, 2009 #9
    I = mk2 = (5/32.2).32 = 9/644 <-- Should i multiply this my 4 since there are 4 wheels

    50W = 1/2(130/32.2)v2 + 1/2 (9/644)w2

    w = v/.5

    50W = (325/161)v2 + (9/1288)(v/.5)2

    50W = 325/161 v2 + 9/322 v2

    50w = 659 / 322 v2

    v = (16100W) / 659 or v = 24.43095599W

    Is this right so far, How do i find W
     
  11. May 15, 2009 #10
    I = mk2 = (5/32.2).32 = 9/644 <-- Should i multiply this my 4 since there are 4 wheels

    50W = 1/2(130/32.2)v2 + 1/2 (9/644)w2

    w = v/.5

    50W = (325/161)v2 + (9/1288)(v/.5)2

    50W = 325/161 v2 + 9/322 v2

    50w = 659 / 322 v2

    v = (16100W) / 659 or v = 24.43095599W

    Is this right so far, How do i find W
     
  12. May 15, 2009 #11
    brain fart

    W = 130 right

    so v = 56.35622 ft/s
     
  13. May 17, 2009 #12
    anyone agree, disagree
     
  14. May 17, 2009 #13

    LowlyPion

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    m*g*h = 50*130 = 6500 = 1/2*m*V2 + 1/2*I*ω2

    I = 4*5/32.2*(.3)2 = .056

    ω = V/.5

    6500 = 1/2*(4.037 + .056)*V2

    V = 56.36

    Looks OK.
     
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