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Im starting to understand.

  1. Aug 26, 2009 #1
    1. The problem statement, all variables and given/known data

    fg6.png

    (b) Fig. Q.6 shows a belt drive system where pulley A (diameter 0.5 m) is transmitting power to pulley B (diameter 0.25 m), mounted on a parallel rotating shaft. The tight and slack-side tensions of the belt are 500 N and 200 N respectively, and pulley A rotates at 300 rev/min.

    Determine
    (i) The linear speed of the belt (in m/s))
    (ii) The rotational speed of pulley B (in rev/min)
    (iii) The torque on pulley B.
    (iv) The power transmitted to pulley B.





    3. The attempt at a solution

    (i)
    To find linear speed in (m/s)

    v= RxW
    V=0.5x(300x(2pi/60))
    v=0.5x31.4159
    v=15.7079m/s

    (II)
    the rotational speed of pully B

    if
    0.5m = 300rev/min
    therefore
    300/0.5 = 600
    600x0.25= 150rev/min

    (iii)
    To find torque

    Tb=(f1-f2)Rb
    Tb=(500-200)0.125
    Tb=(300)0.125
    Tb=37.5Nm

    (iv)
    To find power

    Pb=TbxWb
    Pb=37.5x(150x(2pi/60))
    Pb=589.04watts

    i did my best, can some one tell me if im on the right track or im totaly off
     

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    Last edited: Aug 26, 2009
  2. jcsd
  3. Aug 26, 2009 #2

    Redbelly98

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    You have the right idea. However, the radius of pulley A is not 0.5m.

    That statement makes no sense.

    Not quite. Hint: both pulleys have the same linear speed.

    Yes, that's right.

    You have the right idea, but will need to use the correct value of Wb to get the correct answer here.
     
  4. Aug 26, 2009 #3
    (i)
    To find linear speed in (m/s)

    v= RxW
    V=0.25x(300x(2pi/60))
    v=0.25x31.4159
    v=7.85m/s
    is that corect now?

    i dont get you on the rotational speed of pully B.

    if i get the rotational speed of the pully B, i will be able to have the correct value for Wb yeah?
     
  5. Aug 26, 2009 #4

    djeitnstine

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    [tex]\omega_b = 300 \frac{rev}{min} \left( 2 \frac{\pi}{60} \right) [/tex] not [tex]\omega_b =150 \frac{rev}{min} \left( 2 \frac{\pi}{60} \right) [/tex] (in your power equation.)
     
  6. Aug 27, 2009 #5

    yeah i do know it's 300rev/min, but how did u find the 300rev/min?

    so to answer the question.

    Q. The rotational speed of pully B
    A. Since pully A as a rotational speed of 300rev/min
    and both pully have the same linear speed, so pully b as a rotational speed of 300rev/min?
     
  7. Aug 27, 2009 #6

    Redbelly98

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    Okay. You know that the linear speed v is

    v = r ω​

    And since the linear speed (= the belt speed) is the same for both pulleys,

    rA ωA = vA = vB = rB ωB

    Also, B will have a different rotational speed (not 300 rpm) than A.

    Yes.
     
  8. Aug 27, 2009 #7
    im confused!!
     
  9. Aug 28, 2009 #8

    Redbelly98

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    In other words,

    rA ωA = rB ωB

    (ii) asks for ωB. You know what rA, ωA, and rB are. Solve the equation for ωB.
     
  10. Sep 7, 2009 #9
    So
    Va=RaxWb = Vb=RbxWb

    Since Va=7.85m/s

    Vb=RbxWb
    7.85=0.125xWb
    0.125Wb=7.85
    Wb=7.85/0.125=62.8rev/min
    is that correct?
     
  11. Sep 8, 2009 #10

    Redbelly98

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    Almost, except for the units. It's 62.8 rad/sec.
     
  12. Sep 8, 2009 #11
    how do i convert taht to rev/min.

    i tryed

    62.8/(2pi/60)
    and i got some wierd answer:yuck:
     
  13. Sep 8, 2009 #12

    Redbelly98

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    Good, that is the correct way to convert from rad/s into rev/min.

    I get an entirely reasonable, and correct, answer when I do the calculation.

    p.s.
    If you had posted the answer you actually got, I could now be posting back about whether you did the calculation correctly.
     
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