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I need help can someone carefully check my work

  1. Jan 5, 2010 #1
    A mass is placed on a frictionless incline and attached to a pulley by a light string. The situation is sketched below.

    Rotation Test Figure 2

    The incline angle, theta, is 50.0°, the mass is 4.00 kg, the moment of inertia of the pulley is 0.800 kgm2 and the radius of the pulley is 0.105 m. The mass is released from rest.

    (a) What is the magnitude of the acceleration of the mass?
    _____m/s2

    (b) What is the magnitude of the angular acceleration of the pulley?
    ______rad/s2

    (c) What is the tension in the string?
    ______ N

    For part a I put down 0.438 but it got marked wrong

    T= Tension
    Force from weight going down the ramp= mgsinθ
    a = linear acceleration

    mgsinθ-T=ma
    Iα=T*r

    α= -a/r
    T= -Iα/r = -Ia/r2

    substitute the value of T
    mgsinθ+Ia/r2=ma
    mgsinθ=a(m-I/r2)
    a= mgsinθ/(m-I/r2)= - 0.43798 m/s

    I put 0.43798 because it asked for the magnitude

    2)α = a/r = -4.076 rad/s

    So it should be 4.076 because asked for magnitude

    3)T=Ia/r2= 31.055 N
     
  2. jcsd
  3. Jan 5, 2010 #2

    Redbelly98

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    Homework Helper

    It's a matter of keeping +/- sign conventions consistent.
    From this equation, I gather you are defining down the ramp as a positive direction; we know a is in this direction, therefore a will be positive. Also, rope tension T is positive. Okay so far.

    Here, you are saying that α will be positive (since T, r and I are all positive.) Okay.
    But now you are saying α is negative! This is inconsistent, therefore creating your problem. Instead, use
    α= a/r​
    Fix the "-" sign error, then proceed as you did before.
     
  4. Jan 5, 2010 #3
    thanks your tip helped with part a and part b.
    However, part c was marked wrong.
    Do you mind helping me again?

    here is what i did

    T= Tension
    Force from weight going down the ramp= mgsinθ
    a = linear acceleration

    mgsinθ-T=ma
    Iα=T*r

    α=a/r
    T=Iα/r=Ia/r2

    substitute the value of T
    mgsinθ-Ia/r2=ma
    mgsinθ=a(m+I/r2)
    a=mgsinθ/(m+I/r2)= 0.392

    2)α = a/r = 3.735

    3)T=Ia/r2= (0.8)(0.392)/(0.105^2)= 28.460

    I am confused as to why my answer is wrong.
     
    Last edited: Jan 5, 2010
  5. Jan 6, 2010 #4

    Redbelly98

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    I get the same answer. Perhaps it is marked wrong because you are reporting too many significant figures.
     
  6. Jan 6, 2010 #5
    Nope we were both right. The answer key was wrong. Thanks a lot! You don't know how much you have helped me!
     
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