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Homework Help: Angular motion + Inertia

  1. Jun 15, 2010 #1
    Angular motion + Inertia [SOLVED]

    1. The problem statement, all variables and given/known data

    A uniform beam of length, L, and mass, M, is freely pivoted at one end about an attachment point in a wall. The other end is supported by a horizontal cable also attached to the wall, so that the beam makes an angle phi with the horizontal as shown below. To answer the questions below, find algebraic expressions for the tension in the cable, the angular acceleration of the beam, should the cable break, and the resulting angular velocity as the beam falls through the vertical position.

    1. If L = 1.5 m, M = 10 kg, and phi = 35o, then what is the tension in the cable?
    2. If the cable snaps, what is the angular acceleration about the pivot point?
    3. What is the angular velocity of the falling beam, just as it hits the wall?

    2. Relevant equations

    [tex]\Sigma[/tex]F=m*a
    [tex]\Sigma[/tex][tex]\tau[/tex]=F*d
    mgh=1/2*I*[tex]\omega[/tex]^2

    3. The attempt at a solution

    For 1:

    Force of the weight of the beam:
    Fbeam=10*9.8*cos(35)

    Torque of the beam:
    [tex]\Sigma[/tex][tex]\tau[/tex]=(10*9.8*cos(35))*[tex]\frac{1.5}{2}[/tex]-F*1.5=0
    F = 40.1384 N

    Ftension=40.1384*cos(90-35)=0.888 N​

    For 3:

    h = (1.5/2)sin(35)+1.5/2
    I = 1/3*10*1.5^2*[tex]\omega[/tex]^2

    mgh = 1/2*I*[tex]\omega[/tex]^2
    9.8*((1.5/2)sin(35)+1.5/2)=1/2(1/3*1.5^2*[tex]\omega[/tex]^2)
    [tex]\omega[/tex] = 5.55 rad/s​

    For 2:

    [tex]\alpha[/tex]=r[tex]\omega[/tex]^2=1.5*5.55^2= 23.10 rad/s^2​

    I know that my angular velocity is good, but the rest is incorrect. Any clues?

    (Sorry for the formatting. First post :shy:)
     

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    Last edited: Jun 16, 2010
  2. jcsd
  3. Jun 15, 2010 #2

    tiny-tim

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    Welcome to PF!

    Hi Dalek_Supreme ! Welcome to PF! :smile:

    (have an omega: ω and a degree: º and try using the X2 tag just above the Reply box :wink:)

    For 2:

    Your formula ω2r is a linear acceleration, not an angular one (it's the centripetal acceleration of course).

    You need τ = Iα instead (and I think the question is asking for the initial angular acceleration, just when the string breaks)

    For 3:

    hmm … your answer looks correct to me :confused:

    perhaps someone else can spot what's wrong with it? :smile:
     
  4. Jun 15, 2010 #3
    Re: Welcome to PF!

    Ooh. OK, now I got the angular velocity and acceleration! Thanks a lot :smile:

    I am still having problems with the first part. (the tension in the cable) Any idea what's wrong?
     
  5. Jun 16, 2010 #4

    tiny-tim

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    Hi Dalek_Supreme! :smile:

    (just got up :zzz: …)
    oops! i misread your post, and somehow thought you were saying your first part was correct :redface:

    hmm … no, i can't quite make out what you've done in the first part

    i] if you take moments (torques), you must specify which point you're taking them about

    ii] the "Force of the weight" is just the weight … mg

    iii] "Torque of the beam" makes no sense, that line should start "Torque (or moment) of the weight"

    iv] i don't understand where the -F*1.5 comes from, or what F is supposed to be

    (and v] we usually use T for tension :wink:)

    Try again! :smile:
     
  6. Jun 16, 2010 #5
    OK, let me try again! :smile:

    Torque is [tex]\Sigma\tau=F*d[/tex]

    Where:
    [tex]F[/tex] is the perpendicular force acting on the beam;
    [tex]d[/tex] is the distance from the pivot point.​

    The mass of the beam applies a downwards force on the beam of [tex]mg[/tex], i.e.: [tex]10*9.8=98 N[/tex].

    However, for the torque, one requires the perpendicular force. Therefore, that force would be [tex]10*9.8*\cos(35)[/tex]

    Since the center of mass is in the middle, the distance from the pivot point to this force is [tex]\frac{1.5}{2}[/tex].

    Therefore, using the torque equation, we get:

    [tex]\Sigma\tau=((10*9.8*\cos(35))*\frac{1.5}{2})-(F*1.5)=0[/tex]
    [tex]F=40.1385 N[/tex]

    Here, [tex]F[/tex] represents the perpendicular force acting at the end of the beam in order to hold everything in equilibrium. In order to find the tension [tex]T[/tex], one could use trig:

    [tex]F*\cos(90-35)=T[/tex]
    [tex]T=0.888135 N[/tex]

    I hope this helps clear up things.:smile: See the problem?
     
  7. Jun 16, 2010 #6

    tiny-tim

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    Hi Dalek_Supreme! :smile:
    Yes, that's fine down to here, :smile:

    except that I've never seen it done this way before. :confused:

    We usually say torque = Fd, where d is the perpendicular distance from the point to the line of the force.

    (That's the same as the dot product, F.D, where D is the vector from the point to the point of application of the force, or FDcosθ … since d = Dcosθ, that's the same)

    If your professor has told you to do it your way, then I suppose that's ok (though I find it confusing, and apparently you do too) …

    otherwise you should do it the usual way, and use torque = force times perpendicular distance (not distance times perpendicular force). :wink:
    cos(90º-35º) is correct, but you've put it on the wrong side …

    the perpendicular force (to use your terminology) is always smaller than the actual force …

    since T s the actual force, it should be F = Tcos(90º-35º), or Tsin35º.
     
  8. Jun 16, 2010 #7
    Yay! It works. I can't believe it was something so simple. Thanks a lot for your time! :smile:.

    On a side note, my teacher said that we could do it both ways, but engineers prefer taking the perpendicular distance, but physicists prefer taking the perpendicular force. I suppose it's a matter of choice in the end...:wink:
     
  9. Jun 16, 2010 #8

    tiny-tim

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    Hail Dalek_Supreme! :biggrin:
    No, physicists also take the perpendicular distance.

    I have never even heard of the the phrase "perpendicular force" (except of course, when the whole force is perpendicular to something, eg a normal reaction force, or a force perpendicular to an axle), and indeed a google search for the phrase doesn't show anything in the sense in which you used it (the perpendicular component of the actual force). :wink:
     
  10. Jun 16, 2010 #9
    Yeah, that's probably what he ment. Sorry for the confusion. (I'm horrible with terminology :shy:). Thanks for clearing stuff up! :smile:
     
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