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Homework Help: Angular Momentum of a particle of mass

  1. May 7, 2010 #1
    I need help on solving part A of this problem. This is a homework problem from webassign. I already solved part b. Can you please expain how you get the numbers because I posted on another site and they just put numbers together and I had no idea how they got it? I only have one try left on web assign so I need to be accurate. Thanks.

    1. The problem statement, all variables and given/known data
    A particle of mass 0.300 kg is attached to the 100 cm mark of a meter stick of mass 0.100 kg. The meter stick rotates on a horizontal, frictionless table with an angular speed of 2.00 rad/s.

    (a) Calculate the angular momentum of the system when the stick is pivoted about an axis perpendicular to the table through the 30.0 cm mark.
    ? kg·m^2/s

    (b) What is the angular momentum when the stick is pivoted about an axis perpendicular to the table through the 0 cm mark?
    0.6667 kg·m^2/s
    The answer is correct.

    2. Relevant equations
    L=angular momentum
    I=moment of inertia
    w=angular speed

    3. The attempt at a solution

    Information from problem above:
    (particle) m= 0.300 kg
    D= 100 cm = 1 m
    (meter stick) M=0.100 kg
    w = 2 rad/s

    meter stick:

    additional weight:
    I_w= mD^2

    I=I_m + I_w = (1/12)(.1kg)(1m)^2 + (.3kg)(.3 m)^2=.0353333333= .0353 kg*m^2
    L=Iw=(.0353 kg*m^2)(2 rad/s)= .0706 kg*m^2/s --> incorrect answer

    meter stick:
    I_m=(1/3)(0.100 kg)(1.00m)^2=0.0333333333= .0333 kg*m^2

    I_w=mD^2=(0.300 kg)(1.00 m)^2=0.300 kg*m^2

    adding them together is the rotational inertia:
    I=I_m + I_w=(.0333 kg*m^2 + .3 kg*m^2)=.3333333333= .333 kg*m^2

    angular momentum:
    L= Iw=(.33 kg*m^2)(2 rad/s)=0.6666666666= .6667 kg*m^2/s
  2. jcsd
  3. May 7, 2010 #2
    Because [tex] I = \frac{1}{12}ML^2 [/tex] Is the moment of inertia of the stick if it was rotating around its center of mass (50 cm mark), but it is rotating around the 30cm mark. Also, for the mass on the end of the stick why would its [tex] d[/tex] be 30cm?
    Last edited: May 7, 2010
  4. May 7, 2010 #3
    I know the part that I=(1/12)MD^2 is for the meter stick. I'm confused on the part of the additional particle at the end.

    I= I_m + I_w = (1/12)MD^2 + (I'm confused on this part)

    I'm not sure if I need to use the 30 cm= .3 m.

    This is the work from someone who replied to my topic I made on another site. I don't know if this is right. It looks like he subtracted .3m from 1m. I have no idea how he got .05m.

    (a) MI about the vertical axis thru the 30 cm mark,
    I = m1r12 + m2.r2^2
    = 0.30*0.70*0.70 + 0.10*0.05*0.05
    =0.14725 kg.m2
    Angular momentum,
    L = I.ω = 0.14725*2.0
  5. May 7, 2010 #4
    So I = (ML^2)/(12) is ONLY for a stick rotating around its center of mass ( at the 50 cm mark) it is not so you need to use the parallel axis theorem. As for the particle, the moment of inertia of a particle is just md^2 where d is the distance from the axis of rotation (30 cm mark) to the particle. The total moment of inertia is the sum of the stick and the particle.
  6. May 7, 2010 #5
    D = 100 cm = 1 m
    M = .1 kg
    m = .3 kg
    d = 30 cm = .3 m
    d_p - d_w = 1 m - .3m = .7 m
    w = 2 rad/s

    I_total = I_m + I_w

    I_total = (1/12)MD^2 + md^2

    I_total = (1/12)(.1 kg)(1 m)^2 + (.3 kg)(1 m - .3 m)^2 = .1553333333 kg*m^2

    L=Iw=(.1553333333 kg*m^2)(2 rad/s)= .3106666667 kg*m^2/s

    Am I doing this right? Is that what your basically saying above? What is the parallel axis theorem?
  7. May 7, 2010 #6
    I for the particle is right since .7m (1-0.3)m is its distance from the axis of rotation.


    (1/12)MD^2 IS ONLY VALID IF THE RULER WAS ROTATING AROUND ITS CENTER OF MASS. The Center of mass is at 50cm and its rotating around 30cm. To find the ruler's moment of inertia you must use the parallel axis theorem.
    Last edited: May 7, 2010
  8. May 7, 2010 #7
    I_total = MD^2 + md^2 = (.1 kg)(.05 m)^2 + (.3 kg)(1 m - .3 m)^2= .14725 kg*m^2

    L=Iw=(.14725 kg*m^2)(2 rad/s)= .2945 kg*m^2/s

    Is this right?
  9. May 7, 2010 #8
    No where 0.05 come from?
  10. May 7, 2010 #9
    It cam from 50 cm = .05 m...

    I_total = MD^2 + md^2 = (.1 kg)(1 m)^2 + (.3 kg)(1 m - .3 m)^2= .247 kg*m^2

    L=Iw=(.247 kg*m^2)(2 rad/s)= .494 kg*m^2/s
  11. May 7, 2010 #10
    Please look up parallel axis theorem because you are not understanding.
  12. May 7, 2010 #11
    How can I understand if you don't tell me? You kept mentioning parallel axis theorem with no explanation. What's the point of posting here then. I came here to get an explanation on how to do it. I'm more confused now.


    I_z = I_cm + mr^2
    I_cm is the moment of inertia of the object about an axis passing through its center of mass;
    m is the object's mass;
    r is the perpendicular distance between the two axes.
  13. May 8, 2010 #12
    I am not allowed to give you the answer, only guide you. Besides, me having to mention the Parallel Axis Theorem THREE times without you having the gumption to look up up gives a clue to how willing you are to learn. Take some initiative. Now back to the problem:

    Correct. So [tex]\frac{1}{12}ML^2 = I_{cm} [/tex] and r is the distance from the center of mass (50cm) to the axis of rotation.
  14. May 8, 2010 #13
    I never asked you for the answer, just want to know how to do it. Asking how to do it doesn't mean give me the answer. You can just show or point me to a similar problem.

    You said (1/12)ML^2=I_cm is only valid if it's rotating around the center of mass (50 cm).

    It's rotating at 30cm so I subtract 1m to get .7 m as the center of mass?

    How does getting parallel axis theorem help me?

    I_w = MD^2 = (.3 kg)(1 m - .3 m)^2= .147 kg*m^2
  15. May 9, 2010 #14
    I think this maybe the way to get it? Can someone check please? Thanks.

    I_total = I_m + I_w
    I = (1/12)(.1 kg)(1 m)^2 + (.3 kg)( [1m-.3m]/2)^2)
    I = (1/12)(.1 kg)(1 m)^2 + (.3 kg)(.35m)^2) = .0450833333 kg*m^2
    L = Iw = (.0450833333 kg*m^2)(2 rad/s) .0901666667 kg*m^2/s
  16. May 10, 2010 #15
    Here is an example: Let's say I know that the moment of inertia for a uniform rod rotating around its center of mass (like a helicopter blade) is [tex] I_{cm} = \frac{1}{12}ML^2 [/tex] and I want to know what the moment of inertia if the blade were rotating around one of its ends (as opposed to the middle). |------ as opposed to ---|--- where | is the axis of rotation and ------ is the rod which is rotating into and out of the page. Well instead of doing using the definition of the moment of inertia and doing an integral I can just use the parallel axis theorem since the axis are parallel. so:

    [tex] I = I_{cm} + md^2 [/tex]

    Where d is the distance from the new axis to the center of mass axis = L/2.

    [tex] I = \frac{1}{12}ML^2 + M(\frac{L}{2})^2 = ML^2(\frac{1}{12}+\frac{1}{4}) = \frac{1}{3}ML^2

    [/tex] which is the corect answer. You can use the same theorem in this problem it is just that d is not L/2 anymore.
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