I need help solving this physics question?

Here is a picture of the question with the figure to help. I’m completely stumped on what to do when I’ve tried so many times but can’t seem to friend the right answer. Please help.

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  • oubaas
    Lv 7
    1 month ago

    with just particles 

    J = 0.5^2(3+4) = 7/4 kg*m^2

    ω = V/r = 8/0.5 = 16.0 rad/sec 

    angular momentum L = J*ω   = 16*7/4 = 28.00 kg*m^2/sec 

    with solid spheres  (of identical mass) 

    J' = (0.5+0.0675)^2(3+4) = 2.25 kg*m^2

    ω ' = ω = 16.0 rad/sec

    angular momentum L' = J'*ω' = 16*2.25 = 36.00 kg*m^2/sec

  • Ash
    Lv 7
    1 month ago

    a) For a light rod with point masses at the end,

    Total angular momentum, L = L₁ + L₂

    L = m₁vr + m₂vr

    L = (m₁+m₂)vr

    L = (4.00 + 3.00)(8.00*0.5) = 28 kg m²/s

    This you already found out ! The assumption here is that the masses at the end of the rods are very small, so considered as point masses.

    b)  For a light rod with large spheres at the end,

    Total angular momentum, L = L₁ + L₂

    L = I₁ω + I₂ω

    L = (I₁ + I₂)ω

    L = (I₁ + I₂)(v/r)

    Use parallel axis theorem...

    For sphere 1 with mass 4.00 kg

    I₁ = I(about the centre) + I(parallel axis contribution)

    I₁ = (2/5)m₁R² + m₁r²

    I₁ = m₁[(2/5)R² + r²]

    I₁ = (4.00)[(2/5)(0.135/2)² + (0.500)²]

    I₁ = 1.00729 kg·m²

     

    I₂ = I(about the centre) + I(parallel axis contribution)

    I₂ = (2/5)m₂R² + m₂r²

    I₂ =m₂[(2/5)R² + r²]

    I₂ = (3.00)[(2/5)(0.135/2)² + (0.500)²]

    I₂ = 0.755 kg·m²

    L = (I₁ + I₂)(v/r)

    L = (1.00729 + 0.755)(8.00/0.500)

    L = 28.2 kg m²/s

  • 1 month ago

    From the diagram, it appears that the centre-to-centre distance (not the surface-to-surface distance) between the masses is ℓ =1.00m.  So I’ll assume ℓ is centre-to-centre.

    The speed (8.00m/s) is assumed to be the speed of each mass’s centre.

    ________________________

    Moment of inertia of m1 about its centre (2/5)mr².

    Using the parallel axis theorem, moment of inertia about pivot is I1 = (2/5)mr² + m(ℓ/2)²

    = m(0.4r² + (ℓ/2)²)

    = 4.00(0.4*(0.135/2)² + (1.00/2)²) = 1.00729 kgm²

    Similarly for m2:

    I2 = 3.00 (0.4*(0.135/2)² + (1.00/2)²) = 0.75547 kgm²

    (A quicker way is to note I2 = (3/4)I1 from the ratio of masses.)

    Total moment of inertia I =  1.00729 +  0.75547 = 1.76276 kgm²

    ω = v/( ℓ/2) = 8.00/0.50 = 16 rad/s

    L = Iω =  1.76276 * 16 = 28.20 kgm²/s to 2 d.p.

    (Based on the assumptions stated above.)

    Technically the answer should be given to 3 significant figures (28.2 kgm²/s) but follow the instructions.

  • cosmo
    Lv 7
    1 month ago

    You have to determine the moment of inertia about the pivot point.

    In part a, you can assume the masses are point objects.

    Then, given the speed, determine the angular velocity.

    In part b, the determination of the moment of inertia involves an integral over the spherical shape.

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