A problem about moment of inertia.?

Two uniform spheres, each of mass 500 g and radius 5 cm, mounted at the ends of a 30 cm uniform rod of mass 60 g. (a) Calculate the moment of inertia of the each sphere about an axis perpendicular to the rod and passing through its center using two ways: (i) approximate that the spheres are point particles 20 cm from the axis of rotation (ii) calculate without this approximations. (b) If the spheres retained the same mass but were hollow, would the rotational inertia decrease or increase?

considering the equation

Ip = Ic + Mh^2

2 Answers

Relevance
  • 9 years ago
    Favorite Answer

    a)

    1

    Approximate that the spheres are point particles 20 cm from the axis of rotation M.I of solid sphere .

    M.I = 2 mr^2 = 2*0.5*0.2² = 0.04 kg m²

    2. The moment of inertia of the each sphere about an axis perpendicular to the rod and passing through its center :

    About an axis passing through its center is Ic (2/5)mr ² = (2/5)*0.5*0.05² = 0.0005

    I = 2*Ic + Mh² =2* 0.0005 + 0.5*0.2² = 0.021 kg m²

    b)

    Ic (2/3)mr ² =( 5/3)* 0.0005 = 0.000833

    Ip = 2* 0.000833 + 0.5*0.2² = 0.021666 kg m² decreases

    ======================================

  • 9 years ago

    (i) moment of inertia of each sphere considering them as point mass = 0.500*0.20^2 = 0.02 kg m^2

    (ii) (a) MI of each sphere without approximation = (2/5)*0.5*(0.05^2) + 0.5*0.15^2 =0.0005 + 0.01125 = 0.01175 kg m^2

    So if you approximate the sphere as point mass at its other end you overestimate MI by almost 80 %

    If you consider them as hollow sphere, then MI(hollow) is given by

    MI(hollow)

    (2/3)*0.5*0.05^2 + 0.01125 = 0.00083 + 0.01125 = 0.01258 kg m^2

Still have questions? Get your answers by asking now.