Ask Question
26 May, 08:05

A solid steel sphere of density 7.86 g/cm3 and mass 1 kg spin on an axis through its center with a period of 2.3 s. given vsphere = 4 3 π r3, what is its angular momentum? answer in units of kg m2 / s.

+4
Answers (1)
  1. 26 May, 11:13
    0
    The solution for this problem is:

    density = mass / volume

    7860 = 1 / ((4/3) pi r^3)

    r^3 = 1 / (7860 * 4/3*pi)

    r = (1 / (7860 * 4/3*pi)) ^ (1/3)

    = 0.067 m

    Inertia = (2/5) mr^2

    = (2/5) x 1 x 0.067^2

    = 0.0017956 kg-m^2

    1/2.3 = 0.4348 rev/s

    0.4348 x 2pi = 2.732 rad/s

    Angular momentum = Inertia x rad/s

    0.0017956 x 2.732 = 0.00490557 kg m^2/s
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “A solid steel sphere of density 7.86 g/cm3 and mass 1 kg spin on an axis through its center with a period of 2.3 s. given vsphere = 4 3 π ...” in 📘 Physics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers