Question

10.0 points 012 A solid steel sphere of density 7.75 g/cm3 and mass 0.3 kg spin on an axis through its center with a period of 1.3 s Given Vsphere 3 what is its angular momentum? Answer in units of kg m2/s.

Answer 1

Density of the solid steel sphere = $\rho$ = 7.75 g/cm³ = 7.75 x (1000) kg/m³ = 7750 kg/m³

Mass of the solid steel sphere = M = 0.3 kg

Radius of the solid steel sphere = R

Volume of the solid steel sphere = V

$V = \frac{4\pi R^{3}}{3}$

M = $\rho$V

$M = \rho (\frac{4\pi R^{3}}{3})$

$0.3 = (7750)(\frac{4\pi R^{3}}{3})$

R = 2.098 x 10^-2 m

Moment of inertia of the solid steel sphere = I

$I = \frac{2MR^{2}}{5}$

$I = \frac{2(0.3)(2.098\ast 10^{-2})^{2}}{5}$

I = 5.282 x 10^-5 kg.m²

Time period of rotation = T = 1.3 sec

Angular speed of the solid sphere = $\omega$

$T = \frac{2\pi }{\omega }$

$1.3 = \frac{2\pi }{\omega }$

$\omega$ = 4.833 rad/s

Angular momentum of the solid sphere = L

L = I$\omega$

L = (5.282x10^-5)(4.833)

L = 2.55 x 10^-4 kg.m²/s

Angular momentum of the steel solid sphere = 2.55 x 10^-4 kg.m²/s