A solid sphere of mass 5 kg and radius 10 c | vedclass

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Question

(D) The moment of inertia of a solid sphere about its tangent is given by the parallel axis theorem: $I = I_{cm} + MR^2 = \frac{2}{5}MR^2 + MR^2 = \fr

Vedclass · Accepted Answer

$0.63$

Vedclass · Answer

(D) The moment of inertia of a solid sphere about its tangent is given by the parallel axis theorem: $I = I_{cm} + MR^2 = \frac{2}{5}MR^2 + MR^2 = \frac{7}{5}MR^2$.For the system of two spheres, the total moment of inertia about the common tangent passing through the point of contact is the sum of the moments of inertia of each sphere about that same axis.$I_{total} = I_1 + I_2 = \frac{7}{5}m_1R_1^2 + \frac{7}{5}m_2R_2^2 = \frac{7}{5}[m_1R_1^2 + m_2R_2^2]$.Given: $m_1 = 5 \ kg$, $R_1 = 0.1 \ m$; $m_2 = 10 \ kg$, $R_2 = 0.2 \ m$.$I = \frac{7}{5} [5 	imes (0.1)^2 + 10 	imes (0.2)^2]$.$I = \frac{7}{5} [5 	imes 0.01 + 10 	imes 0.04] = \frac{7}{5} [0.05 + 0.4] = \frac{7}{5} [0.45]$.$I = 7 	imes 0.09 = 0.63 \ kg \cdot m^{2}$.

$A$ solid sphere of mass $5 \ kg$ and radius $10 \ cm$ is kept in contact with another solid sphere of mass $10 \ kg$ and radius $20 \ cm$. The moment of inertia of this pair of spheres about the tangent passing through the point of contact is . . . . . . $kg \cdot m^{2}$.

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