Rotational Motion Question 1
Question 1 - 24 January - Shift 1
Solid sphere A is rotating about an axis PQ. If the radius of the sphere is $5 cm$ then its radius of gyration about PQ will be $\sqrt{x} cm$. The value of $x$ is_______
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Answer: (110)
Solution:
Formula: Moment of inertia, Perpendicular axis theorem,Radius of gyration
$I _{cm}=\frac{2}{5} MR^{2}$
$I _{PQ}=I _{cm}+md^{2}$
$I _{PQ}=\frac{2}{5} mR^{2}+m(10 cm)^{2}$
For radius of gyration
$I _{PQ}=mk^{2}$
$k^{2}=\frac{2}{5} R^{2}+(10 cm)^{2}$
$=\frac{2}{5}(5)^{2}+100$
$=10+100=110$
$k=\sqrt{110} cm$
$x=110$
