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_______

Show Answer

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

$I _{PQ}=mk^{2}$

$k^{2}=\frac{2}{5} R^{2}+(10 cm)^{2}$

$=\frac{2}{5}(5)^{2}+100$

$=10+100=110$

$k=\sqrt{110} cm$

$x=110$