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  4. A solid cylinder rolls down an inclined plane of height 3 m and reaches the bottom of plane with angular velocity, of 2 √ 2 rads -1 . The radius of cylinder must be (Take g = 10 ms -2 )

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Q. A solid cylinder rolls down an inclined plane of height 3 m and reaches the bottom of plane with angular velocity, of 2 $ \sqrt {2} rads ^{-1} $ . The radius of cylinder must be
(Take g = 10 $ ms ^{-2}$ )

KEAMKEAM 2006System of Particles and Rotational Motion

Solution:

$\begin{array}{l} v=\sqrt{\frac{2 g h}{1+k^{2} / R^{2}}}=\sqrt{\frac{2 g h}{1+(1 / 2)}}=\sqrt{\frac{4 g h}{3}}=\sqrt{\frac{4 \times 10 \times 3}{3}} \\ 2 \sqrt{10} \\ \omega=\frac{v}{R} \Rightarrow R=\frac{v}{\omega}=\frac{2 \sqrt{10}}{2 \sqrt{2}}=\sqrt{5} m \end{array}$