Rotational Motion Question 14
Question 14 - 01 February - Shift 1
A solid cylinder is released from rest from the top of an inclined plane of inclination $30^{\circ}$ and length $60 cm$. If the cylinder rolls without slipping, its speed upon reaching the bottom of the inclined plane is $ms^{-1}$. (Given $g=10 ms^{-2}$ )
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Answer: (2)
Solution:
Formula: Radius Of Gyration, Centre of Mass
$v=\sqrt{\frac{2 g h}{1+\frac{k^{2}}{R^{2}}}}$
Where $h=60 \sin 30^{\circ}=30 cm$
$ k^{2}=\frac{R^{2}}{2} $
