Q: A solid sphere of mass 500 gm and radius 10 cm rolls down on an inclined plane with out slipping. The height of the centre of mass of the sphere from the ground is 0.7 m. The translational speed of the centre of mass of the sphere on reaching the bottom of inclined will be
(A) √5 m/s
(B)√6 m/s
(C) √10 m/s
(D) 20 m/s
Click to See Answer :
Ans: (c)
Sol: $\displaystyle m g (0.7) = \frac{1}{2}mv^2 + \frac{1}{2}I_{cm}\omega^2 $
$\displaystyle = \frac{1}{2}mv^2 + \frac{1}{2}\times \frac{2}{5}mR^2 (v/R)^2 $
$ \displaystyle mg(0.7) = \frac{7}{10}mv^2 $
$ \displaystyle v^2 = g $
$ \displaystyle v = \sqrt{10}$