Question

# A disc of mass $$100\ g$$ and radius $$10\ cm$$ has a projection on its circumference. The mass of projection is negligible. A $$20\ g$$ bit of putty moving tangential to the disc with a velocity of $$5\ m\ s^{-1}$$ strikes the projection and sticks to it. The angular velocity of disc is

A
B
C
D

Solution

## The correct option is A $$14.29\ rad\ s_1$$In this case, the angular momentum of  bit of putty about the axis of rotation = angular momentum of system of disc and bit of putty about the axis of rotation. Let:$$M$$ = Mass of puty$$m$$ = Mass of disc$$\therefore MvR=\left( \dfrac{m{{R}^{2}}}{2}+M{{R}^{2}} \right)\omega$$  $$\omega =\dfrac{MvR}{\left( \dfrac{m{{R}^{2}}}{2}+M{{R}^{2}} \right)}$$  Putting all the values  $$\omega =14.298\ m/s$$Physics

Suggest Corrections

0

Similar questions
View More

People also searched for
View More