Question
A small object of uniform density rolls up a curved surface with an initial velocity $$v.$$ It reaches up to a maximum height of $$\frac{{3{v^2}}}{{4g}}$$ with respect to the initial position. The object is
A small object of uniform density rolls up a curved surface with an initial velocity $$v.$$ It reaches up to a maximum height of $$\frac{{3{v^2}}}{{4g}}$$ with respect to the initial position. The object is
A.
Ring
B.
Solid sphere
C.
Hollow sphere
D.
Disc
Answer :
Disc
Solution :
By the concept of energy conservation
$$\eqalign{ & \frac{1}{2}m{v^2} + \frac{1}{2}I{\omega ^2} = mg\left( {\frac{{3{v^2}}}{{4g}}} \right) \cr & \therefore \frac{1}{2}m{v^2} + \frac{1}{2}I\frac{{{v^2}}}{{{R^2}}} = \frac{3}{4}m{v^2}\,\,\,\,\,\left[ {\because v = R\omega } \right] \cr & \therefore \frac{1}{2}I\frac{{{v^2}}}{{{R^2}}} = \frac{3}{4}m{v^2} - \frac{1}{2}m{v^2} = \frac{1}{4}m{v^2} \cr & \Rightarrow I = \frac{1}{2}m{R^2} \cr} $$
This is the formula of the moment of inertia of the disc.
By the concept of energy conservation
$$\eqalign{ & \frac{1}{2}m{v^2} + \frac{1}{2}I{\omega ^2} = mg\left( {\frac{{3{v^2}}}{{4g}}} \right) \cr & \therefore \frac{1}{2}m{v^2} + \frac{1}{2}I\frac{{{v^2}}}{{{R^2}}} = \frac{3}{4}m{v^2}\,\,\,\,\,\left[ {\because v = R\omega } \right] \cr & \therefore \frac{1}{2}I\frac{{{v^2}}}{{{R^2}}} = \frac{3}{4}m{v^2} - \frac{1}{2}m{v^2} = \frac{1}{4}m{v^2} \cr & \Rightarrow I = \frac{1}{2}m{R^2} \cr} $$
This is the formula of the moment of inertia of the disc.
