Question

A small object of uniform density rolls up a curved surface with an initial velocity v′. It reaches up to a maximum height \[\dfrac{{3{v^2}}}{{4g}}\] with respect to the initial position. The object is:

A) Ring
B) Solid sphere
C) Hollow sphere
D) Disc

Answer 1

Hint: The law of conservation of energy states that energy can neither be created nor be destroyed. Although, it may be transformed from one form to another. If you take all forms of energy into account, the total energy of an isolated system always remains constant. All the forms of energy follow the law of conservation of energy.
Here the object is in general motion as we can see that it is going up the curved surface by rolling.
Here object possess three forms of energy
1) Kinetic energy due to rotational motion
2) Kinetic energy due to traversing on the curved path
3) Potential energy due to height it attains on the curved path
As the object rolls up the curved path the sum of its kinetic energies will gradually be converting into potential energy. And at the top point, all the kinetic energy will be transformed into potential energy.
By applying the law of conservation of energy here we can say that
Kinetic Energy due to rotational $ + $Kinetic energy due to transverse motion$ = $potential energy
We know that,
$
  {\left( {K.E} \right)_{transverse}} = \dfrac{1}{2}m{v^2} \\
  {\left( {K.E} \right)_{rotation}} = \dfrac{1}{2}I{\omega ^2} \\
  \left( {P.E} \right) = m \times g \times h \\
 $
So the equation becomes
$\dfrac{1}{2}m{v^2} + \dfrac{1}{2}I{\omega ^2} = mgh$_____________________$\left( 1 \right)$
Here,
$m$: mass of the object
$v$: linear velocity of the object
$\omega $: angular velocity of the object
$h$: height attained by the object
$I$: Moment of inertia
$g$: acceleration due to gravity

Complete step by step solution:
We are given that,
The maximum height up to which the object goes is
$h = \dfrac{{3{v^2}}}{{4g}}$
Initial velocity $v = v`$
We know that angular velocity is given by
$
  v = r\omega \\
  \therefore \omega = \dfrac{v}{r} \\
 $
Substituting all the above value in the equation $\left( 1 \right)$
$
  \dfrac{1}{2}mv{`^2} + \dfrac{1}{2}I{\dfrac{{v`}}{{{r^2}}}^2} = mg \times \dfrac{{3v{`^2}}}{{4g}} \\
   \Rightarrow I = \dfrac{1}{2}m{r^2} \\
 $
Here from the above step we got
$I = \dfrac{1}{2}m{r^2}$
We know that this is the equation for the moment of inertia of a disc.

The correct option is (D), the object is a disk.

Note: Whenever there is a conversion of energy from one form to another law of conservation of energy is always satisfied.
In a general motion, there are two motion transverse and rotational
In general motion, there is only one point of contact