Question

A small object of uniform density rolls up a curved surface with an initial velocity $v\prime$ . It reaches up to a maximum height of $\dfrac{3v^{2}}{4g}$ with the respect to the initial position. The object is
A. Ring
B. Solid sphere
C. Hollow sphere
D. Disc

Answer 1

Hint: Using the law of conservation of energy, we have$mgh=\dfrac{1}{2}mv^{2}+\dfrac{1}{2}I\omega^{2}$, substituting the value of$I=mk^{2}$ and $\omega=\dfrac{v^{2}}{r^{2}}$ . We get $k^{2}$, which is used to denote the object, each object has a specific value of $k$.

Formula used: $v^{2}=\dfrac{2gh}{1+\dfrac{k^{2}}{r^{2}}}$

Complete step by step answer:
We know, from the law of conservation of energy of rotational motion that $mgh=\dfrac{1}{2}mv^{2}+\dfrac{1}{2}I\omega^{2}$
Where $m$ is the mass, $g$ is the acceleration due to gravity, and $h$ is the height, $\omega$ is the angular velocity, and $I$ is the rotational inertia or momentum of inertia.
Moment of inertia is the resistance offered by an object against rotational acceleration.
Also $I=mk^{2}$, where $k$ is the radius of gyration and $\omega=\dfrac{v^{2}}{r^{2}}$
Substituting we get $2mgh=mv^{2}+\dfrac{mk^{2}v^{2}}{r^{2}}$
Or, $2mgh=mv^{2}\left(1+\left(\dfrac{k}{r}\right)^{2}\right)$
Or, $v^{2}=\dfrac{2mgh}{m\left(1 + \dfrac{k^{2}}{r^{2}}\right)}$

And solving we get, $v^{2}=\dfrac{2gh}{1+\dfrac{k^{2}}{r^{2}}}$
Given that height $h=\dfrac{3v^{2}}{4g}$ with $v\prime$ is the initial velocity, then the new velocity $v$ is given by $v=\sqrt{\dfrac{2gh}{1+\dfrac{k^{2}}{r^{2}}}}$
Now substituting the value of $h$ we get
$v^{2}=\dfrac{2g\dfrac{3v^{2}}{4g}}{1+\dfrac{k^{2}}{r^{2}}}$
$1+\dfrac{k^{2}}{r^{2}}=\dfrac{3}{2}$
$\dfrac{k^{2}}{r^{2}}=\dfrac{1}{2}$
$k^{2}=\dfrac{1}{2}r^{2}$
This gives the value of the disc.

So, the correct answer is “Option D”.

Note: Rotational inertia,$I=mk^{2}$, where $k$ is the radius of gyration, this value is unique for each object. Thus it is useful to remember the value of $k$ or $I$ for all the objects. Also, the question might look complex, but it can be solved easily, if the formulas are known.