Question

A solid cylinder of mass $0.1kg$ having radius \[0.2m\] rolls down an inclined plane of height \[0.6m\] without slipping. The linear velocity the cylinder at the bottom of the inclined plane is:
(A) $28m.{s^{ - 1}}$
(B) $2.8m.{s^{ - 1}}$
(C) $280m.{s^{ - 1}}$
(D) $0.28m.{s^{ - 1}}$

Answer 1

Hint: As per the question, if the solid cylinder of the given mass rolls down an inclined plane of height, that indicates the potential energy converted into kinetic energy. And there, we will apply the law of conservation of energy.

Complete answer:
Given that-
Mass of a solid cylinder, $m = 0.1kg$
Radius of a solid cylinder, $r = 0.2m$
Solid cylinder downs an inclined plane of a height of, $h = 0.6m$
So, to find the linear velocity of the cylinder $(v)$ , we should apply the law of conservation of energy:
According to the law of conservation of energy-
$\therefore \dfrac{3}{4}m{v^2} = mgh$
where, $m$ is the mass of the cylinder.
$v$ is the linear velocity of the cylinder, which we have to conclude.
$g$ is the gravitational force, as the cylinder downs an inclined plane,
$h$ is the height at which cylinder downs an inclined plane.
Now, $m$ cancels out from the both sides:
$ \Rightarrow \dfrac{3}{4}{v^2} = gh$
$ \Rightarrow {v^2} = \dfrac{{4gh}}{3}$
Put the constant value of $g$ , $9.8m.{s^{ - 1}}$ .
$\Rightarrow v = \sqrt {\dfrac{{4(9.8) \times 0.6}}{3}} $
$\Rightarrow v = 2.8m.{s^{ - 1}} $
Therefore, the linear velocity of the cylinder at the bottom of the inclined plane is $2.8m.{s^{ - 1}}$ .

Hence, the correct option is (B) $2.8m.{s^{ - 1}}$ .

Note:
As we know, when the solid cylinder descends from an inclined plane, then the potential energy of the cylinder converts into the potential energy (as energy can neither be created nor be destroyed). So, there are two types of kinetic energy formed, translational and rotational kinetic energy.
$\therefore P.E = K.{E_T} + K.{E_R} $
$\Rightarrow mgh = \dfrac{1}{2}m{v^2} + \dfrac{1}{2}I{\omega ^2} $