Question

A thin hollow cylinder open at both ends slides without rotating and then rolls without slipping with the same speed. The ratio of kinetic energies in the two cases is:
A. $1:1$
B. $1:2$
C. $2:1$
D. $1:4$

Answer 1

First of all, we have to find the kinetic energy of the hollow cylinder when it slides without rotating, hence it has only linear motion. Then we have to find the kinetic energy associated with the cylinder when it rolls without slipping. Hence by dividing each other we will find the ratio.

Complete step by step answer:
When the hollow cylinder slides without any rotation, then it has translational kinetic energy only.
Let the kinetic energy for the hollow cylinder when it slides without rotation be $K$ .
$\therefore K = \dfrac{1}{2}m{v^2} - - - - \left( 1 \right)$ where $m =$ mass of the body, $v =$ linear velocity of the body.
Now,
When the hollow cylinder rolls without slipping then it has both translational kinetic energy and rotational kinetic energy.
As the particle has the same velocity according to the question, then the velocity is considered to be $v$ .
Translational kinetic energy possessed by it is $\dfrac{1}{2}m{v^2}$
Rotational kinetic energy possessed by it is $\dfrac{1}{2}I{\omega ^2}$ where $I =$ Moment of inertia and $\omega =$ angular velocity.
We know, $\omega = \dfrac{v}{r}$ where $v =$ velocity or speed and $r =$ radius.
For a hollow cylinder, $I = m{r^2}$
Let the total kinetic energy possessed by the hollow cylinder when it is rolling without slipping be $K'$ .
$\therefore K' = \dfrac{1}{2}m{v^2} + \dfrac{1}{2}I{\omega ^2}$
Substituting the values of $I$ and $\omega$ we get,
$K' = \dfrac{1}{2}m{v^2} + \dfrac{1}{2}\left( {m{r^2}} \right){\left( {\dfrac{v}{r}} \right)^2} = m{v^2}$
Now, the ratio between $K$ and $K'$ is,
$\dfrac{K}{{K'}} = \dfrac{{\dfrac{1}{2}m{v^2}}}{{m{v^2}}} = \dfrac{1}{2}$
Therefore, $K:K' = 1:2$
The correct option is B. $1:2$ .

Note:
Translational kinetic energy refers to that kinetic energy associated with a particle when it moves along a straight line. The kinetic energy associated with a particle when it moves in a circular path or rotates about its mean position is known as rotational kinetic energy. The moment of Inertia is actually the mass of the body while it is in rotational motion. It differs for different bodies.