Question

A sphere of diameter 0.2 m and mass 2 kg is rolling on an inclined plane with velocity $v{\text{ }} = {\text{ }}0.5{\text{ }}m/s$. The kinetic energy of the sphere is:
$\left( A \right)\;0.4J $
$ \left( B \right)\;0.3J $
$ \left( C \right)\;0.6J $
$ \;\left( D \right)0.42J $

Answer 1

Hint
The rolling body that is sphere here exhibits two types of kinetic energy ( translational kinetic energy and rotational kinetic energy). According to the work energy theorem in rotational motion, the change in the rotational kinetic energy of a rigid body is equal to the work done by external torques acting on the body.

Complete step by step solution
The rolling body that is sphere here exhibits two types of kinetic energy (translational kinetic energy and rotational kinetic energy).
Translational kinetic energy will be $ = \dfrac{1}{2}\mathop {mv}\nolimits^2 $
Where m is the mass of sphere and v is velocity ,
Rotational kinetic energy will be $ = \dfrac{1}{2}\mathop {I\omega }\nolimits^2 $
Where I is moment of inertia and $\mathop \omega \nolimits^2 = {\text{ }}angular{\text{ }}speed$
Now, $\omega = \dfrac{v}{r}$ where v is linear velocity
Inertia of sphere will be $\dfrac{2}{3}\mathop {mr}\nolimits^2 $
Now, $r{\text{ }} = {\text{ }}0.1{\text{ }}m{\text{ }},{\text{ }}v{\text{ }} = {\text{ }}0.5{\text{ }}m/s{\text{ }},{\text{ }}m{\text{ }} = {\text{ }}2kg$
Now $kinetic{\text{ }}energy{\text{ }}of{\text{ }}the{\text{ }}sphere{\text{ }}will{\text{ }}be{\text{ }} = {\text{ }}transational{\text{ }}kinetic{\text{ }}energy{\text{ }} + {\text{ }}rotational{\text{ }}kinetic{\text{ }}energy$energy $ = \dfrac{1}{2}\mathop {mv}\nolimits^2 + \dfrac{1}{2}\mathop {I\omega }\nolimits^2 $
$ = \dfrac{1}{2} \times 2 \times \mathop {(0.5)}\nolimits^2 + \dfrac{1}{2} \times \dfrac{2}{3}\mathop {mr}\nolimits^2 \times \dfrac{{\mathop {(0.5)}\nolimits^2 }}{{\mathop {(0.1)}\nolimits^2 }}$
$K.E{\text{ }} = {\text{ }}0.25{\text{ }} + {\text{ }}0.17{\text{ }}J$
$K.E{\text{ }} = {\text{ }}0.42J$
Now let us match this calculated value with given options:
$0.4J$: This value does not match with the calculated value. Thus, this option is not correct.
$0.3{\text{ }}J$: This value does not match with the calculated value. Thus, this option is not correct.
$0.6{\text{ }}J$: This value does not match with the calculated value. Thus, this option is not correct.
$0.42J$: This value exactly matches the calculated value. Thus. this option is correct.
option (D) is the correct answer.

Note
Moment of inertia, $I = 2 \times rotational{\text{ }}K.E.$.
Hence, the moment of inertia of a rigid body about an axis of rotation is numerically equal to twice the rotational kinetic energy of the body when it is rotating with unit angular velocity about that axis. Rotational kinetic energy depends upon the axis of rotation.