A) \[\frac{1}{2}I{{\omega }^{2}}\]
B) \[\frac{1}{2}m{{v}^{2}}\]
C) \[I\omega +mv\]
D) \[\frac{1}{2}I{{\omega }^{2}}+\frac{1}{2}m{{v}^{2}}\]
Correct Answer: D
Solution :
When a sphere is in motion, it has kinetic energy due to rotation (the energy due to rotational motion) and translational (the energy due to motion from one location to another). The translational kinetin energy is . \[{{(KE)}_{T}}=\frac{1}{2}m{{v}^{2}}\] where m is mass of object and v is speed. The rolling kinetic energy is \[{{(KE)}_{R}}=\frac{1}{2}I\,{{\omega }^{2}}\] where \[I\] is moment of inertia and co is angular speed. Hence, total kinetic energy is \[KE={{(KE)}_{T}}+{{(KE)}_{R}}\] \[KE=\frac{1}{2}m{{v}^{2}}+\frac{1}{2}I\,{{\omega }^{2}}\]
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