A) \[0.1\,\,J\]
B) \[0.3\,\,J\]
C) \[0.5\,\,J\]
D) \[0.42\,\,J\]
Correct Answer: D
Solution :
Key Idea: Sphere has rotational and translational kinetic energy. When the spherical shell rolls down and inclined plane it has rotational and translational kinetic energy. The moment of inertia plays the same role in the rotational motion as mass plays in translational motion, that is Translational kinetic energy= \[\frac{1}{2}\,m{{v}^{2}}\] Where \[v\] is velocity of body. Rotational kinetic energy \[\frac{1}{2}I\,{{\omega }^{2}}\] Where \[I\] is moment of inertia and \[\omega \] is angular velocity. \[E=\frac{1}{2}m{{v}^{2}}+\frac{1}{2}I{{\omega }^{2}}\] Given, \[m=2kg,\,v=0.5,\,r=0.1m,\,I=\frac{2}{3}M{{R}^{2}}\] \[\therefore \] \[I=\frac{2}{3}>2\times {{\left( 0.1 \right)}^{2}},\omega =\frac{v}{r}=\frac{0.5}{0.1}\] Thus, \[E=\frac{1}{2}\times 2\times {{\left( 0.5 \right)}^{2}}\] \[+\frac{1}{2}\times \frac{2}{3}\times 2\times {{\left( 0.1 \right)}^{2}}\times {{\left( \frac{0.5}{0.1} \right)}^{2}}\] \[E=0.25+0.17=0.42\,J\]
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