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NEET · Physics · STD 11 - 6. system of particles and rotational motion
A solid sphere of mass \(m\) and radius \(R\) is rotating about its diameter. A solid cylinder of the same mass and same radius is also rotating about its geometrical axis with an angular speed twice that of the sphere. The ratio of their kinetic energies of rotation \(E_{sphere}/E_{cylinder}\) will be
- A \(1:4\)
- B \(3:1\)
- C \(2:3\)
- D \(1:5\)
Answer & Solution
Correct Answer
(D) \(1:5\)
Step-by-step Solution
Detailed explanation
\({{{E_{Sphere}}}}{{{E_{Cylinder}}}} = \frac{{\frac{1}{2}{I_s}\omega _s^2}}{{\frac{1}{2}{I_c}\omega _c^2}} = \frac{{{I_s}\omega _s^2}}{{{I_c}\omega _c^2}}\)
Here,\({I_s} = \frac{2}{5}m{R^2},{I_c} = \frac{1}{2}m{R^2}\)
\({\omega _c} = 2{\omega _s}\)
\({{{E_{Sphere}}}}{{{E_{Cylinder}}}} = \frac{{\frac{2}{5}m{R^2} \times \omega _s^2}}{{\frac{1}{2}m{R^2} \times {{\left( {2{\omega _s}} \right)}^2}}} = \frac{4}{5} \times \frac{1}{4} = \frac{1}{5}\)
Here,\({I_s} = \frac{2}{5}m{R^2},{I_c} = \frac{1}{2}m{R^2}\)
\({\omega _c} = 2{\omega _s}\)
\({{{E_{Sphere}}}}{{{E_{Cylinder}}}} = \frac{{\frac{2}{5}m{R^2} \times \omega _s^2}}{{\frac{1}{2}m{R^2} \times {{\left( {2{\omega _s}} \right)}^2}}} = \frac{4}{5} \times \frac{1}{4} = \frac{1}{5}\)
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