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A body having a moment of inertia about its axis of rotation equal to \( 3 \mathrm{~kg} \mathrm{~m}^{2} \) is rotating with
angular velocity of \( 3 \mathrm{rad} s^{-1} \). Kinetic energy of this rotating body is same as that of a body of
mass \( 27 \mathrm{~kg} \) moving with velocity \( \mathrm{v} \). The value of \( \mathrm{v} \) is
Options:
angular velocity of \( 3 \mathrm{rad} s^{-1} \). Kinetic energy of this rotating body is same as that of a body of
mass \( 27 \mathrm{~kg} \) moving with velocity \( \mathrm{v} \). The value of \( \mathrm{v} \) is
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Verified Answer
The correct answer is:
\( 1 \mathrm{~ms}^{-1} \)
Given, moment of Inertia, \( I=3 \mathrm{kgm}^{2} \); angular velocity, \( \omega=3 \mathrm{rads}^{-1} ; \) mass,, \( \mathrm{m}=27 \mathrm{~kg} \)
Now \( \frac{1}{2} m v^{2}=\frac{1}{2} I \omega^{2} \)
\( \Rightarrow m v^{2}=I \omega^{2} \Rightarrow v^{2}=\frac{I \omega^{2}}{m} \)
Therefore, \( v=\sqrt{\frac{I}{m}} \omega \)
Substituting the values, we get
\( v=\sqrt{\frac{3}{27}} \times 3=\sqrt{\frac{1}{9}} \times 3=\frac{1}{3} \times 3 \)
Thus, velocity of body is \( 1 \mathrm{~ms}^{-1} \)
Now \( \frac{1}{2} m v^{2}=\frac{1}{2} I \omega^{2} \)
\( \Rightarrow m v^{2}=I \omega^{2} \Rightarrow v^{2}=\frac{I \omega^{2}}{m} \)
Therefore, \( v=\sqrt{\frac{I}{m}} \omega \)
Substituting the values, we get
\( v=\sqrt{\frac{3}{27}} \times 3=\sqrt{\frac{1}{9}} \times 3=\frac{1}{3} \times 3 \)
Thus, velocity of body is \( 1 \mathrm{~ms}^{-1} \)
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