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A thin uniform rod of length 2 m, cross sectional area ' A ' and density ' d ' is rotated about an axis passing through the centre and perpendicular to its length with angular velocity \omega. If value of \omega in terms of its rotational kinetic energy E is\sqrt{\frac{\alpha E}{A d}} then value of α is

Option: 1

3


Option: 2

__


Option: 3

__


Option: 4

__


Answers (1)

best_answer

density =d\\ Area =\mathrm{A}

mass \: m=d . A L

m=d A(2)=2 A d ______________(1)

\begin{aligned} & \mathrm{K} . \mathrm{E} \cdot=\frac{1}{2} I \omega^2 \\ & \mathrm{E}=\frac{1}{2} \cdot \frac{\mathrm{m} \ell^2}{12} \cdot \omega^2 \end{aligned}

\begin{aligned} &E=\frac{1}{24} \cdot 2 \operatorname{Ad} \cdot(2)^2 \omega^2\\ &\frac{24 E}{8 A d}=\omega^2 \Rightarrow \omega=\sqrt{\frac{3 E}{A d}} \end{aligned}

\text { Ans. } \alpha=3

 

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shivangi.shekhar

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