One end of a straight uniform 1; m long bar i | vedclass

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(C) Let the length of the bar be $\ell = 1\; m$ and its mass be $m$. The moment of inertia of the bar about the pivot point is $I = \frac{m\ell^2}{3}$

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(C) Let the length of the bar be $\ell = 1\; m$ and its mass be $m$. The moment of inertia of the bar about the pivot point is $I = \frac{m\ell^2}{3}$.Using the principle of conservation of mechanical energy,the initial potential energy of the center of mass of the bar is converted into rotational kinetic energy when it hits the table.The initial height of the center of mass from the table is $h = \frac{\ell}{2} \sin 30^{\circ}$.Initial potential energy $U_i = mgh = mg \left(\frac{\ell}{2}ight) \sin 30^{\circ}$.Final potential energy $U_f = 0$ (at the table level).Initial kinetic energy $K_i = 0$ (released from rest).Final rotational kinetic energy $K_f = \frac{1}{2} I \omega^2$.By conservation of energy: $U_i + K_i = U_f + K_f$$mg \left(\frac{\ell}{2}ight) \sin 30^{\circ} = \frac{1}{2} \left(\frac{m\ell^2}{3}ight) \omega^2$Substituting $\ell = 1\; m$ and $\sin 30^{\circ} = 0.5$:$mg \left(\frac{1}{2}ight) (0.5) = \frac{1}{2} \left(\frac{m(1)^2}{3}ight) \omega^2$$mg \left(\frac{1}{4}ight) = \frac{m}{6} \omega^2$$\frac{g}{4} = \frac{\omega^2}{6} \Rightarrow \omega^2 = \frac{6g}{4} = 1.5g$.Taking $g = 10\; m/s^2$,we get $\omega^2 = 1.5 	imes 10 = 15$.Thus,$\omega = \sqrt{15}\; s^{-1}$.Comparing with $\sqrt{n}$,we get $n = 15$.

One end of a straight uniform $1\; m$ long bar is pivoted on a horizontal table. It is released from rest when it makes an angle $30^{\circ}$ with the horizontal (see figure). Its angular speed when it hits the table is given as $\sqrt{n}\; s^{-1},$ where $n$ is an integer. The value of $n$ is

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