A rod of length 50,cm is pivoted at one end. | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let the length of the rod be $\ell = 0.5\,m$. The center of mass of the rod is at a distance $\frac{\ell}{2}$ from the pivot.When the rod is at an

Vedclass · Accepted Answer

$\sqrt{30}$

Vedclass · Answer

(B) Let the length of the rod be $\ell = 0.5\,m$. The center of mass of the rod is at a distance $\frac{\ell}{2}$ from the pivot.When the rod is at an angle of $30^o$ with the horizontal,the vertical height of the center of mass above the horizontal position is $h = \frac{\ell}{2} \sin(30^o) = \frac{\ell}{2} 	imes \frac{1}{2} = \frac{\ell}{4}$.By the law of conservation of mechanical energy,the loss in potential energy equals the gain in rotational kinetic energy:$mg h = \frac{1}{2} I \omega^2$Where $I = \frac{m\ell^2}{3}$ is the moment of inertia of the rod about the pivot.$mg \left( \frac{\ell}{4} ight) = \frac{1}{2} \left( \frac{m\ell^2}{3} ight) \omega^2$$g \frac{\ell}{4} = \frac{\ell^2}{6} \omega^2$$\omega^2 = \frac{6g}{4\ell} = \frac{3g}{2\ell}$Substituting $g = 10\,ms^{-2}$ and $\ell = 0.5\,m$:$\omega^2 = \frac{3 	imes 10}{2 	imes 0.5} = \frac{30}{1} = 30$$\omega = \sqrt{30}\,rad\,s^{-1}$.

$A$ rod of length $50\,cm$ is pivoted at one end. It is raised such that it makes an angle of $30^o$ from the horizontal as shown and released from rest. Its angular speed when it passes through the horizontal (in $rad\,s^{-1}$) will be $(g = 10\,ms^{-2})$.

Similar Questions

$A$ uniform rod of mass $M$ and length $L$ is hinged at one end as shown in the figure. Initially,the rod is kept horizontal by a massless string. When the string is cut,the angular acceleration of the rod is:

$A$ stationary horizontal disc is free to rotate about its axis. When a torque is applied on it,its kinetic energy as a function of $\theta,$ where $\theta$ is the angle by which it has rotated,is given as $K\theta^2.$ If its moment of inertia is $I,$ then the angular acceleration of the disc is

$A$ wheel having a moment of inertia of $2 \, kg \cdot m^2$ about its vertical axis rotates at the rate of $60 \, rpm$ about the axis. The torque required to stop the wheel's rotation in one minute would be

In the following figure,$r_1 = 5 \, cm$ and $r_2 = 30 \, cm$. If the moment of inertia of the wheel is $5100 \, kg \cdot m^2$,then its angular acceleration will be:

$A$ solid sphere of mass $2 \,kg$ and radius $1 \,m$ is free to rotate about an axis passing through its centre. $A$ constant tangential force $F$ is required to rotate the sphere with $10 \,rad/s$ in $2 \,s$ starting from rest. The value of $F$ is . . . . . . (in $\,N$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module