A thin rod MN,free to rotate in the vertical | vedclass

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

Question

(D) Let $l$ be the length of the rod and $m$ be its mass.When the rod is horizontal,its potential energy is taken as zero.When the rod makes an angle

Vedclass · Accepted Answer

$\sqrt{\sin \alpha}$

Vedclass · Answer

(D) Let $l$ be the length of the rod and $m$ be its mass.When the rod is horizontal,its potential energy is taken as zero.When the rod makes an angle $\alpha$ with the horizontal,the center of mass of the rod (at distance $l/2$ from $N$) descends by a vertical distance $h = (l/2) \sin \alpha$.By the law of conservation of energy,the loss in potential energy equals the gain in rotational kinetic energy:$mg(l/2) \sin \alpha = \frac{1}{2} I \omega^2$Here,$I$ is the moment of inertia of the rod about the fixed end $N$,which is $I = \frac{ml^2}{3}$.Substituting $I$ into the energy equation:$mg(l/2) \sin \alpha = \frac{1}{2} (\frac{ml^2}{3}) \omega^2$$mg(l/2) \sin \alpha = \frac{ml^2}{6} \omega^2$$\omega^2 = \frac{3g \sin \alpha}{l}$$\omega = \sqrt{\frac{3g \sin \alpha}{l}}$The linear speed $v$ of the end $M$ is given by $v = \omega l$:$v = l \sqrt{\frac{3g \sin \alpha}{l}} = \sqrt{3gl \sin \alpha}$Thus,the speed $v$ is proportional to $\sqrt{\sin \alpha}$.

$A$ thin rod $MN$,free to rotate in the vertical plane about the fixed end $N$,is held horizontal. When the end $M$ is released,the speed of this end,when the rod makes an angle $\alpha$ with the horizontal,will be proportional to (see figure):

Similar Questions

$A$ thin uniform metal rod of mass $M$ and length $L$ is swinging about a horizontal axis passing through its end. Its maximum angular velocity is $\omega$. Its centre of mass rises to a maximum height of (where $g$ is the acceleration due to gravity):

$A$ thin rod of length $L$ and mass $M$ is held vertically with one end on the floor and is allowed to fall. Find the velocity of the other end when it hits the floor,assuming that the end on the floor does not slip.

The power $(P)$ is supplied to a rotating body having moment of inertia '$I$' and angular acceleration '$\alpha$'. Its instantaneous angular velocity is

$A$ tangential force $F$ is applied on the rim of a ring of radius $R$. If the ring rotates by an angle $\theta$,what is the work done by the force?

$A$ light rod carries three equal masses $A$,$B$,and $C$ as shown in the figure. If the rod is released from the horizontal position,what will be the velocity of mass $B$ when the rod reaches the vertical position?

Vedclass Test Series

Exam Paper Generator

Online Exam Module