The end B of the rod AB which makes an angle | vedclass

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

Question

(A) Let the coordinates of $A$ be $(0, y)$ and $B$ be $(x, 0)$. The length of the rod is $l = \sqrt{x^2 + y^2}$.Since $x = l \cos 	heta$ and $y = l \

Vedclass · Accepted Answer

velocity of end $A$ is $\frac{4}{3} v_0$ downwards

Vedclass · Answer

(A) Let the coordinates of $A$ be $(0, y)$ and $B$ be $(x, 0)$. The length of the rod is $l = \sqrt{x^2 + y^2}$.Since $x = l \cos 	heta$ and $y = l \sin 	heta$,we have $v_0 = \frac{dx}{dt} = -l \sin 	heta \frac{d	heta}{dt}$.Thus,the angular velocity $\omega = |\frac{d	heta}{dt}| = \frac{v_0}{l \sin 	heta}$.For end $A$,$v_A = |\frac{dy}{dt}| = |l \cos 	heta \frac{d	heta}{dt}| = l \cos 	heta (\frac{v_0}{l \sin 	heta}) = v_0 \cot 	heta$.At $	heta = 37^o$,$\cot 37^o = \frac{4}{3}$,so $v_A = \frac{4}{3} v_0$.Also,$\omega = \frac{v_0}{l \sin 37^o} = \frac{v_0}{l (3/5)} = \frac{5}{3} \frac{v_0}{l}$.Since $v_A$ and $\omega$ depend on $	heta$,they are not constant. Thus,option $A$ is correct.

The end $B$ of the rod $AB$ which makes an angle $\theta$ with the floor is being pulled with a constant velocity $v_0$ as shown. The length of the rod is $l$. At the instant when $\theta = 37^o$,then:

Similar Questions

$A$ proton of mass $1.6 \times 10^{-27} \, kg$ moves in a circular orbit of radius $0.10 \, m$ under a centripetal force of $4 \times 10^{-13} \, N$. The frequency of revolution of the proton is about:

$A$ stone of mass $16 \, kg$ is attached to a string $144 \, m$ long and is whirled in a horizontal circle. The maximum tension the string can withstand is $16 \, N$. The maximum velocity of revolution that can be given to the stone without breaking it will be ....... $m/s$.

$A$ mass is supported on a frictionless horizontal surface. It is attached to a string and rotates about a fixed centre at an angular velocity $\omega_0$. If the length of the string and angular velocity are doubled,the tension in the string which was initially $T_0$ is now

$A$ ball of mass $0.25 \,kg$ attached to the end of a string of length $1.96 \,m$ is moving in a horizontal circle. The string will break if the tension is more than $25 \,N$. What is the maximum speed with which the ball can be moved (in $\,m/s$)?

$A$ particle of mass $M$ moves with constant speed along a circular path of radius $r$ under the action of a force $F$. Its speed is

Vedclass Test Series

Exam Paper Generator

Online Exam Module