A frictionless wire AB is fixed on a circle o | vedclass

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

Question

(C) In the circle of radius $R$,$AC$ is the diameter,so $AC = 2R$. In the right-angled triangle $ABC$,the length of the wire $AB$ is given by $AB = AC

Vedclass · Accepted Answer

$2\sqrt{\frac{R}{g}}$

Vedclass · Answer

(C) In the circle of radius $R$,$AC$ is the diameter,so $AC = 2R$. In the right-angled triangle $ABC$,the length of the wire $AB$ is given by $AB = AC \cos	heta = 2R \cos	heta$. The acceleration of the bead along the wire $AB$ due to gravity $g$ is $a = g \cos	heta$. Using the equation of motion $s = ut + \frac{1}{2}at^2$,where $u = 0$ (starting from rest),we have: $AB = 0 + \frac{1}{2} (g \cos	heta) t^2$. Substituting $AB = 2R \cos	heta$: $2R \cos	heta = \frac{1}{2} g \cos	heta t^2$. Canceling $\cos	heta$ from both sides: $2R = \frac{1}{2} g t^2$. $t^2 = \frac{4R}{g}$. $t = 2\sqrt{\frac{R}{g}}$.

$A$ frictionless wire $AB$ is fixed on a circle of radius $R$. $A$ very small bead slips on this wire. The time taken by the bead to slip from $A$ to $B$ is:

Similar Questions

$A$ perfectly smooth sphere $A$ of mass $2 \ kg$ is in contact with a rectangular block $B$ of mass $4 \ kg$ and a vertical wall as shown in the figure. All surfaces are smooth. Find the normal reaction exerted by the vertical wall on sphere $A$ (in $N$). Take $g = 10 \ m/s^2$.

If $M_1 = M_2$ and $\theta = 30^o$,then the acceleration is ........... $ms^{-2}$.

Two bodies of masses $m_{1}=5\,kg$ and $m_{2}=3\,kg$ are connected by a light string going over a smooth light pulley on a smooth inclined plane as shown in the figure. The system is at rest. The force exerted by the inclined plane on the body of mass $m_{1}$ will be $....N$ [Take $g=10\,ms^{-2}$]

$A$ body of mass $5 \ kg$ is suspended by a spring balance on an inclined plane as shown in the figure. The spring balance measures (in $N$):

$A$ block of mass $m$ is released on a smooth inclined plane of inclination $\theta$ with the horizontal. The force exerted by the plane on the block has a magnitude

Vedclass Test Series

Exam Paper Generator

Online Exam Module