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

$A$ body sliding on a smooth inclined plane requires $4$ seconds to reach the bottom starting from rest at the top. How much time does it take to cover one-fourth distance starting from rest at the top?

$A$ frictionless wire $AB$ is fixed on a sphere of radius $R$. $A$ very small spherical ball slips on this wire. The time taken by this ball to slip from $A$ to $B$ is

Three blocks of masses $2m$,$4m$,and $6m$ are placed as shown in the figure. If $\sin 37^{\circ}=\frac{3}{5}$ and $\sin 53^{\circ}=\frac{4}{5}$,the acceleration of the system is:

As shown in the figure,two spheres are released from point $A$ along paths $AB$ and $AC$ respectively. If both surfaces are smooth,find the time taken by both spheres to reach the bottom of the incline. (Take $g = 10 \, m/s^2$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module