A ball attached to one end of a string swings | vedclass

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

Question

(C) At point $A$ (extreme position),the velocity is zero. The acceleration is purely tangential,given by $a_A = g \sin 	heta$.At point $B$ (mean posi

Vedclass · Accepted Answer

$\cos^{-1}(3/5)$

Vedclass · Answer

(C) At point $A$ (extreme position),the velocity is zero. The acceleration is purely tangential,given by $a_A = g \sin 	heta$.At point $B$ (mean position),the acceleration is purely radial (centripetal),given by $a_B = \frac{v^2}{l}$.Using the conservation of energy between $A$ and $B$,the change in potential energy equals the change in kinetic energy: $mgl(1 - \cos 	heta) = \frac{1}{2}mv^2$,which implies $v^2 = 2gl(1 - \cos 	heta)$.Substituting this into the expression for $a_B$: $a_B = \frac{2gl(1 - \cos 	heta)}{l} = 2g(1 - \cos 	heta)$.Given that $a_A = a_B$,we have $g \sin 	heta = 2g(1 - \cos 	heta)$.Dividing by $g$: $\sin 	heta = 2 - 2 \cos 	heta$.Squaring both sides: $\sin^2 	heta = (2 - 2 \cos 	heta)^2 \Rightarrow 1 - \cos^2 	heta = 4(1 - \cos 	heta)^2$.$1 - \cos^2 	heta = 4(1 - 2 \cos 	heta + \cos^2 	heta) \Rightarrow 1 - \cos^2 	heta = 4 - 8 \cos 	heta + 4 \cos^2 	heta$.$5 \cos^2 	heta - 8 \cos 	heta + 3 = 0$.Solving the quadratic equation for $\cos 	heta$: $\cos 	heta = \frac{8 \pm \sqrt{64 - 4(5)(3)}}{2(5)} = \frac{8 \pm \sqrt{64 - 60}}{10} = \frac{8 \pm 2}{10}$.This gives $\cos 	heta = 1$ (which corresponds to $	heta = 0$,the mean position) or $\cos 	heta = 6/10 = 3/5$.Thus,$	heta = \cos^{-1}(3/5)$.

$A$ ball attached to one end of a string swings in a vertical plane such that its acceleration at point $A$ (extreme position) is equal to its acceleration at point $B$ (mean position). The angle $\theta$ is :-

Similar Questions

$A$ small body of mass $m$ slides down from the top of a hemisphere of radius $r$. The surfaces of the block and the hemisphere are frictionless. The height at which the body loses contact with the surface of the sphere is

$A$ particle tied to one end of a string is being rotated in a vertical circle with constant frequency. The tension in the string at points $A, B, C$ and $D$ are $T_1, T_2, T_3$ and $T_4$ respectively. Then:

$A$ particle is moving in a vertical circle. The tensions in the string when passing through two positions at angles $30^\circ$ and $60^\circ$ from the vertical (lowest position) are $T_1$ and $T_2$ respectively. Then:

$A$ body of mass $0.4\, kg$ is whirled in a vertical circle making $2\, rev/sec$. If the radius of the circle is $2\, m$,then tension in the string when the body is at the top of the circle,is ......... $N$.

$A$ rod of length $L$ is hung from one of its ends,and a mass $m$ is attached to its free end. What tangential velocity must be imparted to $m$ so that it reaches the top of the vertical circle? ($g$ = acceleration due to gravity)

Vedclass Test Series

Exam Paper Generator

Online Exam Module