A rod of length L is hung from one of its end | vedclass

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(B) For a mass attached to a rigid rod to complete a vertical circle,the minimum velocity at the bottom point is determined by the conservation of mec

Vedclass · Accepted Answer

$2 \sqrt{gL}$

Vedclass · Answer

(B) For a mass attached to a rigid rod to complete a vertical circle,the minimum velocity at the bottom point is determined by the conservation of mechanical energy.At the bottom point,the potential energy is $0$ (taking the bottom as the reference level).At the top point,the height $h = 2L$ and the velocity $v_{top}$ can be $0$ because the rod provides support (unlike a string).Using the law of conservation of energy:$E_{bottom} = E_{top}$$\frac{1}{2} mv^2 = mg(2L) + \frac{1}{2} m(v_{top})^2$For the minimum velocity,we set $v_{top} = 0$:$\frac{1}{2} mv^2 = 2mgL$$v^2 = 4gL$$v = 2 \sqrt{gL}$

List-$I$ (Speed at lowest point)	List-$II$ (Possible situation)
$(P) v_0 = \sqrt{5 g \ell}$	$(1)$ Tension at lowest point $= 6 mg$
$(Q) v_0 = \sqrt{g \ell}$	$(2)$ String will slack for a finite time
$(R) v_0 = 2 \sqrt{g \ell}$	$(3)$ Bob will oscillate
$(S) v_0 = 3 \sqrt{g \ell}$	$(4)$ Tension at highest point $= 4 mg$

$A$ 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)

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