A particle moves along a vertical circle of radius r with a velocity - 8rg at Y- T A- T B- T x- T y respresent tension at A- B- X and Y- respectively- then

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Work Energy Theorem Application

A particle mo...

Question

A particle moves along a vertical circle of radius r with a velocity $\sqrt{8 r g}$ at Y. $T_{A}$ , $T_{B}$ , $T_{x}$ , $T_{y}$ respresent tension at A, B, X and Y, respectively, then

T_{A} = T_{B}

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T_{X} - T_{Y} = 6 m g

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T_{Y} - T_{X} = 6 m g

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T_{Y} > T_{X} \neq 6 m g

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Solution

The correct options are
A $T_{A} = T_{B}$
C $T_{Y} - T_{X} = 6 m g$
When the string is horizontal only tension provides the radial acceleration, thus, $T_{A} = T_{B} = \frac{m v^{2}}{r} T_{X} + m g = \frac{m {v_{X}}^{2}}{r} a n d T_{Y} - m g = \frac{m {v_{Y}}^{2}}{r}$
by conservation of $M E : \frac{m {v_{X}}^{2}}{2} - \frac{m {v_{Y}}^{2}}{2} = - 2 m g r \Rightarrow {v_{X}}^{2} - {v_{Y}}^{2} = - 4 g r T h u s, T_{X} - T_{Y} + 2 m g = \frac{m}{r} ({v_{X}}^{2} - {v_{Y}}^{2}) = - 4 m g \Rightarrow T_{X} - T_{Y} = - 6 m g$

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A particle moves along a vertical circle of radius r with a velocity √8rg at Y. TA, TB, Tx, Ty respresent tension at A, B, X and Y, respectively, then

The correct options are A TA=TB C TY−TX=6mgWhen the string is horizontal only tension provides the radial acceleration, thus,TA=TB=mv2rTX+mg=mvX2r and TY−mg=mvY2rby conservation of ME:mvX22−mvY22=−2mgr⇒vX2−vY2=−4grThus, TX−TY+2mg=mr(vX2−vY2)=−4mg⇒TX−TY=−6mg

A particle moves along a vertical circle of radius r with a velocity $\sqrt{8 r g}$ at Y. $T_{A}$ , $T_{B}$ , $T_{x}$ , $T_{y}$ respresent tension at A, B, X and Y, respectively, then