Question

# A particle of mass m is tied to one end of a string of length l. The particle is held horizontal with the string taut. It is then projected upward with a velocity u. The tension in the string is $$\frac{mg}{2}$$ when it is inclined at an angle $$30^0$$ to the horizontal. Find the value of $$u$$.

Solution

## $$T=\frac { mg }{ 2 } \\ but\quad T=\frac { m{ u }^{ 2 } }{ r } \\ \therefore \quad \frac { mg }{ 2 } =\frac { m{ u }^{ 2 } }{ r } \\ ={ u }^{ 2 }=\frac { rg }{ 2 } \\ using\quad energy\quad balance,\\ \frac { 1 }{ 2 } m{ v }^{ 2 }=\frac { 1 }{ 2 } m{ u }^{ 2 }+mg(rsin{ 30 }^{ \circ })\\ \therefore \quad { v }^{ 2 }={ u }^{ 2 }+2gr\times \frac { 1 }{ 2 } \\ =\frac { rg }{ 2 } +rg=\frac { 3rg }{ 2 } \\ \therefore \quad v=\sqrt { \frac { 3rg }{ 2 } }$$Physics

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