Question

A small mass m is tied to a string of length L and is whirled in vertical circular motion. The speed of the mass v is such that the ratio of the string tension at the top of the circle to that at the bottom of the circle is FtopT/FbotT = 0.5. Derive an expression for the speed v.

Answer 1

Answer:

(mv^2/R)/(mg)=1/2

v^2=R/2g

Answer 2

The expression for the speed of the mass is $v = \sqrt{\frac{gL}{3} }$

The given parameters;

mass of the string, = m

length of the string, = L

speed of the string, = v

The tension of the string at the top of the circle;

$T_{top} = \frac{mv_t^2}{r} - mg$

The tension of the string at the bottom of the string;

$T_{bottom} = \frac{mv_b^2}{r} + mg$

The ratio of the two force is given as;

$\frac{T_{top}}{T_{bottom}} = \frac{mg - \frac{mv_t^2}{r} }{mg + \frac{mv_b^2}{r}} = 0.5\\\\0.5(mg + \frac{mv_b^2}{r}) = mg - \frac{mv_t^2}{r}\\\\mg + \frac{mv_b^2}{r} = 2(mg - \frac{mv_t^2}{r})\\\\mg + \frac{mv_b^2}{r} = 2mg - 2\frac{mv_t^2}{r}\\\\2\frac{mv_t^2}{r} + \frac{mv_b^2}{r} = 2mg - mg\\\\2\frac{mv_t^2}{r} + \frac{mv_b^2}{r} = mg\\\\assuming \ v_t = v_b\\\\3\frac{mv^2}{r}= mg\\\\\frac{v^2}{r} = \frac{g}{3} \\\\v^2 = \frac{rg}{3} \\\\v = \sqrt{\frac{rg}{3}}$

$v = \sqrt{\frac{gL}{3} }$

Thus, the expression for the speed of the mass is $v = \sqrt{\frac{gL}{3} }$

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