A stone of mass 1 kg is tied to a string 4 m long and is rotated at constant speed of 40 $ms^{- 1}$ in a vertical circle. The ratio of the tension at the top and the bottom is

Question

Dr. Abubacker · Accepted Answer

(b) $T_{top} = \frac{mv^{2}}{r} - mg$ …(i)
$T_{bottom} = \frac{mv^{2}}{r} + mg$ …(ii)
$\frac{T_{top}}{T_{bottom}} = \frac{\frac{v^{2}}{r} - g}{\frac{v^{2}}{r} + g} = \frac{\frac{40 	imes 40}{4} - 10}{\frac{40 	imes 40}{4} + 10}$
$= \frac{400 - 10}{400 + 10} = \frac{390}{410} = \frac{39}{41}$

Q. A stone of mass 1 kg is tied to a string 4 m long and is rotated at constant speed of 40 $ms^{- 1}$ in a vertical circle. The ratio of the tension at the top and the bottom is

Explanation

Reviewed by Dr. Abubacker

Want more practice?