Question

The maximum and minimum tension in the string whirling in a vertical circle of the radius $2.5\,m$ with constant speed are in the ratio $5:3$ then its velocity is $\left(\right.g=9.8\,ms^{- 2}\left.\right)$

• A. $\sqrt{98} \, m \, s^{- 1}$
• B. $7 \, m \, s^{- 1}$
• C. $\sqrt{490} \, m \, s^{- 1}$
• D. $\sqrt{4.9} \, m \, s^{- 1}$

Answer 1

Answer: (A)

In this problem, it is assumed that particle although moving in a vertical loop, but its speed remains constant
Tension at lowest point $T_{m a x}=\frac{m v^{2}}{r}+m$ g
Tension at highest point $T_{m i n}=\frac{m v^{2}}{r}-mg$
$\frac{T_{m a x}}{T_{m i n}}=\frac{\frac{m v^{2}}{r} + m g}{\frac{m v^{2}}{r} - m g}=\frac{5}{3}$
$\frac{3 m v^{2}}{r}+3mg=\frac{5 m v^{2}}{r}-5mg$
$\frac{2 m v^{2}}{r}=8mg$
$v=\sqrt{4 r g}$
$=\sqrt{4 \times 2.5 \times 9.8}$
$=\sqrt{98} \, ms^{- 1}$