Q. A string is whirling in a vertical circle of radius $2.5m$ with constant velocity. The maximum and minimum tension in the string are in the ratio $5:3$ . Find the velocity.
NTA AbhyasNTA Abhyas 2020
Solution:
In vertical circular motion, the tention at any point on the vertical circular path is $T=\frac{m v_{\theta }^{2}}{r}+mgcos\theta $
at lowest point $\theta =0^{0}$ ,tension is maximum
$T_{m a x}=\frac{m v_{0}^{2}}{r}+mgcos0^{0}=\frac{m v_{0}^{2}}{r}+mg$
at highest point is $\theta =180^{0}$ the tension is minimum $T_{m i n}=\frac{m v^{2}_{180}}{r}+mgcos180^{0}=\frac{m v^{2}_{180}}{r}-mg$
In the given problum speed is constant, then $v_{0}^{2}=v^{2}_{180}=v^{2}$
$\frac{T_{m a x}}{T_{m i n}}=\frac{\frac{m v^{2}}{r}+mg}{\frac{m v^{2}}{r}-mg}\frac{5}{3}=\frac{v^{2} + r g}{v^{2} - r g}2v^{2}=8rgv=\sqrt{4 r g}=\sqrt{4 \times 2 . 5 \times 10}=10m/s$
therefore the speed of particle is $10m/s$
at lowest point $\theta =0^{0}$ ,tension is maximum
$T_{m a x}=\frac{m v_{0}^{2}}{r}+mgcos0^{0}=\frac{m v_{0}^{2}}{r}+mg$
at highest point is $\theta =180^{0}$ the tension is minimum $T_{m i n}=\frac{m v^{2}_{180}}{r}+mgcos180^{0}=\frac{m v^{2}_{180}}{r}-mg$
In the given problum speed is constant, then $v_{0}^{2}=v^{2}_{180}=v^{2}$
$\frac{T_{m a x}}{T_{m i n}}=\frac{\frac{m v^{2}}{r}+mg}{\frac{m v^{2}}{r}-mg}\frac{5}{3}=\frac{v^{2} + r g}{v^{2} - r g}2v^{2}=8rgv=\sqrt{4 r g}=\sqrt{4 \times 2 . 5 \times 10}=10m/s$
therefore the speed of particle is $10m/s$