Question

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.

• A. $10m/s$
• B. $7m/s$
• C. $15m/s$
• D. $5m/s$

Answer 1

Answer: (A)

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
$T_{max}=\frac{m{v}_0^2}{r}+mgcos{0}^0=\frac{m{v}_0^2}{r}+mg$
at highest point is $\theta =180^0$ the tension is minimum $T_{min}=\frac{m{v}_{180}^2}{r}+mgcos180^0=\frac{m{v}_{180}^2}{r}-mg$
In the given problum speed is constant, then $v_0^2=v_{180}^2=v^2$
$\frac{T_{max}}{T_{min}}=\frac{\frac{m{v}^2}{r}+mg}{\frac{m{v}^2}{r}-mg}\frac{5}{3}=\frac{v^2+rg}{v^2-rg}2v^2=8rgv=\sqrt{4rg}=\sqrt{4\times 2.5\times 10}=10m/s$
therefore the speed of particle is $10m/s$