A bob is whirled in a horizontal plane by means of a string with an initial speed of $\omega \mathrm

Question

A bob is whirled in a horizontal plane by means of a string with an initial speed of $\omega \mathrm{rpm}$. The tension in the string is $T$. If speed becomes $2 \omega$ while keeping the same radius, the tension in the string becomes

Professor PC Thomas Classes · Answer

(b) : When the bob is moving with speed ' $\omega$ ', then $F B D$ is, Let ' $r$ ' be the radius of the circle. Applying Newton's second law on mass of the bob ' $m$ ' along centripetal direction, we have \ $ T=m r \omega^2 $ \ When speed becomes $2 \omega$, the FBD is  \ $ \begin{aligned} & T^{\prime}=m r(2 \omega)^2=m r\left(4 \omega^2ight) \ & 	ext { or } m r=\frac{T^{\prime}}{4 \omega^2} \end{aligned} \ $ Using value of equation (ii) in equation in (i), $ T=\frac{T^{\prime} \omega^2}{4 \omega^2}=\frac{T^{\prime}}{4} \ $ Hence, $T^{\prime}=4 T$