Question

Find speed given to particle at lowest point so that tension in string at point A becomes zero. 

• A. $\sqrt{\dfrac{7g\ell}{2}}$
• B. $\sqrt{3g\ell}$
• C. $\sqrt{\dfrac{9g\ell}{4}}$
• D. $\sqrt{\dfrac{g\ell}{2}}$

Hint:

When tension becomes zero in circular motion, gravity alone provides the centripetal force.

Answer 1

The Correct Option is A

Step 1: Condition for zero tension at point A.
At point A, when tension becomes zero, the centripetal force is provided only by the component of weight.
\[ mg\cos 60^\circ = \dfrac{mv^2}{\ell} \]
Step 2: Calculating speed at point A.
\[ \dfrac{mg}{2} = \dfrac{mv^2}{\ell} \Rightarrow v^2 = \dfrac{g\ell}{2} \]
Step 3: Applying conservation of mechanical energy.
At lowest point, initial speed is $u$. Using M.E.C.:
\[ \dfrac{1}{2}mu^2 = mg\big(\ell + \ell\cos60^\circ\big) + \dfrac{1}{2}mv^2 \]
Step 4: Substituting values.
\[ \dfrac{1}{2}u^2 = mg\left(\ell + \dfrac{\ell}{2}\right) + \dfrac{1}{2}\cdot \dfrac{g\ell}{2} \]
\[ \dfrac{1}{2}u^2 = \dfrac{3g\ell}{2} + \dfrac{g\ell}{4} = \dfrac{7g\ell}{4} \]
Step 5: Final speed at lowest point.
\[ u^2 = \dfrac{7g\ell}{2} \Rightarrow u = \sqrt{\dfrac{7g\ell}{2}} \]