Question

If a particle of mass $M$ is tied to a light inextensible string fixed at point $P$ and particle is projected at A with velocity $V_A=\sqrt{4gL}$ as shown. Find tension in the string at $B$. (Assume particle is projected in the vertical plane.)

• A. $Mg$
• B. $3Mg$
• C. $2Mg$
• D. $7Mg$

Answer 1

The correct option is A $Mg$

Taking point A as reference where gravitational potential energy is zero,

Total energy of particle at point A is, $E_A=\frac{1}{2}M{V}_A^2=2MgL$

Let velocity at point B as $V_B$.

So, Total energy at point B is, $E_B=mgL+\frac{1}{2}M{V}_B^2$

By conservation of energy, $E_A=E_B$

$⟹\frac{1}{2}M{V}_B^2=mgL⟹\frac{M{V}_B^2}{L}=2Mg$

So centripetal force point B is $2Mg$

At point B, $T+Mg=\frac{M{V}_B^2}{L}=2Mg⟹T=Mg$

So Tension in the string is $Mg$, at point B.