Question

A body crosses the topmost point of a vertical circle at critical speed. What will be its centripetal acceleration (in ${\text{m/s}}^2$) when the string becomes horizontal?

• A. $g$
• B. $2g$
• C. $3g$
• D. $5g$

Answer 1

The correct option is C $3g$


At highest point, if body is having critical speed $v$ then tension $T=0$ i.e condition for string becoming just slacked.
From newton's $2^{nd}$ law towards centre ${}^{\prime }{\text{o}}^{\prime }$:
$mg+T=\frac{m{v}^2}{R}.....\left(i\right)$
Putting $T=0$ in Eq.$\left(i\right)$
$\Rightarrow \frac{m{v}^2}{R}=mg$
$\therefore v=\sqrt{Rg}$

Applying law of conservation of energy between point B & C,
$P{E}_B+K{E}_B=P{E}_C+K{E}_C$
$\Rightarrow mg\left(2R\right)+\frac{1}{2}m{v}^2=mg\left(R\right)+\frac{1}{2}m{v}_c^2......\left(ii\right)$
Substituting $v=\sqrt{gR}$
$\therefore v_c=\sqrt{3gR}$
Now centripetal acceleration at horizontal position C:
$a_c=\frac{v_c^2}{R}=\frac{(\sqrt{3gR}{)}^2}{R}=3g$
$\therefore a_c=3g{\text{m/s}}^2$
Hence option C is the correct answer