Question

A simple pendulum is oscillating in a vertical plane. If resultant acceleration of bob of mass $m$ at a point $A$ is in horizontal direction, find the tangential force at this point in terms of tension $\left(T\right)$ and weight of pendulum $\left(mg\right)$.

• A. $T+mg$
• B. $\frac{mg}{T}\sqrt{T^2-(mg{)}^2}$
• C. $\frac{mg}{T}\sqrt{(mg{)}^2+T^2}$
• D. $\frac{T}{mg}\sqrt{(mg{)}^2+T^2}$

Answer 1

The correct option is B $\frac{mg}{T}\sqrt{T^2-(mg{)}^2}$

When the net acceleration of bob is in horizontal direction at point $A$ , net vertical force on the bob will be zero.
$T\cos \theta -mg=0$
$\Rightarrow \cos \theta =\frac{mg}{T}$ ........(1)


The tangential force at that instant is
$F=mg\sin \theta =mg\sqrt{1-{\cos }^2\theta }$
$F=\frac{mg}{T}\sqrt{T^2-(mg{)}^2}$ [ from (1) ]