Question

A simple pendulum of length $L$ carries a bob of mass $m$. When the bob is at its lowest position, it is given that the minimum horizontal speed necessary for it to move in a vertical circle about the point of suspension. When the string is horizontal, the net force on the bob is

• A. $\sqrt{10} \,mg$
• B. $\sqrt{5} \,mg$
• C. $4\, mg$
• D. $1\, mg$

Answer 1

Answer: (A)

According to the question, the bob is moving in a vertical circle, so the velocity at lowest point $ v=\sqrt{5 r g}$
Now, applying conservation of mechanical energy, we get
$\frac{1}{2} m \times 5 r g =\frac{1}{2} m v^{2}+m g r $
or, $\frac{5}{2} r g =\frac{v^{2}}{2}+r g $ or, $\frac{3}{2} r g=\frac{v^{2}}{2}$
or, $v^{2}=3 r g\,\,\,...(i)$
Now, Horizontal force on the bob, $F x=F_{c}=\frac{m v^{2}}{r}$
$=\frac{m \times 3 r g}{r}=3 \,m g\,\,$ [from Eq. (i)]
Vertical force on the bob, $F_{y}=m g$
$\therefore F_{\text {net }}=\sqrt{F_{x}^{2}+F_{y}^{2}} $
$=\sqrt{(3 m g)^{2}+(m g)^{2}}=\sqrt{10} \,m g$