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MHT CET · Physics · Laws of Motion

In a conical pendulum the bob of mass ' \(\mathrm{m}\) ' moves in a horizontal circle of radius ' \(r\) ' with uniform speed ' \(V\) '. The string of length ' \(L\) ' describes a cone of semi vertical angle ' \(\theta\) '. The centripetal force acting on the bob is ( \(\mathrm{g}=\) acceleration due to gravity)

  1. A \(\frac{\mathrm{mgr}}{\sqrt{\mathrm{L}^2-\mathrm{r}^2}}\)
  2. B \(\frac{\mathrm{mgr}}{\left(\mathrm{L}^2-\mathrm{r}^2\right)}\)
  3. C \(\frac{\sqrt{\mathrm{L}^2-\mathrm{r}^2}}{\mathrm{mgL}}\)
  4. D \(\frac{\mathrm{mgL}}{\sqrt{\mathrm{L}^2-\mathrm{r}^2}}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{\mathrm{mgr}}{\sqrt{\mathrm{L}^2-\mathrm{r}^2}}\)

Step-by-step Solution

Detailed explanation

\(\begin{aligned} & \mathrm{T} \cos \theta=\mathrm{mg} \\ & \sin \theta=\frac{\mathrm{r}}{\mathrm{L}} \\ & \cos \theta=\frac{\sqrt{\mathrm{L}^2-\mathrm{r}^2}}{\mathrm{~L}}\end{aligned}\)


\(\begin{aligned} \therefore \quad \mathrm{T} & =\frac{\mathrm{mg}}{\cos \theta}=\frac{\mathrm{mgL}}{\sqrt{\mathrm{L}^2-\mathrm{r}^2}} \\ \mathrm{mr} \omega^2 & =\mathrm{T} \sin \theta \\ \frac{\mathrm{T} \times \mathrm{r}}{\mathrm{L}} & =\mathrm{m}^2 ...(\sin \theta= \frac{r}{L}) \\ \omega^2 & =\frac{\mathrm{g}}{\sqrt{\mathrm{L}^2-\mathrm{r}^2}}\end{aligned}\)
\(\therefore \quad\) The centripetal force is
\(\mathrm{F}=\frac{\mathrm{mgr}}{\sqrt{\mathrm{L}^2-\mathrm{r}^2}}\)
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