MHT CET · Physics · Laws of Motion
When the bob of mass ' \(m\) ' moves in a horizontal circle of radius ' \(r\) ' with uniform speed ' \(v\) ' having length of string ' \(L\) ' describes a cone of semi vertical angle ' \(q\) ' the centripetal force acting on the bob is given by
[g = acceleration due to gravity.]
- A \(\frac{\mathrm{mgL}}{\sqrt{\mathrm{L}^2-\mathrm{r}^2}}\)
- B \(\frac{\sqrt{\mathrm{L}^2-\mathrm{r}^2}}{\mathrm{mgL}}\)
- C \(\frac{\operatorname{mgr}}{\sqrt{\mathrm{L}^2-\mathrm{r}^2}}\)
- D \(\frac{\mathrm{mgr}}{\mathrm{L}^2-\mathrm{r}^2}\)
Answer & Solution
Correct Answer
(C) \(\frac{\operatorname{mgr}}{\sqrt{\mathrm{L}^2-\mathrm{r}^2}}\)
Step-by-step Solution
Detailed explanation
Consider the figure below:
On taking the ratio of equation (1) and (2):
Centripetal force \(=\mathrm{mg} \tan \theta\)
From geometry, \(\tan \theta=\frac{\mathrm{r}}{\sqrt{\mathrm{L}^2-\mathrm{r}^2}}\)
\(\therefore\) Centripetal force \(=\frac{\mathrm{mgr}}{\sqrt{\mathrm{L}^2-\mathrm{r}^2}}\)
On taking the ratio of equation (1) and (2):
Centripetal force \(=\mathrm{mg} \tan \theta\)
From geometry, \(\tan \theta=\frac{\mathrm{r}}{\sqrt{\mathrm{L}^2-\mathrm{r}^2}}\)
\(\therefore\) Centripetal force \(=\frac{\mathrm{mgr}}{\sqrt{\mathrm{L}^2-\mathrm{r}^2}}\)
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