Question 255: Work, Energy and Power & Circular Motion

Question

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

Accepted Answer

$$\begin{aligned}
& \mathrm{T} \cos 	heta=\mathrm{mg} \
& \sin 	heta=\frac{\mathrm{r}}{\mathrm{L}} \
& \cos 	heta=\frac{\sqrt{\mathrm{L}^2-\mathrm{r}^2}}{\mathrm{~L}} \
& 	herefore \quad \mathrm{T}=\frac{\mathrm{mg}}{\cos 	heta}=\frac{\mathrm{mgL}}{\sqrt{\mathrm{L}^2-\mathrm{r}^2}} \
& \mathrm{mr} \omega^2=\mathrm{T} \sin 	heta
\end{aligned}$$
$$\begin{aligned}
& \frac{\mathrm{T} 	imes \mathrm{r}}{\mathrm{L}}=\mathrm{m} \omega^2 \quad ...(\sin	heta=\frac{r}{L})\
& \omega^2=\frac{\mathrm{g}}{\sqrt{\mathrm{L}^2-\mathrm{r}^2}}
\end{aligned}$$
$$	herefore \quad$$ The centripetal force is
$$F=\frac{m g r}{\sqrt{L^2-r^2}}$$

Answer

$$\frac{\mathrm{mgr}}{\sqrt{\mathrm{L}^2-\mathrm{r}^2}}$$

Answer

$$\frac{\mathrm{mgr}}{\left(\mathrm{L}^2-\mathrm{r}^2\right)}$$

Answer

$$\frac{\sqrt{\mathrm{L}^2-\mathrm{r}^2}}{\mathrm{mgL}}$$

Answer

$$\frac{\mathrm{mgL}}{\sqrt{\mathrm{L}^2-\mathrm{r}^2}}$$

Work, Energy and Power & Circular Motion

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

Choose the correct answer: