Question

A bead of mass m stays at point$\mathrm{P}(\mathrm{a},\mathrm{b})$ on a wire bent in the shape of a parabola $\mathrm{y}=4{\mathrm{Cx}}^2$ and rotating with angular speed $\text{'}\mathrm{\omega }\text{'}$(see figure). The value of $\text{'}\mathrm{\omega }\text{'}$is (neglect friction):

• A. $\sqrt{\frac{2\mathrm{g}}{\mathrm{C}}}$
• B. $2\sqrt{\mathrm{gC}}$
• C. $\sqrt{\frac{2\mathrm{gC}}{\mathrm{ab}}}$
• D. $2\sqrt{2\mathrm{gC}}$

Answer 1

The correct option is D

$2\sqrt{2\mathrm{gC}}$

Step 1: Given data and required figure

The equation of a parabola is, $\mathrm{y}=4{\mathrm{Cx}}^2$

Mass of a bead$=\mathrm{m}$

Angular speed of a bead $=\mathrm{\omega }$

The coordinate point is given as $\mathrm{P}(\mathrm{a},\mathrm{b})$

Acceleration due to gravity$=\mathrm{g}$

The required figure for given problem is shown below:

Step 2: Finding the value of $\text{'}\mathrm{\omega }\text{'}$

We have the equation of parabola $\mathrm{y}=4{\mathrm{Cx}}^2$

Differentiating with respect to $\mathrm{x}$ we get,

$$\begin{gathered} \frac{d\mathrm{y}}{d\mathrm{x}}=8\mathrm{Cx} \\ \Rightarrow \tan \left(\mathrm{\theta }\right)=8\mathrm{Cx}\left(1\right) \end{gathered}$$

Now resolving the forces on a bead we get,

$\mathrm{Ncos}\left(\mathrm{\theta }\right)=\mathrm{mg}(\mathrm{N}=\mathrm{normal}\mathrm{force})$

And also:

$\mathrm{Nsin}\left(\mathrm{\theta }\right)={\mathrm{m\omega }}^2\mathrm{a}(\because \mathrm{x}=\mathrm{a})$

Now dividing both the forces, we get

$$\begin{gathered} \frac{\mathrm{Nsin}\left(\mathrm{\theta }\right)}{\mathrm{Ncos}\left(\mathrm{\theta }\right)}=\frac{{\mathrm{m\omega }}^2\mathrm{a}}{\mathrm{mg}} \\ \Rightarrow \tan \left(\mathrm{\theta }\right)=\frac{{\mathrm{\omega }}^2\mathrm{a}}{\mathrm{g}} \\ \Rightarrow 8\mathrm{Cx}=\frac{{\mathrm{\omega }}^2\mathrm{a}}{\mathrm{g}} \\ \Rightarrow 8\mathrm{Ca}=\frac{{\mathrm{\omega }}^2\mathrm{a}}{\mathrm{g}}(\because \mathrm{x}=\mathrm{a}) \\ \Rightarrow {\mathrm{\omega }}^2=8\mathrm{Cg} \\ \Rightarrow \mathrm{\omega }=2\sqrt{2\mathrm{Cg}} \end{gathered}$$

Therefore the angular speed is equal to $2\sqrt{2\mathrm{Cg}}$

Hence, option(D) is the correct answer.