Question

A rod is kept making an angle $\theta$ with the vertical as shown, there is a small bead of mass m at a distance r from end A. End A of rod is hinged. The rod starts rotating about end A making an angle $\theta$f with the vertical with an angular acceleration $\alpha$ and makes a cone. If the coefficient of a static friction between the ring and rod is $\mu ,$ find the time after which ring will start slipping on the rod. (consider the system is placed in gravity free space)

• A. $\sqrt{\frac{\mu }{\alpha (si{n}^2\theta -{\mu }^{2}co{s}^2\theta {)}^{\frac{1}{2}}}}$
• B. $\sqrt{\frac{\mu }{\alpha (si{n}^2\theta -\mu cos\theta {)}^{\frac{1}{2}}}}$
• C. $\sqrt{\frac{\mu }{\alpha (si{n}^2\theta -{\mu }^{2}co{s}^2\theta )}}$
• D. $\sqrt{\frac{2\mu }{\alpha (si{n}^2\theta -{\mu }^{2}co{s}^2\theta )}}$

Answer 1

The correct option is B $\sqrt{\frac{\mu }{\alpha (si{n}^2\theta -\mu cos\theta {)}^{\frac{1}{2}}}}$

Mass$=m$

distance$=r$

Angular acceleration$=\alpha$

And makes a cone

The rod starts rotating about end A making an angle $=\theta$

If the coefficient of a static friction between the ring and rod$=\mu$

The time after which ring will start shipping on the rod

With angular acceleration $=\alpha$

And makes a cone

(consider the system is placed in gravity free space)

When the inclined plane acceleration of a body then

$f=g(g\sin \theta -\mu \cos \theta )f=\alpha (\sin \theta -\mu \cos \theta )$

Now putting all the value and we get

$\sqrt{\frac{\mu }{\alpha ({\sin }^2\theta -\mu \cos \theta {)}^{1/2}}}$