Question

A block of mass m is kept on a horizontal ruler. The friction coefficient between the ruler and the block is μ. The ruler is fixed at one end and the block is at a distance L from the fixed end. The ruler is rotated about the fixed end in the horizontal plane through the fixed end. If the angular speed of the ruler is uniformly increased from zero at an angular acceleration α, at what angular speed will the block slip?

• A.
• B.
• C.
• D. None of these

Answer 1

The correct option is C

It has a radial acceleration of ${\omega }^2$ L and a tangential acceleration of $\alpha$ L.

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$\therefore$ Its magnitude of net acceleration ,$a_{net}=\sqrt{{\omega }^4{L}^2+{\alpha }^2{L}^2}$

As friction is the force that provides this acceleration,

$f=m\sqrt{{\omega }^4{L}^2+{\alpha }^2{L}^2}$

Now, for more value of $\omega$ , f = $\mu N$

$\mu N=m\sqrt{{\omega }^4{L}^2+{\alpha }^2{L}^2}$
$\Rightarrow \mu mg=m\sqrt{{\omega }^4{L}^2+{\alpha }^2{L}^2}$
$\Rightarrow \mu g=\sqrt{{\omega }^4{L}^2+{\alpha }^2{L}^2}$
$\Rightarrow {\mu }^2{g}^2={\omega }^4{L}^2+{\alpha }^2{L}^2$
$\Rightarrow \omega ={[{\left(\frac{\mu g}{L}\right)}^2-{\alpha }^2]}^{\frac{1}{4}}$