A triangular wedge of mass m is placed on a frictionless table its inclined face makes an angle $θ$ with the horizontal. A solid cylinder of mass m and radius r rolls down the inclined face without sliding or slipping if $a_{r}$ , ‘a’ denote acceleration of cylinder with respect to wedge and wedge with respect to ground respectively then which of the following are correct.

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$a_{r} \cos θ = 2 a$

$a = \frac{g \sin θ \cos θ}{2 + \sin^{2} θ}$

Normal force between wedge and incline is in less than $m g \cos θ$ .

Minimum coefficient of friction required to support rolling, $μ = \frac{1}{2} \tan θ$ cylinder along the incline is

answer is A, B, C, D.

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Detailed Solution

${(a_{C M})}_{x} = 0 \Rightarrow a_{r} \cos θ - a = a$

$a_{r} \cos θ = 2 a$ …………………….1)
$m g \sin θ - f = m (a_{r} - a \cos θ)$ ……………..2)
$f R = m \frac{R^{2}}{2} α$ …………………..3)
$a_{r} = R α$ ……………..4)
$\Rightarrow f = \frac{m a_{r}}{2} \Rightarrow m g \sin θ + m a \cos θ = \frac{3}{2} m a_{r} = \frac{3}{2} m \frac{2 a}{\cos θ}$
$\Rightarrow m g \sin θ = m a [\frac{3 - \cos^{2} θ}{\cos θ}] \Rightarrow a = \frac{g \sin θ \cos θ}{2 + \sin^{2} θ}$
$m g \sin θ - N = m a \sin θ \Rightarrow N = m g \cos θ [\frac{2}{2 + \sin^{2} θ}]$
$\Rightarrow \frac{m}{2} \frac{2 g \sin θ}{2 + \sin^{2} θ} \leq μ \frac{2 m g \cos θ}{2 + \sin^{2} θ} μ \geq \frac{\tan θ}{2}$