A square plate of side length 2m has a groove made in the shape of two quarter circles joining at the centre of the plate. The plate is free to rotate about vertical axis passing through its centre. The moment of inertia of the plate about this axis is $4 kg - m^{2}$ A small block of mass 1 kg enters the groove at end A travelling with a velocity of 2m/s parallel to the side of the square plate. The block move along the frictionless groove of the horizontal plate and comes out at the other end B with speed V. Find V assuming that width of the groove is negligible

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$\frac{6}{5} m / s$

$\frac{5}{6} m / s$

$\frac{4}{5} m / s$

$\frac{5}{4} m / s$

answer is A.

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Detailed Solution

Let $ω$ = = angular speed of square plate when the block moves out Angular momentum conservation about rotation axis of the plate gives
$1 \times 2 \times 1 = - 1 \times V \times 1 + 4 ω \Rightarrow ω = \frac{V + 2}{4} \dots \dots \dots \dots \dots (i)$
Energy conservation
$\frac{1}{2} \times 1 \times 2^{2} = \frac{1}{2} \times 1 \times V^{2} + \frac{1}{2} \times 4 \times ω^{2} V^{2} + 4 ω^{2} = 4$
From (I) $V^{2} + 4 {(\frac{V + 2}{4})}^{2} = 4$
$5 V^{2} + 4 V - 12 = 0 \Rightarrow V = \frac{6}{5} m / s$