Question

A solid wood door (???? = 1 3 ????(width) 2 ) 1.00 m wide and 2.00 m high is hinged along one side and has a total mass of 44.0 kg. Initially open and at rest, the door is struck at its center by a handful of sticky mud with mass 0.700 kg , traveling perpendicular to the door at 14.0 m/s just before impact. Find the final angular speed of the door in (rad/s).

Answer 1

Answer:

$\omega =0.33\ rad/s$

Explanation:

given,

size of door = 1 m x 2 m

total mass = 44 Kg

mud mass attached = 0.700 Kg

speed of door = 14 m/s

angular speed of door = ?

conservation of angular momentum

initial angular momentum is equal to final momentum

        $L_i = L_f$

        $mvr = (I_{door}+I_{clay})\times \omega$

$I_{door} = \dfrac{1}{3}MW^2$

$I_{door} = \dfrac{1}{3}\times 44 \times 1^2$

$I_{door} =14.67\ kg.m^2$

$I_{clay} = mr^2$

$I_{clay} = 0.7\times 0.5^2$

$I_{clay} =0.175\ kg.m^2$

        $0.7\times 14\times \dfrac{1}{2} = (14.67+0.175)\times \omega$

        $\omega =\dfrac{4.9}{14.845}$

        $\omega =0.33\ rad/s$