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

`\omega =0.33\ rad/s`

Step-by-step 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))* \omega`

`I_(door) = (1)/(3)MW^2`

`I_(door) = (1)/(3)* 44 * 1^2`

`I_(door) =14.67\ kg.m^2`

`I_(clay) = mr^2`

`I_(clay) = 0.7* 0.5^2`

`I_(clay) =0.175\ kg.m^2`

`0.7* 14* (1)/(2) = (14.67+0.175)* \omega`

`\omega =(4.9)/(14.845)`

`\omega =0.33\ rad/s`