Question

A solid wood door 1.00m wide and 2.00m high is hinged along one side and has a total mass of 43.0kg. Initially open and at rest, the door is struck at its center by a handful of sticky mud with mass 0.700kg, traveling perpendicular to the door at 13.0m/s just before impact.
A) Find the angular speed of the door.
B) Does the mud make a significant contribution to the moment of inertia?

Answer 1

Answer:

$\omega_f = 0.314\ rad/s$

Explanation:

given,

door dimension  = 1 m x 2 m

mass of = 43 Kg

Mass of dust = 0.7 Kg

speed of the door = 13 m/s

$I_{total} =I_{door} + I_{mud}$

$I_{total} =\dfrac{1}{3}MW^2 + m(\dfrac{W}{2})^2$

a) from conservation of angular momentum  

$L_i = L_f$

$mv\dfrac{W}{2} = I_{total}\omega_f$

$mv\dfrac{W}{2}= (\dfrac{1}{3}MW^2 + m(\dfrac{W}{2})^2)\omega_f$

$\omega_f=\dfrac{mv\dfrac{W}{2}}{\dfrac{1}{3}MW^2 + m(\dfrac{W}{2})^2}$

$\omega_f= \dfrac{\dfrac{mv}{2}}{\dfrac{MW }{3}+(\dfrac{mW}{4})}$

$\omega_f= \dfrac{\dfrac{0.7\times 13}{2}}{\dfrac{43\times 1 }{3}+(\dfrac{0.7\times 1}{4})}$

$\omega_f = 0.314\ rad/s$

b) angular speed without considering mud

$\omega_f= \dfrac{\dfrac{mv}{2}}{\dfrac{MW }{3}+(\dfrac{mW}{4})}$

$\omega_f= \dfrac{\dfrac{0.7\times 13}{2}}{\dfrac{43\times 1 }{3}}$

$\omega_f = 0.317\ rad/s$

there is no significant contribution of mud in moment of inertia .