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?

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wagonbelleville wagonbelleville · Answer 1 · 2019-12-06T07:58:45+01:00

Answer:

[tex]\omega_f = 0.314\ rad/s[/tex]

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

[tex]I_{total} =I_{door} + I_{mud}[/tex]

[tex]I_{total} =

\dfrac{1}{3}MW^2 + m(\dfrac{W}{2})^2[/tex]

a) from conservation of angular momentum

[tex]L_i = L_f[/tex]

[tex] mv\dfrac{W}{2} = I_{total}\omega_f[/tex]

[tex] mv\dfrac{W}{2}= (\dfrac{1}{3}MW^2 + m(\dfrac{W}{2})^2)\omega_f[/tex]

[tex]\omega_f=\dfrac{mv\dfrac{W}{2}}{\dfrac{1}{3}MW^2 + m(\dfrac{W}{2})^2}[/tex]

[tex]\omega_f= \dfrac{\dfrac{mv}{2}}{\dfrac{MW }{3}+(\dfrac{mW}{4})}[/tex]

[tex]\omega_f= \dfrac{\dfrac{0.7\times 13}{2}}{\dfrac{43\times 1 }{3}+(\dfrac{0.7\times 1}{4})}[/tex]

[tex]\omega_f = 0.314\ rad/s[/tex]

b) angular speed without considering mud

[tex]\omega_f= \dfrac{\dfrac{mv}{2}}{\dfrac{MW }{3}+(\dfrac{mW}{4})}[/tex]

[tex]\omega_f= \dfrac{\dfrac{0.7\times 13}{2}}{\dfrac{43\times 1 }{3}}[/tex]

[tex]\omega_f = 0.317\ rad/s[/tex]

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