A door d=1.0m wide, Of mass 15 kg, can rotate freely about a vertical axis throu

Question

A door d=1.0m wide, Of mass 15 kg, can rotate freely about a vertical axis through its hinges. A bullet with a mass of 10 g and a speed of 400 n/s strikes the center of the door, in a direction perpendicular to the plane of the door, and embeds itself there. The inertia of the door about the axis going through the hinges is 1/3 Md2. (a) Find the door's angular speed after the strike. (b) What is the total kinetic energy (door and bullet) before and after the strike? A door d=1.0m wide, Of mass 15 kg, can rotate freely about a vertical axis through its hinges. A bullet with a mass of 10 g and a speed of 400 n/s strikes the center of the door, in a direction perpendicular to the plane of the door, and embeds itself there. The inertia of the door about the axis going through the hinges is 1/3 Md2. (a) Find the door's angular speed after the strike. (b) What is the total kinetic energy (door and bullet) before and after the strike? A door d=1.0m wide, Of mass 15 kg, can rotate freely about a vertical axis through its hinges. A bullet with a mass of 10 g and a speed of 400 n/s strikes the center of the door, in a direction perpendicular to the plane of the door, and embeds itself there. The inertia of the door about the axis going through the hinges is 1/3 Md2. (a) Find the door's angular speed after the strike. (b) What is the total kinetic energy (door and bullet) before and after the strike?

Explanation / Answer

(A)Applying angular momentum conservation about the hinge axis,

Li = Lf

(m v0 (d/2)) + ( I x 0) = I w + (m d^2 w )

and I = M d^2 / 3

m v0 d / 2 = M d^2 w / 3 + m d^2 w

0.010 x 400 / 2 = (15/3 + 0.010) (1)(w)

w = 0.40 rad/s ............Ans

(b) Ki = m v0^2 /2

= (0.010)(400^2) / 2 = 800 J

Kf = (I + m d^2) w^2 / 2

= (15/3 + 0.01)(1^2) (0.4^2) / 2

= 0.40 J

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