Question #161485

a gun barrel weight 300 kg and has a recoil spring of stiffness 250 Newton per mm the barrel is equal to 0.8m on firing determine i the critical recoil velocity of the gun .

Expert's answer

(i) The natural frequency is

𝜔 = "\\sqrt \\frac{\ud835\udc58 }{\ud835\udc5a }"

= "\\sqrt \\frac{250000}{300}"

=28.87 𝑟𝑎𝑑/𝑠 .

From conservation of energy, we have

"\\frac{ \ud835\udc5a\ud835\udc632}{2}" = "\\frac{\ud835\udc58\ud835\udc652}{2 }"

𝑣 = 𝜔𝑥 = (28.87)(0.8)

=23.096𝑚/𝑠 .

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