Question

. Стержень длиной 1,0 м и массой 7 кг может свободно вращаться вокруг горизонтальной оси, проходящей через его верхний конец. В нижний конец стержня ударяет пуля массой m = 10,0 г, летящая в горизонтальном направлении со скоростью V0 = 500 м/с, и пробивает его. С какой угловой скоростью ω начнет двигаться стержень сразу после удара, если пуля вылетает из стержня со скоростью V1 = 200 м/с?

Answer 1

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The problem can be solved by using the conservation of angular momentum. Before the collision, the angular momentum of the system (the rod and the bullet) is zero, since both are moving horizontally. After the collision, the angular momentum of the system is equal to the sum of the angular momenta of the rod and the bullet. The rod has an angular momentum of M*L^2*ω/3, where M is the mass of the rod, L is the length of the rod, and ω is the angular velocity of the rod. The bullet has an angular momentum of m*(L/2)*(V1 - ω*L/2), where m is the mass of the bullet, V1 is the velocity of the bullet after the collision, and ω is the same angular velocity of the rod. Therefore, we can write:

0 = M*L^2*ω/3 + m*(L/2)*(V1 - ω*L/2)

Solving for ω, we get:

ω = 6*m*V1/(M*L + 3*m*L)

Substituting the given values, we get:

ω = 6*0.01*200/(7*1 + 3*0.01*1)

ω = 1.71 rad/s

Therefore, the angular velocity of the rod after the collision is 1.71 rad/s.