A bullet with mass m and a velocity of v is shot perpendicular to the edge of a uniform rod with length L and mass 4m which is free to move and rotate on a frictionless horizontal surface. The bullet is lodged into the rod and move together after the collision. a) Calculate the linear velocity of the center of mass of the system (rod-bullet) after the collision. b) Find the position (on the rod) of center of mass of the system (rod-bullet) after the collision. c) Find the moment of inertia of the system around the axis through the center of mass. d) Calculate the angular velocity ω of the system after the collision. (The moment of inertia of a uniform rod about its center of mass is mL2/12.)
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To solve this problem, we can apply the principles of conservation of linear momentum and conservation of angular momentum.
a) To calculate the linear velocity of the center of mass of the system after the collision, we can use the principle of conservation of linear momentum.
Before the collision, the bullet has a momentum of mv, and the rod has a momentum of 0 since it is at rest. After the collision, they move together, so the total momentum of the system is (m + 4m)(v') where v' is the linear velocity of the center of mass of the system.
According to the principle of conservation of linear momentum, the initial momentum is equal to the final momentum:
mv = (m + 4m)(v')
Simplifying the equation, we get:
v' = mv / (m + 4m)
v' = v / 5
Therefore, the linear velocity of the center of mass of the system after the collision is v/5.
b) To find the position of the center of mass of the system after the collision, we can use the principle of conservation of linear momentum from a different perspective.
Before the collision, the bullet is located at one end of the rod, so the center of mass of the system is at a distance of L/2 from the end of the rod. After the collision, the bullet becomes lodged into the rod and they move together.
Let d be the distance from the end of the rod to the new center of mass of the system after the collision. Since the bullet and rod move together, their combined mass is (m + 4m) and the total momentum of the system is (m + 4m)(v').
According to the principle of conservation of linear momentum, the initial momentum is equal to the final momentum:
mv = (m + 4m)(v')
Using this equation, we can solve for d:
mv = (m + 4m)(v')
md = 4m(v') - mv'
md = 4mv' - mv'
d = (4mv' - mv') / m
d = 3v'
Substituting the value of v' we found in part a), we get:
d = 3(v/5)
d = 3v/5
Therefore, the position of the center of mass of the system after the collision is 3v/5 units from the end of the rod.
c) To find the moment of inertia of the system around the axis through the center of mass, we can use the parallel axis theorem.
The moment of inertia of the uniform rod about its center of mass is mL^2/12. Since the bullet is lodged into the rod, the effective mass distribution of the system changes. However, the total mass of the system remains the same, which is 5m.
The parallel axis theorem states that the moment of inertia about any axis parallel to the axis through the center of mass is equal to the moment of inertia about the center of mass plus the product of the total mass and the square of the perpendicular distance between the two axes.
Using this theorem, we can calculate the moment of inertia of the system as follows:
I = (mL^2/12) + (5m)(3v/5)^2
I = mL^2/12 + 15mv^2/25
Therefore, the moment of inertia of the system around the axis through the center of mass is mL^2/12 + 15mv^2/25.
d) To calculate the angular velocity ω of the system after the collision, we can use the principle of conservation of angular momentum.
The initial angular momentum is 0 as the rod is at rest and the bullet is moving perpendicular to its edge.
The final angular momentum is given by the product of the moment of inertia (found in part c)) and the angular velocity (ω) of the system.
Using the principle of conservation of angular momentum, we can set up the equation:
0 = (mL^2/12 + 15mv^2/25) ω
Solving for ω, we get:
ω = 0
Therefore, the angular velocity of the system after the collision is 0.