A 2.0 g bullet is shot with an initial velocity of 200.0 m/s into a board (I = 12.0 kg m2). The board is held still by a frictionless pivot a third away from its bottom end, as shown below. Assume that you can treat the board as a uniform rod.
1. What is the initial angular momentum of the bullet with respect to the axle?
2. What is the distance between the axis of rotation and the center of mass of the board?
3.If the board is initially uniform, what is its mass?
4. Once the bullet enters the board, what is its contribution to the moment of inertia?
5. What is the angular velocity of the board + bullet system after the collision?
1. To determine the initial angular momentum of the bullet with respect to the axle, we need to know the moment of inertia and the angular velocity. The formula for angular momentum is L = Iω, where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.
Since we are given the initial velocity of the bullet but not the angular velocity, we need to calculate the latter. The angular velocity can be found using the equation v = ωr, where v is the linear velocity, ω is the angular velocity, and r is the distance from the axis of rotation (pivot) to the point of interest. In this case, the distance from the pivot to the center of mass of the bullet is the point of interest.
Given:
- Linear velocity of the bullet (v) = 200.0 m/s
- Distance from pivot to bullet center of mass (r) = 1/3 of the length of the board
By substituting the values into the equation, we can solve for the angular velocity (ω):
200.0 m/s = ω * (1/3 of the length of the board)
Once we determine ω, we can calculate the initial angular momentum (L) using the formula: L = Iω, where I is the moment of inertia of the bullet.
2. To find the distance between the axis of rotation and the center of mass of the board, we need to know the length of the board. The distance is equal to 2/3 of the length of the board since the pivot is a third away from the bottom end and the center of mass is at the middle.
3. If the board is initially uniform, its mass can be calculated using the equation for the moment of inertia of a uniform rod: I = (1/3) * m * L^2, where I is the moment of inertia, m is the mass, and L is the length of the board. Rearranging the equation, we can solve for the mass (m) when given the moment of inertia (I) and the length (L).
4. Once the bullet enters the board, its contribution to the moment of inertia can be calculated using the formula for the moment of inertia of a point mass: I_bullet = m_bullet * r^2, where I_bullet is the moment of inertia of the bullet, m_bullet is the mass of the bullet, and r is the distance from the axis of rotation to the center of mass of the bullet.
5. To find the angular velocity of the board + bullet system after the collision, we need to consider the conservation of angular momentum. The initial angular momentum of the system (consisting of the board and the bullet) is equal to the final angular momentum of the system. We can calculate the final angular momentum using the formula L = I_total * ω_f, where I_total is the total moment of inertia of the system, and ω_f is the final angular velocity. Rearranging the equation, we can solve for ω_f.