A small 5.00 kg brick is released from rest 3.00 m above a horizontal seesaw on a fulcrum at its center, with a radius of 1.60 m. Find the angular momentum of this brick about a horizontal axis through the fulcrum and perpendicular to the plane of the figure the instant the brick is released. Then, find the angular momentum of this brick about a horizontal axis through the fulcrum and perpendicular to the plane of the figure the instant before it strikes the seesaw.
What progress and effort have you made on this? Is the brick really released above or on the fulcrum, as the problem states? The instant the brick is released, there is no momentum.
To find the angular momentum of the brick about a horizontal axis through the fulcrum and perpendicular to the plane of the figure, we need to consider the conservation of angular momentum.
1. At the instant the brick is released:
- The brick is initially at rest, so its initial angular momentum is zero.
- The angular momentum is given by the product of the rotational inertia (I) and the angular velocity (ω): L = I * ω.
- In this case, the brick is a point mass and its rotational inertia is given by I = m * r^2, where m is the mass of the brick and r is the distance from the fulcrum (radius).
- So, the angular momentum at the instant the brick is released is L1 = m * r^2 * ω1, where ω1 is the initial angular velocity.
2. Instant before the brick strikes the seesaw:
- To solve this, we need to calculate the final angular velocity of the brick right before it strikes the seesaw.
- Since the brick is falling vertically, it will gain linear velocity due to gravitational acceleration (g).
- The linear velocity at the instant before striking the seesaw can be found using the equation v = √(2 * g * h), where h is the vertical distance fallen.
- The final angular velocity can be calculated using the relation v = ω * r, where r is the radius.
- So, ω2 = v2 / r.
Now we can calculate the angular momentum at the instant before the brick strikes the seesaw:
1. Calculate the linear velocity:
- v2 = √(2 * g * h) = √(2 * 9.8 m/s^2 * 3.00 m) ≈ 7.67 m/s
2. Calculate the angular velocity:
- ω2 = v2 / r = 7.67 m/s / 1.60 m ≈ 4.79 rad/s
3. Calculate the angular momentum:
- L2 = m * r^2 * ω2 = 5.00 kg * (1.60 m)^2 * 4.79 rad/s ≈ 76.93 kg·m^2/s
Therefore:
- The angular momentum of the brick about a horizontal axis through the fulcrum and perpendicular to the plane of the figure at the instant it is released is 0 kg·m^2/s.
- The angular momentum of the brick about the same axis at the instant before it strikes the seesaw is approximately 76.93 kg·m^2/s.