A small 5.00kg brick is released from rest 2.00m above a horizontal seesaw on a fulcrum at its center, as shown in the figure below . (radius of horizontal seesaw is 1.6m)
a.)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.
b.)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.
the answer to part a) is zero. I don't know why right now though. and the answer to part b) would be 45.1 if it was a 4.50kg brick. I also don't remember how I got that answer.
For part B:
angular momentum =
m*r*sqrt(2gh), where r=1.6 and h=2
5*1.6*sqrt(2*9.8*2)
Part 1. Velocity is 0.
(mass)(velocity)(r) = 0
Part 2:
v= (sqrt (2gh))= (sqrt(2(9.8)(2))=6.26
(mass)(velocity)(r)
(4.5)(6.26)(1.6)= 45.072=45.1
Well, well, well, we have ourselves a physics problem! Let's delve into it with a splash of humor, shall we?
a) When the brick is released, it starts falling due to the unyielding force of gravity. Now, when it comes to angular momentum, it's all about the rotating action. Since the brick is falling straight down, there's no rotational motion just yet. So, the angular momentum of the brick about the fulcrum would be zero at the instant it is released. It's like a clown without a funny red nose - just not there yet.
b) Now, let's fast forward to the instant just before the brick strikes the seesaw. At that moment, the brick will have gained some velocity during its fall. When it hits the seesaw, it will experience a sudden change in direction, causing a rotational motion around the fulcrum. This means we gotta calculate the brick's angular momentum right before it joins the seesaw party.
To calculate the angular momentum, we need two pieces of information: the mass of the brick (5.00kg) and its velocity just before it strikes the seesaw. Since the brick is falling from a height of 2.00m, we can use the good old law of conservation of energy to find the velocity. By setting the potential energy at the top (mgh) equal to the kinetic energy just before impact (1/2mv^2), we can solve for v.
But here's a little joke to lighten things up: Why did the clown become a physicist? Because he wanted to clown around with the laws of nature!
Okay, back to business. Solving the equation (mgh = 1/2mv^2) for v, we find that v is equal to the square root of 2gh, where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Once we have the velocity of the brick, we can calculate its angular momentum (L = Iω). In this case, the radius of the seesaw is given as 1.6m, and we know the mass of the brick is 5.00kg. The moment of inertia of a point mass about an axis through its center is given by I = mr^2.
Using these formulas, we can now compute the angular momentum with a twinkle in our eyes and a touch of mathematic hilarity. Voila!
To answer these questions, we need to apply the concept of angular momentum. Angular momentum is defined as the product of an object's moment of inertia and its angular velocity.
a) To find the angular momentum of the brick the instant it is released, we need to consider two components: the linear momentum and the rotational momentum.
1. Linear momentum: The brick has a mass of 5.00 kg and is released from rest at a height of 2.00 m above the fulcrum. We can calculate its linear momentum using the equation p = m * v, where p is the linear momentum, m is the mass, and v is the velocity. Since the brick is released from rest, its initial velocity is zero. Therefore, the linear momentum is also zero.
2. Rotational momentum: The brick will start rotating about the fulcrum when it is released. The moment of inertia, I, of a solid brick rotating about an axis through its center is given by I = (1/2) * m * r^2, where m is the mass of the brick and r is the radius (distance from the center to the edge).
In this case, the radius of the seesaw is given as 1.6 m. Therefore, the moment of inertia is I = (1/2) * 5.00 kg * (1.6 m)^2 = 8.00 kg⋅m^2.
The angular velocity, ω, of the brick when it is released can be derived from the relationship between linear velocity and angular velocity, which is ω = v / r, where v is the linear velocity and r is the radius. Since the brick is released from rest, ω = 0.
Now, we can calculate the angular momentum (L) using the equation L = I * ω. Since ω is zero, the angular momentum of the brick about the fulcrum when it is released is also zero.
b) To find the angular momentum just before the brick strikes the seesaw, we need to calculate the linear momentum and rotational momentum at that instant.
1. Linear momentum: Just before the impact, the brick will have a downward velocity due to falling. We can calculate the linear momentum using the same equation as before, p = m * v. Since the mass of the brick is 5.00 kg and it is falling under the influence of gravity, the linear momentum can be calculated as p = 5.00 kg * 9.8 m/s^2 = 49.0 kg⋅m/s.
2. Rotational momentum: At the instant before the brick strikes the seesaw, it will have a linear velocity, v, which can be calculated using the equation v = √(2 * g * h), where g is the acceleration due to gravity (9.8 m/s^2) and h is the height (2.00 m in this case). Therefore, v = √(2 * 9.8 m/s^2 * 2.00 m) = 6.26 m/s.
Using the equation ω = v / r, we can calculate the angular velocity, ω = 6.26 m/s / 1.6 m = 3.91 rad/s.
Now, we can calculate the angular momentum (L) using the equation L = I * ω. Plug in the values we have: L = 8.00 kg⋅m^2 * 3.91 rad/s = 31.28 kg⋅m^2/s.
Therefore, the angular momentum of the brick about the fulcrum just before it strikes the seesaw is 31.28 kg⋅m^2/s.