A projectile of mass m is launched with an initial velocity vector v i making an angle θ with the horizontal as shown below. The
projectile moves in the gravitational field of the Earth. Find the angular momentum of the projectile about the origin when the particle is at the following locations. (Use the following as necessary: vi, θ, m, and g for the acceleration due to gravity.) (a) at the origin
L with arrow =
rmvcos(θ)
Incorrect: Your answer is incorrect.
(b) at the highest point of its trajectory
L with arrow =
The angular momentum of the projectile about the origin when the particle is at the following locations is zero
<h3>Explanation:
</h3>
A projectile of mass m is launched with an initial velocity making an angle θ with the horizontal as shown below (attached). The projectile moves in the gravitational field of the Earth. The angular momentum of the projectile about the origin when the particle is at the following locations. (Use the following as necessary: vi, θ, m, and g for the acceleration due to gravity.)
Angular momentum is zero because there is no r. Since the angular momentum depends on then cross product of the position vector and momentum vector which is in the same direction as velocity, therefore the formula for the angular momentum can be written as , where the cross product component of r vector along/parallel to velocity/momentum is 0.
Angular momentum is given by the length of the arm to the object, multiplied by the momentum of the object, times the cosine of the angle that the momentum vector makes with the arm. From your illustration, that will be: <span>L = R * m * vi * cos(90 - theta) </span>
<span>cos(90 - theta) is just sin(theta) </span> <span>and R is the distance the projectile traveled, which is vi^2 * sin(2*theta) / g </span>
<span>so, we have: L = vi^2 * sin(2*theta) * m * vi * sin(theta) / g </span>
<span>We can combine the two vi terms and get: </span>
<span>L = vi^3 * m * sin(theta) * sin(2*theta) / g </span>
<span>What's interesting is that angular momentum varies with the *cube* of the initial velocity. This is because, not only does increased velocity increase the translational momentum of the projectile, but it increase the *moment arm*, too. Also note that there might be a trig identity which lets you combine the two sin() terms, but nothing jumps out at me right at the moment. </span>
<span>Now, for the first part... </span>
<span>There are a few ways to attack this. Basically, you have to find the angle from the origin to the apogee (highest point) in the arc. Once we have that, we'll know what angle the momentum vector makes with the moment-arm because, at the apogee, we know that all of the motion is *horizontal*. </span>
<span>Okay, so let's get back to what we know: </span>
<span>L = d * m * v * cos(phi) </span>
<span>where d is the distance (length to the arm), m is mass, v is velocity, and phi is the angle the velocity vector makes with the arm. Let's take these one by one... </span>
<span>m is still m. </span> <span>v is going to be the *hoizontal* component of the initial velocity (all the vertical component got eliminated by the acceleration of gravity). So, v = vi * cos(theta) </span> <span>d is going to be half of our distance R in part two (because, ignoring friction, the path of the projectile is a perfect parabola). So, d = vi^2 * sin(2*theta) / 2g </span>
<span>That leaves us with phi, the angle the horizontal velocity vector makes with the moment arm. To find *that*, we need to know what the angle from the origin to the apogee is. We can find *that* by taking the arc-tangent of the slope, if we know that. Well, we know the "run" part of the slope (it's our "d" term), but not the rise. </span>
<span>The easy way to get the rise is by using conservation of energy. At the apogee, all of the *vertical* kinetic energy at the time of launch (1/2 * m * (vi * sin(theta))^2 ) has been turned into gravitational potential energy ( m * g * h ). Setting these equal, diving out the "m" and dividing "g" to the other side, we get: </span>
<span>h = 1/2 * (vi * sin(theta))^2 / g </span>
<span>So, there's the rise. So, our *slope* is rise/run, so </span>
<span>slope = [ 1/2 * (vi * sin(theta))^2 / g ] / [ vi^2 * sin(2*theta) / g ] </span>
<span>The "g"s cancel. Astoundingly the "vi"s cancel, too. So, we get: </span>
<span>(It's not too alarming that slope-at-apogee doesn't depend upon vi, since that only determines the "magnitude" of the arc, but not it's shape. Whether the overall flight of this thing is an inch or a mile, the arc "looks" the same). </span>
<span>Okay, so... using our double-angle trig identities, we know that sin(2*theta) = 2*sin(theta)*cos(theta), so... </span>
<span>Okay, so the *angle* (which I'll call "alpha") that this slope makes with the x-axis is just: arctan(slope), so... </span>
<span>alpha = arctan( tan(theta) / 4 ) </span>
<span>Alright... last bit. We need "phi", the angle the (now-horizontal) momentum vector makes with that slope. Draw it on paper and you'll see that phi = 180 - alpha </span>
Multiply each side by 0.09 m³ : (4,000 kg/m³) x (0.09 m³) = mass
mass = 360 kg .
Force of gravity = (mass) x (acceleration of gravity)
= (360 kg) x (9.8 m/s²)
= (360 x 9.8) kg-m/s²
= 3,528 newtons .
That's the force of gravity on this block, and it doesn't matter what else is around it. It could be in a box on the shelf or at the bottom of a swimming pool . . . it's weight is 3,528 newtons (about 793.7 pounds).
Now, it won't seem that heavy when it's in the water, because there's another force acting on it in the upward direction, against gravity. That's the buoyant force due to the displaced water.
The block is displacing 0.09 m³ of water. Water has 1,000 kg of mass in a m³, so the block displaces 90 kg of water. The weight of that water is (90) x (9.8) = 882 newtons (about 198.4 pounds), and that force tries to hold the block up, against gravity.
So while it's in the water, the block seems to weigh
But again ... it's not correct to call that the "force of gravity acting on the block in water". The force of gravity doesn't change, but there's another force, working against gravity, in the water.
The natural factor that is most likely to support the formation of an oceanic island is the rise of magma from the seafloor. Oceanic islands are also otherwise known as volcanic islands. When volcanoes erupt, they create layers of lava that break the surface of the water. When the tops of the volcanoes emerge, an island is created.<span> </span>