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77julia77 [94]
1 year ago
12

A uniform disk of mass 3.25 kg has a radius of 0.100 m and spins with a frequency of 0.550 rev/s. What is its angular momentum?

Physics
1 answer:
shutvik [7]1 year ago
7 0

ANSWER:

0.0562 J

STEP-BY-STEP EXPLANATION:

Angular momentum is expressed in terms of moment of inertia and angular velocity. This is expressed as follows:

L=I\cdot\omega

Here, I is the angular momentum and ω is the angular velocity.

Angular momentum is mass time the square of the radius of the object. Moment of inertia for a uniform disk is given as,

I=\frac{1}{2}m\cdot r^2

Here, m is the mass of the disk and r is the radius of the disk.

Replacing:

L=\frac{1}{2}m\cdot r^2\cdot\omega

Convert the units of angular velocity into rad/s.

\begin{gathered} \omega=0.55\text{ rev/s}\cdot\frac{2\pi\text{ rad}}{1\text{ rev}} \\ \omega=3.46\text{ rad/s} \end{gathered}

We replace each data to calculate the angular momentum:

\begin{gathered} L=\frac{1}{2}\cdot3.25\cdot0.1^2\cdot3.46 \\ L=0.0562 \end{gathered}

The angular momentum of the uniform disk is 0.0562 J

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The velocity of the block when t == 2 s is 60.7 ft./sec.

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The velocity of the block as a function of t can be determined by

integrating dv = adt using the initial condition v = 4 ft./s at t = 0.

\int_{ 4 ft/s }^{ v } dv = \int_0^t 3.22(8t^2 – 2)dt \\ \begin{aligned} v – &4 = 3.22 (\frac 8 3 t^3 – 2t) \\ & v = \{8.5867t^3 – 6.44t + 4 \} ft/s \end{aligned}

The displacement as a function of t can be determined by integrating

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the initial condition s = 0 at t = 0

\int_0^s ds = \int_0^t (8.5867t^3 – 6.44t + 4)dt \\ s = \{2.1467t^4 – 3.22t^2 + 4t \} ft

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s = 30 ft.

Thus, at s = 30 ft.,

\begin{aligned} v &= 8.5867(2.0089^3) – 6.44(2.0089) + 4 \\ &= 60.67 ft/s \\ &= 60.7 ft/s \end{aligned}

Kinematics is a subfield of physics, developed in classical mechanics, that describes the motion of points, bodies (objects), and systems of bodies (groups of objects) without considering the forces that cause them to move.

Kinematics, as a field of study, is often referred to as the "geometry of motion" and is occasionally seen as a branch of mathematics. A kinematics problem begins by describing the geometry of the system and declaring the initial conditions of any known values of position, velocity and/or acceleration of points within the system.

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Scenario
Anvisha [2.4K]

Answer:

1) t = 23.26 s,  x = 8527 m, 2)   t = 97.145 s,  v₀ = 6.4 m / s

Explanation:

1) First Scenario.

After reading your extensive problem, we are going to solve it, for this exercise we must use the parabolic motion relationships. Let's carry out an analysis of the situation, for deliveries the planes fly horizontally and we assume that the wind speed is zero or very small.

Before starting, let's reduce the magnitudes to the SI system

         v₀ = 250 miles/h (5280 ft / 1 mile) (1h / 3600s) = 366.67 ft/s

         y = 2650 m

Let's start by looking for the time it takes for the load to reach the ground.

         y = y₀ + v_{oy} t - ½ g t²

in this case when it reaches the ground its height is zero and as the plane flies horizontally the vertical speed is zero

         0 = y₀ + 0 - ½ g t2

          t = \sqrt{ \frac{2y_o}{g} }

          t = √(2 2650/9.8)

          t = 23.26 s

this is the horizontal scrolling time

          x = v₀ t

          x = 366.67  23.26

          x = 8527 m

the speed at the point of arrival is

         v_y = v_{oy} - g t = 0 - gt

         v_y = - 9.8 23.26

         v_y = -227.95 m / s

Module and angle form

        v = \sqrt{v_x^2 + v_y^2}

         v = √(366.67² + 227.95²)

        v = 431.75 m / s

         θ = tan⁻¹ (v_y / vₓ)

         θ = tan⁻¹ (227.95 / 366.67)

         θ = - 31.97º

measured clockwise from x axis

We see that there must be a mechanism to reduce this speed and the merchandise is not damaged.

2) second scenario. A catapult located at the position x₀ = -400m y₀ = -50m with a launch angle of θ = 50º

we look for the components of speed

           cos θ = v₀ₓ / v₀

           sin θ = v_{oy} / v₀

            v₀ₓ = v₀ cos θ

            v_{oy} = v₀ sin θ

we look for the time for the arrival point that has coordinates x = 0, y = 0

            y = y₀ + v_{oy} t - ½ g t²

            0 = y₀ + vo sin θ t - ½ g t²

            0 = -50 + vo sin 50 t - ½ 9.8 t²

            x = x₀ + v₀ₓ t

            0 = x₀ + vo cos θ t

            0 = -400 + vo cos 50 t

podemos ver que tenemos un sistema de dos ecuación con dos incógnitas

          50 = 0,766 vo t – 4,9 t²

          400 =   0,643 vo t

resolved

          50 = 0,766 ( \frac{400}{0.643 \ t}) t – 4,9 t²

          50 = 476,52 t – 4,9 t²

          t² – 97,25 t + 10,2 = 0

we solve the quadratic equation

         t = [97.25 ± \sqrt{97.25^2 - 4 \ 10.2}] / 2

         t = 97.25 ±97.04] 2

         t₁ = 97.145 s

         t₂ = 0.1 s≈0

the correct time is t1 the other time is the time to the launch point,

         t = 97.145 s

let's find the initial velocity

         x = x₀ + v₀ cos 50 t

         0 = -400 + v₀ cos 50 97.145

         v₀ = 400 / 62.44

         v₀ = 6.4 m / s

5 0
3 years ago
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