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viktelen [127]
3 years ago
10

A thin rod (length = 2.97 m) is oriented vertically, with its bottom end attached to the floor by means of a frictionless hinge.

The mass of the rod may be ignored, compared to the mass of the object fixed to the top of the rod. The rod, starting from rest, tips over and rotates downward. (a) What is the angular speed of the rod just before it strikes the floor? (Hint: Consider using the principle of conservation of mechanical energy.)(b) What is the magnitude of the angular acceleration of the rod just before it strikes the floor?
Physics
1 answer:
nlexa [21]3 years ago
3 0

Answer:

a)  w = 2.57 rad / s , b)   α = 3.3 rad / s²

Explanation:

a) Let's use the conservation of mechanical energy, we will write it in two points the highest and when touching the ground

Initial. Higher

       Em₀ = U = m g h

Final. Touching the ground

       Em_{f} = K = ½ I w²

How energy is conserved

       Em₀ = Em_{f}

       mg h = ½ I w2

The moment of specific object inertia

        I = m L²

We replace

       m g h = ½ (mL²) w²

       w² = 2g h / L²

The height of the object is the length of the bar

        h = L

        w = √ 2g / L

       w = √ (2 9.8 / 2.97)

       w = 2.57 rad / s

b) the angular acceleration can be found from Newton's second rotational law

       τ = I α

       W L = I α

       mg L = (m L²) α

       α = g / L

       α = 9.8 / 2.97

       α = 3.3 rad / s²

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The work-energy principle states that the work done by all the __________ forces acting on an object (or system of objects) caus
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The work-energy principle states that the work done by all the non-conservative forces acting on an object (or system of objects) causes a change in the total mechanical energy of the object or system.


What is the work-energy principle?
The work-energy principle states that the total work done on a system is equal to the change in kinetic energy of the system. It is given as:

W.D = ΔK.E

       = K.E₁ - K.E₂

where K.E₁ is the initial kinetic energy of the system

            K.E₂ is the final kinetic energy of the system



What is meant by non-conservative forces?

Non-conservative forces as the name suggests are not conserved i.e. these forces cause a loss of mechanical energy from the system. A prime example of non-conservative forces is friction.

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Learn more about work and energy here:

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2 years ago
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4 years ago
Object is dropped from the top of a building that is 1125 m tall at the point when it’s speed is 30 m/s how far has it fallen
FrozenT [24]

-- Gravity makes a falling object fall 9.8 m/s faster every second.

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-- The distance an object falls from rest is D = 1/2 (acceleration) (time)²

D = 1/2 (9.8 m/s²) (3.06 sec)²

D = (4.9 m/s²) (9.37 sec²)

<em>D = 45.8 meters</em>

Notice that we don't care how high the building is.  The problem works just as long as the object can reach 30 m/s before it hits the ground.  That  turns out to be anything higher than 45.8 meters for the drop . . . maybe something like 13 floors or more.

Now I'll go a little farther for you !  Writing the last paragraph made me a little curious and uncomfortable.  So I went and looked up the world's tallest buildings . . . and I found out that this problem could never happen !

The tallest building in the world now is the Burj Khalifa, in  Dubai.  It has 163 floors, and it's 828 meters high !  That's 2,717 feet.  It's gonna be a long time before there's a building that's 1125 meters tall, like this problem says.  That's close to 3700 feet . . . I've had flying lessons where I wasn't that far off the ground !

4 0
3 years ago
A transverse sinusoidal wave is moving along a string in the positive direction of an x axis with a speed of 89 m/s. At t = 0, t
earnstyle [38]

Answer:

The answer is below

Explanation:

Given that:

y = transverse displacement = 4.2 cm = 0.042 m at x = 0 and t = 0.

Speed = v = 89 m/s, maximum transverse speed of the string particle = u_m = 16 m/s.

ω = u_m / y_m = 16 / 0.42 = 380.95 rad/s

a) Frequency = ω/2π = 380.95 / 2π = 60.63 Hz

b) Wavelength (λ) = speed / frequency

λ = v / f = 89/63.66= 1.468 m

c) Using the wave equation:

y=y_msin(kx \pm wt \pm \phi)\\y=0.042,t=0,x=0\\\\Hence\\y_m=0.042\ m

d) Wave number k is given by:

k = 2π / λ = 2π / 1.468 = 4.28 rad/s

e) The angular velocity is given by:

ω = u_m / y_m = 16 / 0.42 = 380.95 rad/s

f)  Using the wave equation:

y=y_msin(kx \pm wt \pm \phi)\\\\y=0.042,t=0,x=0,y_m=0.042\\\\Hence\\0.042=0.042sin(4.28(0)\pm 380.95(0)\pm \phi)\\\\sin\phi=1\\\\\phi=\frac{\pi}{2} \\\\y=0.042sin(4.28x\pm 380.95t\pm \frac{\pi}{2})

g) Since the wave is in the positive x direction, hence ω is negative

4 0
3 years ago
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