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anzhelika [568]
2 years ago
10

A small sphere with mass m is attached to a massless rod of length L that is pivoted at the top, forming a simple pendulum. The

pendulum is pulled to one side so that the rod is at an angle θ from the vertical, and released from rest.
a. Show the pendulum just after it is released. Draw vectors representing the forces acting on the small sphere and the acceleration of the sphere. Accuracy counts! At this point, what is the linear acceleration of the sphere?
b. Repeat part (a) for the instant when the pendulum rod is at an angle 9/2 from the vertical.
c. Repeat part (a) for the instant when the pendulum rod is vertical. At this point, what is the linear speed of the sphere?

Physics
1 answer:
USPshnik [31]2 years ago
4 0

Answer:

a) see attached, a = g sin θ

b)

c)   v = √(2gL (1-cos θ))

Explanation:

In the attached we can see the forces on the sphere, which are the attention of the bar that is perpendicular to the movement and the weight of the sphere that is vertical at all times. To solve this problem, a reference system is created with one axis parallel to the bar and the other perpendicular to the rod, the weight of decomposing in this reference system and the linear acceleration is given by

          Wₓ = m a

          W sin θ = m a

          a = g sin θ

b) The diagram is the same, the only thing that changes is the angle that is less

                θ' = 9/2  θ

             

c) At this point the weight and the force of the bar are in the same line of action, so that at linear acceleration it is zero, even when the pendulum has velocity v, so it follows its path.

The easiest way to find linear speed is to use conservation of energy

Highest point

            Em₀ = mg h = mg L (1-cos tea)

Lowest point

          Emf = K = ½ m v²

          Em₀ = Emf

          g L (1-cos θ) = v² / 2

              v = √(2gL (1-cos θ))

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A roller coaster travels 200 feet horizontally and then rises 135 feet at an angle of 30 degrees above the ground. What is the m
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8 0
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You know that you sound better when you sing in the shower. This has to do with the amplification of frequencies that correspond
luda_lava [24]

Answer:

a) L = 0.75m   f₁ = 113.33 Hz , f₃ = 340 Hz, b) L=1.50m   f₁ = 56.67 Hz ,  f₃ = 170 Hz

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In this configuration we have a node at the closed end and a belly at the open end whereby the wavelength

With 1  node         λ₁ = 4 L

With 2 nodes      λ₂ = 4L / 3

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The general term would be      λ_n= 4L / n         n = 1, 3, 5, ((2n + 1)

The speed of sound is

         v = λ f

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Let's consider each length independently

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         f₁ = 113.33 Hz

        f₃ = 113.33   3

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L = 1.5 m

       f₁ = 340 n / 4 1.5 = 56.67 n

       f₁ = 56.67 Hz

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8 0
2 years ago
How, if at all, would the equations written in Parts C and E change if the projectile was thrown from the cliff at an angle abov
sveta [45]

Answer:

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Explanation:

This is a projectile launch exercise, in this case we will write the equations for the x and y axes

Let's use trigonometry to find the components of the initial velocity

              sin θ = v_{oy} / v₀

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now let's write the equations of motion

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7 0
2 years ago
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loris [4]

Answer: 36.86\°

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Clearing \theta:

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\theta=36.86\° This is the angle at which Ferdinand the frog jumps between lily pads

4 0
2 years ago
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