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DIA [1.3K]
3 years ago
15

A rhinoceros beetle rides the rim of a disk rotating like a merry-go-round counterclockwise.

Physics
1 answer:
Pie3 years ago
8 0

Answer:

a) The angular momentum remains constant because the system is isolated

b)  if the beetle walks

- In the direction of rotation of the disc, the angular velocity of the disc decreases.

-   If the beetle goes in the opposite direction] to the disk the angular velocity increases

Explanation:

The angular momentum is

           L = I w

For an isolated system the angular momentum must be preserved.

Let us define the system as formed by the beetle plus the disk, in this case when the beetle walks, the forces are an internal force (action and reaction), therefore the angular momentum is conserved. Let's write in two moments

Initial. Before the beetle movement

           L₀ = I w₀

Final. When the beetle is moving

          L_{f} = I w + I₂ w₂

The moment is preserved

           L₀ =  L_{f}

           I w₀ = I w + I₂ w₂

 

We cleared the final angular speed of the disk

          w = w₀ - w₂ I₂ / I

In general, the moment of inertia of the beetle is less than the moment of inertia of the disc and its angular velocity is also small.

Let's examine the questions.

a) The angular momentum remains constant because the system is isolated

b) The angular velocity of the system changes according to the previous equation, if the beetle walks

- In the direction of rotation of the disc, the angular velocity of the disc decreases.

-   If the beetle goes in the opposite direction] to the disk the angular velocity increases

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Diagram shows the velocity time graph for a particle moving in 16 seconds.
Leya [2.2K]

Answer:

The detailed calculations are shown below;

Explanation:

a)The maximum acceleration of the particle:

  It is seen that the maximum change in velocity is at the time between 8s to 10s.

              Maximum acceleration: \frac{V}{t}

                                                    = \frac{20}{2}

                                                    = 10 m/s^{2}

b) The deceleration of the particle

          The velocity of particle is decreased after 10s so,

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                                           = - 6.67 m/s^{2}

c)The total distance traveled by the particle = Area under the curve

                               = \frac{1}{2}* 4*20 + 4*20 +  \frac{1}{2}* 2*20+ 2*20+ \frac{1}{2}* 40*16

                                                       = 290 m

d)The average velocity of the particle = \frac{Area under the curve}{Total time}

                                                               = \frac{290}{16}

                                                               = 18.12 m/s

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3 years ago
A 332 kg mako shark is moving in the positive direction at a constant velocity of 2.30 m/s along the bottom of a sea when it enc
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To solve this problem we will apply the concepts related to the conservation of momentum. By definition we know that the initial moment must be equivalent to the final moment of the two objects therefore

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