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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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A 1.0 kg copper rod rests on two horizontal rails 1.0 m apart and carries a current of 50 A from one rail to the other.
vagabundo [1.1K]

Answer

given,

mass of copper rod = 1 kg

horizontal rails = 1 m

Current (I) = 50 A

coefficient of static friction = 0.6

magnetic force acting on a current carrying wire is

           F = B i L

Rod is not necessarily vertical

F_x =i L B_d

F_y= i L B_w

the normal reaction N = mg-F y

static friction       f = μ_s (mg-F y )

horizontal acceleration is zero

F_x-f = 0

iLBd = \mu_s(mg-F_y )

 B_w = B sinθ

 B_d = B cosθ

iLB cosθ= μ_s (mg- iLB sinθ)

B = \dfrac{\mu_smg}{i(cos\theta +\mu_s sin\theta)}

\theta =tan{-1}{\mu_s}

\theta =tan{-1}{0.6}

\theta = 31^0

B = \dfrac{0.6\times 1 \times 9.8}{50(cos31^0 +0.6 sin31^0)}

       B = 0.1 T

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Explanation:

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