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Andreyy89
3 years ago
12

Consider a long rod of mass, m, and length, l, which is thin enough that its width can be ignored compared to its length. The ro

d is connected at its end to frictionless pivot.
a) Find the angular frequency of small oscillations, w, for this physical pendulum.
b) Suppose at t=0 it pointing down (0 = 0) and has an angular velocity of 120 (that is '(t = 0) = 20) Note that 20 and w both have dimensions of time-1. Find an expression for maximum angular displacement for the pendulum during its oscillation (i.e. the amplitude of the oscillation) in terms of 20 and w assuming that the angular displacement is small.
Physics
1 answer:
ehidna [41]3 years ago
4 0

Answer:

Explanation:

The rod will act as pendulum for small oscillation .

Time period of oscillation

T=2\pi\sqrt{\frac{l}{g} }

angular frequency ω = 2π / T

= \omega=\sqrt{\frac{g}{l} }

b )

ω = 20( given )

velocity = ω r = ω l

Let the maximum angular displacement in terms of degree be θ .

1/2 m v ² = mgl ( 1 - cosθ ) ,

[ l-lcosθ is loss of height . we have applied law of conservation of mechanical energy .]

.5 (  ω l )² = gl( 1 - cos θ )

.5 ω² l = g ( 1 - cosθ )

1 - cosθ  = .5 ω² l /g

cosθ = 1 - .5 ω² l /g

θ can be calculated , if value of l is given .

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when a metal ball is heated through 30°c,it volume becomes 1.0018cm^3 if the linear expansivity of the material of the ball is 2
Vlad1618 [11]

Answer:

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Explanation:

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We have;

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Where;

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From the volume of the ball, V₂, we have;

V₂ = 1.0018 cm³ = (4/3)×π×r₂³

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r₂ = The new radius = d₂/2

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∴ d₂ = ∛(2³ × 1.0018 cm³/((4/3) × π)) ≈ 1.241445 cm

d₁ = d₂/(1 + α·ΔT)

∴ d₁ ≈ 1.241445 cm/(1 + 2.0 × 10⁻⁵·K × 30 K) ≈ 1.24070058 cm

The original volume of the metal ball, V₁ = (4/3)×π×(d₁/2)³

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kolbaska11 [484]
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<span>The second part required you find the relative distance between the atoms of the compounds. Really, the lattice energy is stronger with smaller atoms, an indirect relationship. </span>
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