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cricket20 [7]
3 years ago
13

A torsional pendulum consists of a 5 kg uniform disk with a diameter of 50 cm attached at its center to a rod 1.5 m in length. T

he torsional spring constant is 0.625 N-m/rad. Disregarding the mass of the rod, what is the natural frequency of the torsional pendulum?
Engineering
1 answer:
IRISSAK [1]3 years ago
7 0

Answer:

natural frequency of the torsional pendulum =  1.4 rad/s

Explanation:

given data

mass = 5 kg

diameter = 50 cm = 0.50 m

so radius R = 0.25 m

length =  1.5 m

torsional spring constant = 0.625 N-m/rad

solution

first we get here moment of inertia that is

moment of inertia = mR²  .........1

moment of inertia = 5 × 0.25²

moment of inertia  = 0.3125 kg-m²

so now we get here natural frequency of the torsional pendulum that is

natural frequency of the torsional pendulum = \sqrt{\frac{Kt}{I} }    ..................2

here Kt is torsional spring constant  and I is moment of inertia  

natural frequency of the torsional pendulum = \sqrt{\frac{0.625}{3125} }  

natural frequency of the torsional pendulum =  1.4 rad/s

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Answer:

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Explanation:

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6 0
3 years ago
The diameter of an extruder barrel = 85 mm and its length = 2.00 m. The screw rotates at 55 rev/min, its channel depth = 8.0 mm,
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Answer:

Qx = 9.109.10^5 \times 10^{-6} m³/s  

Explanation:

given data

diameter = 85 mm

length = 2 m

depth = 9mm

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pressure p = 11 × 10^6 Pa

viscosity n = 100 Pas

angle = 18°

so  Qd will be

Qd = 0.5 × π² ×D²×dc × sinA × cosA   ..............1

put here value and we get

Qd = 0.5 × π² × ( 85 \times 10^{-3} )²× 9  \times 10^{-3}  × sin18 × cos18

Qd = 94.305 × 10^{-6} m³/s

and

Qb = p × π × D × dc³ × sin²A ÷  12  × n × L    ............2

Qb = 11 × 10^{6} × π × 85 \times 10^{-3}  × ( 9  \times 10^{-3} )³ × sin²18 ÷  12  × 100 × 2

Qb = 85.2 × 10^{-6} m³/s

so here

volume flow rate Qx = Qd - Qb   ..............3

Qx =  94.305 × 10^{-6}  - 85.2 × 10^{-6}  

Qx = 9.109.10^5 \times 10^{-6} m³/s  

8 0
3 years ago
A cylinder fitted with a frictionless piston contains 2 kg of R-134a at 3.5 bar and 100 C. The cylinder is now cooled so that th
inna [77]

Answer:

The answer to the question is

The heat transferred in the process is -274.645 kJ

Explanation:

To solve the question, we list out the variables thus

R-134a = Tetrafluoroethane

Intitial Temperaturte t₁ = 100 °C

Initial pressure = 3.5 bar = 350 kPa

For closed system we have m₁ = m₂ = m

ΔU = m×(u₂ - u₁) = ₁Q₂ -₁W₂

For constant pressure process we have

Work done = W = \int\limits^a_b P \, dV  = P×ΔV = P × (V₂ - V₁) = P×m×(v₂ - v₁)

From the tables we have

State 1 we have h₁ = (490.48 +489.52)/2 = 490 kJ/kg

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Therefore Q₁₂ = m×(u₂ - u₁) + W₁₂ = m × (u₂ - u₁) + P×m×(v₂ - v₁)

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photoshop1234 [79]

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