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kvasek [131]
3 years ago
9

A single-engine helicopter has two rotors; a main rotor and a tail rotor. The main rotor has a diameter of 14.4 m and rotates at

the rate of 450 rev/min while the tail rotor with a diameter of 1.6 m rotates at 4100 rev/min. What are the speeds, in m/s, of the tips of each rotor? main rotor speed m/s tail rotor speed m/s
Physics
1 answer:
Ber [7]3 years ago
6 0

Answer:

339.12 m/s, 343.31 m/s

Explanation:

For main rotor

diameter, d = 14.4 m

radius, r = 7.2 m

f = 450 rev/min = 450 / 60 = 7.5 rps

Angular speed, ω = 2 x π x f = 2 x 3.14 x 7.5 = 47.1 rad/s

linear speed, v = r x ω = 7.2 x 47.1 = 339.12 m/s

For tail rotor

diameter, d = 1.6 m

radius, r = 0.8 m

f = 4100 rev/min = 4100 / 60 = 68.33 rps

Angular speed, ω = 2 x π x f = 2 x 3.14 x 68.33 = 429.13 rad/s

linear speed, v = r x ω = 0.8 x 429.13 = 343.31 m/s

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F' = W-F ........................ Equation 1

Where F' = Net force on the bag, W = gravitational force on the bag, F = Force acting upward on the bag

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2 years ago
Two metal disks, one with radius R1 = 2.45 cm and mass M1 = 0.900 kg and the other with radius R2 = 5.00 cm and mass M2 = 1.60 k
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Answer:

part (a) a_1\ =\ 2.9\ kg

Part (b) a_2\ =\ 6.25\ kg

Explanation:

Given,

  • mass of the smaller disk = M_1\ =\ 0.900\ kg
  • Radius of the smaller disk = R_1\ =\ 2.45\ cm\ =\ 0.0245\ m
  • mass of the larger disk = M_2\ =\ 1.6\ kg
  • Radius of the larger disk =R_2\ =\ 5.0\ cm\ =\ 0.05\ m
  • mass of the hanging block = m = 1.60 kg

Let I be the moment of inertia of the both disk after the welding,\therefore I\ =\ I_1\ +\ I_2\\\Rightarrow I\ =\ \dfrac{1}{2}(M_1R_1^2\ +\ M_2R_2^2)\\\Rightarrow I\ =\ 0.5\times (0.9\times 0.0245^2\ +\ 1.6\times 0.05^2)\\\Rightarrow I\ =\ 2.27\times 10^{-3}\ kgm^2

part (a)

A block of mass m is hanging on the smaller disk,

From the f.b.d. of the block,

Let 'a' be the acceleration of the block and 'T' be the tension in the string.

mg\ -\ T\ =\ mg\\\Rightarrow T\ =\ mg\ -\ ma\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,eqn (1)

Net torque on the smaller disk,

\therefore \tau\ =\ I\alpha\\\Rightarrow TR_1\ =\ \dfrac{Ia}{R_1}\\\Rightarrow T\ =\ \dfrac{Ia}{R_1^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,enq (2)

From eqn (1) and (2), we get,

mg\ -\ ma\ =\ \dfrac{Ia}{R_1^2}\\\Rightarrow a\ =\ \dfrac{mg}{\dfrac{I}{R_1^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}}{0.027^2}\ +\ 1.60}\\\Rightarrow a\ =\ 2.91\ m/s^2

part (b)

In this case the mass is rapped on the larger disk,

From the above expression of the acceleration of the block, acceleration is only depended on the radius of the rotating disk,

Let 'a_2' be the acceleration of the block in the second case,

From the above expression,

\therefore a\ =\ \dfrac{mg}{\dfrac{I}{R_1^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}}{0.05^2}\ +\ 1.60}\\\Rightarrow a\ =\ 6.25\ m/s^2

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