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iris [78.8K]
3 years ago
5

The two shafts of a Hooke’s coupling have their axes inclined at 20°.The shaft A revolves at a uniform speed of 1000 rpm. The sh

aft B carries a flywheel of mass 30 kg. If the radius of gyration of the flywheel is 100 mm, find the maximum torque in shaft B.
Engineering
1 answer:
lapo4ka [179]3 years ago
4 0

Answer:

33.429 N-m

Explanation:

Given :

Inclination angle of two shaft, α = 20°

Speed of shaft A, N_{A} = 1000 rpm

Mass of flywheel, m = 30 kg

Radius of Gyration, k =100 mm

                                   = 0.1 m

Now we know that for maximum velocity,

\frac{N_{B}}{N_{A}} = \frac{cos\alpha }{1 - sin^{2}\alpha }

\frac{N_{B}}{1000} = \frac{cos20}{1 - sin^{2}20 }

N_{B} = 1064.1 rpm

Now we know

Mass of flywheel, m = 30 kg

Radius of Gyration, k =100 mm

                                   = 0.1 m

Therefore moment of inertia of flywheel, I = m.k^{2}

                                                                      =30 X 0.1^{2}

                                                                     = 0.3 kg-m^{2}

Now torque on the output shaft

T₂ = I x ω

    = 0.3 X 1064.2 rpm

    = 0.3\times \frac{2\pi \times 1064.1}{60}

     = 33.429 N-m

Torque on the Shaft B is 33.429 N-m

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Answer:

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ii) when 5V output voltage is applied :  R1 = 1 kΩ , R2 =  1 kΩ

iii) when 2.5 v output voltage is applied : R1 = 1500 Ω,  R2 = 500 Ω

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attached below is the sketch of the circuit

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R1 = 1500 Ω

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iv) when the load is connected to each tap one at a time

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when: R1 = 1 kΩ , R2 =  1 kΩ  is connected in parallel output voltage < 5 V

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