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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.

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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