Answer:
33.429 N-m
Explanation:
Given :
Inclination angle of two shaft, α = 20°
Speed of shaft 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 }](https://tex.z-dn.net/?f=%5Cfrac%7BN_%7BB%7D%7D%7BN_%7BA%7D%7D%20%3D%20%5Cfrac%7Bcos%5Calpha%20%7D%7B1%20-%20sin%5E%7B2%7D%5Calpha%20%7D)
![\frac{N_{B}}{1000} = \frac{cos20}{1 - sin^{2}20 }](https://tex.z-dn.net/?f=%5Cfrac%7BN_%7BB%7D%7D%7B1000%7D%20%3D%20%5Cfrac%7Bcos20%7D%7B1%20-%20sin%5E%7B2%7D20%20%7D)
= 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}](https://tex.z-dn.net/?f=k%5E%7B2%7D)
=30 X ![0.1^{2}](https://tex.z-dn.net/?f=0.1%5E%7B2%7D)
= 0.3 kg-![m^{2}](https://tex.z-dn.net/?f=m%5E%7B2%7D)
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}](https://tex.z-dn.net/?f=0.3%5Ctimes%20%5Cfrac%7B2%5Cpi%20%5Ctimes%201064.1%7D%7B60%7D)
= 33.429 N-m
Torque on the Shaft B is 33.429 N-m