Answer:
22.90 × 10⁸ kg
Explanation:
Given:
Diameter, d = 0.02 m
ωₙ = 0.95 rad/sec
Time period, T = 0.35 sec
Now, we know
T=
where, L is the length of the steel cable
g is the acceleration due to gravity
0.35=
or
L = 0.0304 m
Now,
The stiffness, K is given as:
K =
Where, A is the area
E is the elastic modulus of the steel = 2 × 10¹¹ N/m²
or
K =
or
K = 20.66 × 10⁸ N
Also,
Natural frequency, ωₙ =
or
mass, m =
or
mass, m =
mass, m = 22.90 × 10⁸ kg
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,
= 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.
=30 X
= 0.3 kg-
Now torque on the output shaft
T₂ = I x ω
= 0.3 X 1064.2 rpm
=
= 33.429 N-m
Torque on the Shaft B is 33.429 N-m