The equations are based on the following assumptions
1) The bar is straight and of uniform section
2) The material of the bar is has uniform properties.
3) The only loading is the applied torque which is applied normal to the axis of the bar.
4) The bar is stressed within its elastic limit.
Nomenclature
T = torque (Nm)
l = length of bar (m)
J = Polar moment of inertia.(Circular Sections) ( m^4)
J' = Polar moment of inertia.(Non circluar sections) ( m^4 )
K = Factor replacing J for non-circular sections.( m^4)
r = radial distance of point from center of section (m)
ro = radius of section OD (m)
τ = shear stress (N/m^2)
G Modulus of rigidity (N/m^2)
θ = angle of twist (radians)
Answer:
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Answer:
μ = 0.136
Explanation:
given,
velocity of the car = 20 m/s
radius of the track = 300 m
mass of the car = 2000 kg
centrifugal force


F c = 2666. 67 N
F f= μ N
F f = μ m g
2666.67 = μ × 2000 × 9.8
μ = 0.136
so, the minimum coefficient of friction between road surface and car tyre is equal to μ = 0.136
C. Route and roadways defined as class I highways
Explanation:
Thermodynamics system :
Thermodynamics system is a region or space in which study of matters can be done.The system is separated from surroundings by a boundary this boundary maybe flexible or fixed it depends on situations.The out side the system is called surroundings.
Generally thermodynamics systems are of three types
1.Closed system(control mass system)
Only energy transfer take place ,no mass transfer take place.
2.Open system(control volume system)
Both mass as well as energy transfer take place.
3.Isolated system
Neither mass or nor energy transfer take place.
At steady state ,property is did not changes with respect to time.