Concept of centrifugal tension:
Explain concept of centrifugal tension in any belt drive. What is main consideration for taking maximum tension?
Sol: The belt continuously runs over both the pulleys. In tight side and slack side of belt tension is increased because of presence of centrifugal Tension in belt. At lower speeds centrifugal tension can be ignored but at the higher speed its effect is considered.
The tension caused in the running belt by the centrifugal force is known as centrifugal tension. Whenever particle of mass 'm' is rotated in circular path of radius 'r' at uniform velocity 'v', a centrifugal force is acting outward radially and its magnitude is equal to mv^{2}/r
i.e., F_{c} = mv^{2}/r
The centrifugal tension in belt can be computed by considering forces acting on an elemental length of the belt(that is length MN) subtending an angle dq at the center as shown in the given figure
Let
v = Velocity of belt in m/s
r = Radius of pulley over which belt run. M = Mass of elemental length of belt.
m = Mass of belt per meter length
T_{1} = Tight side tension
T_{c} = Centrifugal tension acting at points M and N tangentially
F_{c} = Centrifugal force acting radially outwards
The centrifugal force R which is acting radially outwards is balanced by components of Tc acting radially inwards. The elemental length of belt is
MN = r. δθ
Mass of belt MN = Mass per meter length X Length of MN M = m X r X δθ
Centrifugal force = F_{c} = M X v^{2}/r = m.r.δθ.v^{2}/r
Now by resolving force horizontally, we get
T_{c}.sinδθ/2 + T_{c}.sinδθ/2 = F_{c}
Or 2T_{c}.sinδθ/2 = m.r.δθ.v^{2}/r
At the angle δθ is very small, thus = sinδθ/2 = δθ/2
Then above equation becomes as
2T_{c}.δθ/2 = m.r.δθ.v^{2}/r or
T_{c}= m.v^{2}