Question

If T and m represent the maximum tension and mass per unit length of a belt, then the maximum permissible speed of the belt is given by
1. \(\sqrt {\frac{T}{{3m}}} \)
2. \(\sqrt {\frac{3T}{{m}}} \)
3. \(\sqrt {\frac{2T}{{3m}}} \)
4. \(\sqrt {\frac{T}{{m}}} \)

Answer 1

Correct Answer - Option 1 : \(\sqrt {\frac{T}{{3m}}} \)

Concept:

Power transmitted by a belt:

P = (T1 – T2)v

For maximum power:

T1 = T/3

where T1 = Tension in the tight side, T = Maximum tension to which the belt is subjected.

The velocity (maximum) of the belt for maximum power:

\(v = \sqrt {\frac{T}{{3m}}} \)

where m is mass of belt per unit length.

m = Area × length × density = b.t.l.ρ 

Centrifugal Tension (Tc):

Since the belt continuously runs over the pulleys, therefore some centrifugal force is caused, whose effect is to increase the tension on both the tight as well as the slack sides.

The tension caused by this centrifugal force is called centrifugal tension.

Condition for Maximum power transmitted by the belt is T = 3TC