Question

The power transmitted by a belt drive is maximum when running at a velocity equal to
1. \(\sqrt {\frac{T_m}{3W}}\;\)
2. \(\sqrt {\frac{T_m}{3Wg}}\;\)
3. \(\sqrt {\frac{T_mg}{W}}\;\)
4. \(\sqrt {\frac{T_mg}{3W}} \;\)

Answer 1

Correct Answer - Option 4 : \(\sqrt {\frac{T_mg}{3W}} \;\)

Concept:

Centrifugal Tension: 

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 the centrifugal force is called centrifugal tension (Tc) 

Condition for Maximum power transmitted by the belt: 

When T_m = 3Tc i.e. power transmitted will be maximum when tension is equal to three-time centrifugal tension or it shows that when the power transmitted is maximum, 1/3rd of the maximum tension is absorbed as centrifugal tension.

Power transmitted by a belt:

P = (T1 – T2)V

where

T1 is the tension on the tight side (N), T2 is the tension on the slack side (N), and V is the velocity of the belt in (m/s).

The belt velocity and centrifugal tension is given by:

Tc = mv2

Therefore, v = \(\sqrt{\frac{T_c}{m}}=\sqrt{\frac{T_m}{3m}}\)

weight of the belt W = mg

or \(m =\frac{W}{g}\)

The power transmitted by a belt drive is maximum when running at a velocity, \(v=\sqrt {\dfrac{T_mg}{3W}}\)