In a pulley and belt drive, if m is mass of belt per unit length the velocity of belt for maximum power transmission is equal to

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ISRO URSC Technical Assistant Mechanical 24 March 2019 Official Paper
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  1. \(\sqrt {\frac Tm}\)
  2. \(\sqrt {\frac T{2m}}\)
  3. \(\sqrt {\frac {2T}m}\)
  4. \(\sqrt {\frac T{3m}}\)

Answer (Detailed Solution Below)

Option 4 : \(\sqrt {\frac T{3m}}\)
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Detailed Solution

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Explanation:

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 is: T = 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 = (T– Ts).v

where T the tension on the tight side (N), Ts is the tension on the slack side (N), V is the velocity of the belt in m/s.

The belt velocity is given by, 

Tc = mv2

\(\frac T3 = mv^2 \)

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

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