# Considering the effect of centrifugal tension in a flat belt drive with T1 = tight side tension and Tc = centrifugal tension and m = mass per unit len

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Considering the effect of centrifugal tension in a flat belt drive with T1 = tight side tension and Tc = centrifugal tension and m = mass per unit length of the belt, the velocity of the belt for maximum power transmission is given by:
1. $V\; = \;\sqrt {\dfrac{{{T_1}}}{{3m}}}$
2. $V\; = \;\sqrt {\dfrac{{{T_c}}}{{3m}}}$
3. $V\; = \;\sqrt {\dfrac{{\left( {{T_1} - \;{T_c}} \right)}}{{3m}}}$
4. $V\; = \;\sqrt {\dfrac{{\left( {{T_1} + \;{T_c}} \right)}}{{3m}}}$
5.

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Correct Answer - Option 4 : $V\; = \;\sqrt {\dfrac{{\left( {{T_1} + \;{T_c}} \right)}}{{3m}}}$

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).

When the centrifugal tension in the flat belt is considered. Then maximum tension in the belt,

⇒ T = T1 + Tc

Power transmitted by a belt is given by:

P = (T1 – T2) × v      …(1)

For Flat Belt:

⇒ T1 = T2eμθ

Also, Tc = mV2      …(2)

where, T1 = friction tension on the tight side, T2 = friction tension on the slack side

Using equation (1) and (2),

$P = {T_1}\left( {1 - \dfrac{1}{{{e^{\mu \theta }}}}} \right)V = \left( {T - {T_c}} \right)VK$

Where,

$K = \left( {1 - \dfrac{1}{{{e^{\mu \theta }}}}} \right)\;,$

⇒ P = (T – mV2) VK

For Maximum power transmitted by the belt,

$\Rightarrow \dfrac{{dP}}{{dV}} = 0$

T = 3TC      ..(3)

So, the 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.

For maximum power transmission,

Using equation (2) and (3),

$\Rightarrow V = \sqrt {\dfrac{T}{{3m}}} = \sqrt {\dfrac{{\left( {{T_1} + {T_c}} \right)}}{{3m}}}$