The centrifugal tension in belts:

Explanation:

Centrifugal effects on belts:

While in motion, as a belt passes over a pulley, the centrifugal effect due to its own weight tends to lift the belt from the pulley. Owing to symmetry, the centrifugal force produces equal tensions on the two sides of the belt i.e. on the tight side as well as on the slack side.

The magnitude of centrifugal force is T_c = mv²

Thus, we can say that the centrifugal tension is independent of the tight and slack side tensions and depends only on the velocity of the belt over the pulley.

Total tension on tight side T ⇒ T₁ + T_c

Total tension on tight side T ⇒ T₂ + T_c

Considering centrifugal tension:

Friction tension on tight side = T - T_c = T₁

Let T₂ be the friction tension on the slack side

\(\frac{T_1}{T_2}=e^{\mu \theta}=k\) or \(T_2=\frac{T_1}{k}\)

We know that power P = (T1 - T2)v

Putting the value of T₂ in power

\(P=T_1\left(1-\frac{1}{k}\right)v\)

Neglecting centrifugal tension:

Friction tension on tight side = T

Let T2' be the tension on the slack side.

\(\frac{T_1}{T_2^{'}}=e^{\mu \theta}=k\) or \(T_2^{'}=\frac{T_1}{k}\)

We know that power P = (T1 - T2')v

\(P=T\left(1-\frac{1}{k}\right)v\)

As T₁ is lesser than T, the power transmitted is less when centrifugal tension is taken into account.