The moment equation of a pulley with a rope applying tension on both sides is as follows:

$$I\alpha = f + T_1R - T_2R$$

( $I$ - moment of inertia; $\alpha$ - angular acceleration; $f$ - friction between axle and the axle holder; $T$ - Tension; $R$ - perpendicular distance from centre of axle to tension)

The L.H.S and the first term of R.H.S equates to zero as the pulley is 'massless and frictionless'.

But there is no mention about friction between rope and pulley. Why is that ?

If we consider a pulley to have friction between itself and the ropes, how would its moment equation be ?

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    $\begingroup$ Your equation is dimensionally incorrect. $\endgroup$ – Farcher May 11 at 15:35

It is the friction between the pulley and rope due to which the factor $T_1R-T_2R$ appears in the equation. Due to friction,the tensions in the ropes on either ends cannot be same[ The relation is actually $T_1= T_2 e^{\mu \theta }; \mu \rightarrow$ coefficient of friction, $\theta \rightarrow$ angle of wrap of rope around pulley] The friction prevents slipping between rope and pulley and actually causes pulley to turn with the rope. This implies these tensions can't be same. In absence of this friction, the pulley wouldn't rotate. The rope would just slip over the pulley and the tension is same throughout the rope. The friction at the axle, on the other hand, opposes the rotation of pulley - a completely opposite effect as compared to that by the friction between rope and pulley.

  • $\begingroup$ The first line would suffice as a concise answer. But the rest couldn't be grasped. Both tensions are always same for a 'massless and frictionless (between axle and axle holder)' pulley. But then why the mention, 'Due to friction (between pulley and rope), the tensions in the ropes on either ends cannot be same' ? $\endgroup$ – Zam May 12 at 11:31
  • $\begingroup$ The friction on the surface of pulley causes tension to vary in rope along the circumference. The mass of pulley has no effect. $\endgroup$ – Tojrah May 12 at 11:37
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    $\begingroup$ The equation you wrote is correct only if there is kinetic friction (relative slip between the rope and the pulley). With static friction, that is not the case, and the two tensions must be equal (which follows directly from the overall moment balance). $\endgroup$ – Chet Miller May 12 at 14:00
  • $\begingroup$ If the pulley rotates, the tensions are unequal $\endgroup$ – Tojrah May 12 at 14:01

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