Problem — Pulley with two masses

Problem

A solid cylindrical pulley of $M = 1\;\mathrm{kg}$, $R = 0{,}10\;\mathrm{m}$. Two masses $m_1 = 2\;\mathrm{kg}$ and $m_2 = 3\;\mathrm{kg}$ hang from the two ends of the rope. Calculate the acceleration of the system, including the effect of the pulley’s mass.

Solution

The solid pulley contributes $I = \tfrac{1}{2}MR^2$; in the denominator it appears as $\tfrac{1}{2}M$: $a = \frac{(m_2 - m_1)g}{m_1 + m_2 + \tfrac{1}{2}M} = \frac{(3-2)\cdot 9{,}81}{2 + 3 + 0{,}5} = \frac{9{,}81}{5{,}5}$ $\ev{a \approx 1{,}78\;\mathrm{m/s^2}}$ Neglecting the pulley’s mass would give $a = 9{,}81/5 = 1{,}96\;\mathrm{m/s^2}$: the massive pulley slows the system down, because part of the masses’ action goes into setting it into rotation.

Links

Topics: Rotational dynamics Concepts: Torque · Moment of inertia Objects: Pulley · Atwood machine