Problem — Atwood machine with a massive pulley

Problem

Two masses $m_1 = 4\;\mathrm{kg}$ and $m_2 = 6\;\mathrm{kg}$ are connected by a rope passing over a solid cylindrical pulley ($m_p = 2\;\mathrm{kg}$, $R = 0{,}1\;\mathrm{m}$, $I_p = \tfrac{1}{2}m_p R^2$). Find the acceleration of the system and the two tensions.

Solution

Three equations: Newton’s law for each mass and the rotation of the pulley, with the constraint $a = \alpha R$. $m_2 g - t_2 = m_2 a, \qquad t_1 - m_1 g = m_1 a, \qquad (t_2 - t_1)R = I_p\,\alpha = \tfrac{1}{2}m_p R^2\cdot\frac{a}{R}$ The last one becomes $t_2 - t_1 = \tfrac{1}{2}m_p a$. Adding the three: $a = \frac{(m_2 - m_1)g}{m_1 + m_2 + \tfrac{1}{2}m_p} = \frac{(6-4)\cdot 9{,}81}{4 + 6 + 1} = \frac{19{,}62}{11} \approx \ev{1{,}78\;\mathrm{m/s^2}}$ Tensions: $t_1 = m_1(g + a) = 4(9{,}81 + 1{,}78) \approx \ev{46{,}4\;\mathrm{N}}$ $t_2 = m_2(g - a) = 6(9{,}81 - 1{,}78) \approx \ev{48{,}2\;\mathrm{N}}$ The two tensions are different ($t_2 > t_1$): it is their difference that makes the massive pulley rotate. With an ideal pulley they would be equal.

Links

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