Problem — Dual-radius pulley

Problem

A pulley has radii $r_1 = 0{,}05\;\mathrm{m}$ and $r_2 = 0{,}1\;\mathrm{m}$, with moment of inertia $I = \tfrac{1}{2}m_p r_2^2$. A mass hangs from each radius: $m_1 = 3\;\mathrm{kg}$ on $r_1$ and $m_2 = 5\;\mathrm{kg}$ on $r_2$. Derive the angular acceleration, starting from the constraint $a_i = \alpha\,r_i$.

Solution

The pulley rotates with a single $\alpha$, but the two masses have different linear accelerations, $a_i = \alpha\,r_i$. We write Newton’s law for the two masses and the pulley’s rotation, and substitute the constraint. The general result is $\ev{\alpha = \frac{m_2\,g\,r_2 - m_1\,g\,r_1}{I + m_1\,r_1^2 + m_2\,r_2^2}}$ Numerically the numerator is $g(m_2 r_2 - m_1 r_1) = 9{,}81\,(5\cdot 0{,}1 - 3\cdot 0{,}05) = 9{,}81\cdot 0{,}35 \approx 3{,}4\;\mathrm{N\cdot m}$, while the denominator is $I + m_1 r_1^2 + m_2 r_2^2 = I + 0{,}0575\;\mathrm{kg\,m^2}$ (with $I = \tfrac{1}{2}m_p r_2^2$ depending on the pulley’s mass). The positive sign of the numerator confirms that $m_2$, hanging from the larger radius, prevails.

Links

Topics: Rotational dynamics Concepts: Torque · Angular acceleration Skills: Symbolic setup Objects: Pulley