Work and Rotational Kinetic Energy — Question 5

Question 5

Pulley rotation consumes part of a falling system energy release

Masses $3$ and $1\,\mathrm{kg}$ start at rest on opposite sides of a fixed pulley. Its radius is $0.20\,\mathrm m$ and inertia $0.080\,\mathrm{kg\,m^2}$. A light cord does not slip and there is no friction. The heavier mass descends $0.40\,\mathrm m$. Use $g=9.8\,\mathrm{m/s^2}$.

1. Find the common mass speed and pulley angular speed.

2. Find the fraction of the released potential energy stored in pulley rotation.

Question 5 - Solution

Pulley rotation consumes part of a falling system energy release

(a) Derivation and calculation

The gravitational energy release is shared by both masses and the pulley: $$\Delta E_g=(3-1)(9.8)(0.40)=7.84\ \mathrm J.$$ $$\Delta E_g=\tfrac12(m_h+m_l)v^2+\tfrac12I\left(\frac vR\right)^2 =\tfrac12(4+2)v^2=3v^2.$$ $$\boxed{v=\sqrt{7.84/3}=1.617\ \mathrm{m/s}},\qquad \boxed{|\omega|=v/0.20=8.083\ \mathrm{rad/s}}.$$

(b) Result and interpretation

The rotational share is $$\boxed{\frac{K_{\mathrm{pulley}}}{\Delta E_g} =\frac{I/R^2}{m_h+m_l+I/R^2}=\frac{2}{6}=\frac13}.$$ Thus $K_{\mathrm{pulley}}=7.84/3=2.613$ J. Omitting this energy would overestimate the falling speed.

Physical check

The remaining two thirds is the combined translational energy of the masses.