Conservation of Angular Momentum — Question 8

Question 8

A pivot conserves angular momentum while changing linear momentum

A uniform rod of mass $3\,\mathrm{kg}$ and length $2\,\mathrm m$ lies along $+x$ on a horizontal frictionless plane and is hinged at its left end about vertical $z$. A $0.10\,\mathrm{kg}$ projectile moving at $100\hat y\,\mathrm{m/s}$ sticks to it at $x=1.5\,\mathrm m$. The rod starts at rest; hinge angular impulse about $z$ is zero.

1. Find angular speed immediately afterward.

2. Find the hinge impulse along $y$ on the rod-plus-projectile system.

Question 8 - Solution

A pivot conserves angular momentum while changing linear momentum

(a) Derivation and calculation

About the hinge, the projectile brings $$L_i=mub=.10(100)(1.5)=15.0\ \mathrm{kg\,m^2/s}.$$ The final axial inertia is $$I_f=\frac{ML^2}{3}+mb^2=\frac{3(2)^2}{3}+.10(1.5)^2=4.225\ \mathrm{kg\,m^2}.$$ Hence $$\boxed{\omega_f=\frac{15}{4.225}=\frac{600}{169}=3.550\ \mathrm{rad/s}}.$$

(b) Result and interpretation

Immediately after capture, the rod center and projectile speeds are $(L/2)\omega_f$ and $b\omega_f$ along $+y$. Thus $$P_{y,f}=[M(L/2)+mb]\omega_f=3.15\frac{600}{169}=\frac{1890}{169}.$$ Subtract the initial $P_{y,i}=.10(100)=10$: $$\boxed{J_{{\mathrm{hinge}},y}=P_{y,f}-P_{y,i}=\frac{200}{169}=1.183\ \mathrm{N\,s}}.$$ The hinge supplies external linear impulse with zero moment about its own axis.

Physical check

Angular momentum is conserved about the hinge while total linear momentum changes.