Equilibrium — Question 3

Question 3

Balanced forces can leave a free body rotating faster

A free horizontal bar has inertia $0.50\,\mathrm{kg\,m^2}$ about its mass center. Opposite vertical forces of $4\,\mathrm N$ act at its ends, separated by $1.0\,\mathrm m$. The bar starts at rest. Ignore gravity for this planar test.

1. Find center acceleration and initial angular acceleration.

2. Is force balance sufficient for equilibrium of an extended body?

Question 3 - Solution

Balanced forces can leave a free body rotating faster

(a) Derivation and calculation

Equal and opposite end forces give $$\sum\vec F=\vec0\quad\Longrightarrow\quad\boxed{\vec a_C=\vec0}.$$ Their perpendicular separation is $d=1.0$ m, so the initial couple is $$|\tau_C|=Fd=4(1.0)=4\ \mathrm{N\,m}.$$ $$\boxed{|\alpha(0)|=|\tau_C|/I_C=4/.50=8\ \mathrm{rad/s^2}}.$$

(b) Result and interpretation

For planar rigid-body equilibrium both independent conditions are needed: $$\sum\vec F=m\vec a_C=\vec0,\qquad \sum\tau_C=I_C\alpha=0.$$ Here only the first holds. The center can remain at rest while the bar begins rotating; a point-particle model would omit this rotational degree of freedom.

Physical check

The couple is independent of the torque origin because its force resultant is zero.