Newton's Second Law for Rotation — Question 5

Question 5

The hinge force is not the rod weight during release

A uniform $4\,\mathrm{kg}$ rod of length $1.2\,\mathrm m$ is hinged at one end and released from rest horizontally to the right. Its inertia about the hinge is $ML^2/3$. Use $g=9.8\,\mathrm{m/s^2}$.

1. Find angular acceleration immediately after release.

2. Find the hinge-force components at that instant.

Question 5 - Solution

The hinge force is not the rod weight during release

(a) Derivation and calculation

Immediately after release, gravity supplies the hinge moment: $$\tau_z=-Mg\frac L2,\qquad I_H=\frac{ML^2}{3}.$$ $$\boxed{\alpha=\frac{\tau_z}{I_H}=-\frac{3g}{2L} =-\frac{3(9.8)}{2(1.2)}=-12.25\ \mathrm{rad/s^2}}.$$

(b) Result and interpretation

Initially $\omega=0$, so the center has no radial acceleration: $$a_{c,x}=0,\qquad a_{c,y}=\alpha\frac L2=-\frac{3g}{4}=-7.35\ \mathrm{m/s^2}.$$ $$H_x=Ma_{c,x}=\boxed{0},\qquad H_y-Mg=Ma_{c,y}\Rightarrow\boxed{H_y=\frac{Mg}{4}=9.8\ \mathrm N}.$$ The hinge supports only part of the weight because the center accelerates downward.

Physical check

The net vertical force is $9.8-39.2=-29.4$ N, equal to $4(-7.35)$ N.