Relating the Linear and Angular Variables — Question 2

Question 2

Two acceleration components reveal the angular state

A point at radius $0.50\,\mathrm m$ moves counterclockwise around a fixed axis. Its acceleration has inward component $8.0\,\mathrm{m/s^2}$ and a tangential component $-1.5\,\mathrm{m/s^2}$ relative to positive counterclockwise motion.

1. Infer angular velocity and angular acceleration.

2. Could the radial acceleration alone determine the rotation direction?

Question 2 - Solution

Two acceleration components reveal the angular state

(a) Derivation and calculation

Use the inward magnitude and the signed tangential component separately: $$a_{\mathrm{in}}=r\omega^2\Rightarrow |\omega|=\sqrt{\frac{8.0}{0.50}}=4\ \mathrm{rad/s}.$$ The stated counterclockwise sense selects $$\boxed{\omega=+4\ \mathrm{rad/s}},\qquad \boxed{\alpha=\frac{a_\theta}{r}=\frac{-1.5}{0.50}=-3\ \mathrm{rad/s^2}}.$$

(b) Result and interpretation

The radial measurement is invariant under sign reversal: $$r(+4)^2=r(-4)^2=8.0\ \mathrm{m/s^2}.$$ It cannot determine clockwise versus counterclockwise motion. Here $\omega\alpha=-12<0$, so the known positive rotation is slowing.

Physical check

The direction information comes from the stated motion, not from the centripetal magnitude.